Physics-Study Guide



Kinematics is he study of the motion of objects through time and space, Kinematic equations govern the relationship between position, velocity and acceleration of objects over time.

When solving kinematic problems it is important to carefully consider three critical points:

  1. When the object first begins moving or becomes interesting
  2. When the object changes direction
  3. When the object finally comes to rest or otherwise stops being of interest to you

Knowledge about these three points will allow you to use the kinematic equations to get information about the points (in time ans space) in between.

To begin the study of kinematics, we define velocity and acceleration:

(1)   \begin{equation*}v=\frac{x-x_0}{t-t_0}=\frac{\Delta x}{\Delta t}\end{equation*}

(2)   \begin{equation*}a=\frac{v-v_0}{t-t_0}=\frac{\Delta v}{\Delta t}\end{equation*}

Students familiar with differential calculus will appreciate that velocity  is the time derivative of position and acceleration is the time derivative of velocity.

We can then use these definitions to derive other useful relationships:

(3)   \begin{equation*}x=x_0+v_0 *t+(1/2)(at^2)\end{equation*}

(4)   \begin{equation*}v=v_0+at\end{equation*}

(5)   \begin{equation*}v^2=v^2_0 +2a(x-x_0)\end{equation*}

Problem Statement:

You are golfing at the edge of a cliff that stands 500 feet high. You hit the ball with an initial velocity of 90 miles per hour (132 feet/sec) at an angle of 30 degrees. How far does the ball travel before it hits the ground? Assume gravitational acceleration to be 32.2 ft/sec^2. Neglect air resistance.


The golf ball starts out at a point A with a velocity in the positive horizontal and vertical directions. After a short while, at point B, the ball stops gaining height and begins to fall to the ground because of the effect of gravity. During the entire process the ball has a positive horizontal velocity that gets the ball to point C.


It is helpful to break this problem into its horizontal and vertical components because the horizontal distance that the ball travels is limited by how long it is in the air.

Begin by looking at the y-component. Determine how much time it takes to get from ya to yb. At point b the ball is transitioning from rising to falling, it is instantaneously hanging in midair – so it has a velocity of zero in the y direction.

The equation we will use to relate velocities, acceleration and time is the following:

(6)   \begin{equation*}v=v_0+at\end{equation*}

Applied to find the time to go from point A to point B:

(7)   \begin{equation*}v_B_y=v_A_y + a_y *t_A_B\end{equation*}

(8)   \begin{equation*}t_A_B=\frac{v_B_y-v_A_y}{a_y}\end{equation*}

(9)   \begin{equation*}t_A_B=\frac{0-132*sin(30)}{-32.2}=2.05 sec\end{equation*}

Next, determine how high point B is using  the following relationship:

(10)   \begin{equation*}y=y_0+v_0t+\frac{1}{2}at^2\end{equation*}

Applied to find the height at point B;

(11)   \begin{equation*}y_B=y_A+v_A_y*t_A_B+\frac{1}{2}a_yt^2_A_B\end{equation*}

Now we can calculate how long it will take the ball to fall 568 ft and add that to the time it took to get up to that height and we will know the total time of flight:

(12)   \begin{equation*}y_C=y_B+v_B_yt_B_C+\frac{1}{2}a_yt^2_B_C\end{equation*}

(13)   \begin{equation*}0-568+0(t_B_C)+\frac{1}{2}(-32.2)t^2_B_C\end{equation*}

(14)   \begin{equation*}t_B_C=\frac{568.2}{32.2}=35.3sec\end{equation*}

Now we can determine the total time of flight:

(15)   \begin{equation*}t_A_C=t_A_B+t_B_C=2.05+35.3=37.4sec\end{equation*}

To determine the final distance, we use this relationship between distance, rate and time:

(16)   \begin{equation*}x=v_xt_A_C=132cos(30)\frac{ft}{sec}37.4sec=4280ft\end{equation*}

Kinematics & Impulse Momentum


An impulse is the sum of all forces applied to an object over a designated time period. For those familiar with integral calculus, an impulse is defined as the integral of force over time:

(17)   \begin{equation*}I=\int(F) dt\end{equation*}

For those not familiar with integral calculus, impulse can be understood as the area under the curve on a Force vs. Time graph:

The impulse depicted here is the gray area:

(18)   \begin{equation*}I=\frac{1}{2}(5)(0.1)+(0.22-0.1)(5)=0.85N*sec\end{equation*}

When F is a constant, the integral equation can be simplified to:

(19)   \begin{equation*}I=F\Delta t\end{equation*}

The Impulse Momentum Theorem states that the impulse on an object is equal to its change in linear momentum:

(20)   \begin{equation*}F\Delta t=m\Delta v\end{equation*}

Problem Statement:

You are standing at home plate and wish to hit a home run over the 10 foot tall center field wall, 390 feet away. The pitcher throws a 90 mile per hour fast ball that your bat makes contact 4 feet above the ground. The ball was in contact  for 10.0 milliseconds and leaves at an angle of 45 degrees. The ball has a mass of 0.33 lb. With what minimum force must you hit the ball/



This problem has a lot of information given, begin by making a list of the unknown variables and their values:

(21)   \begin{equation*}y_0=4ft\end{equation*}

(22)   \begin{equation*}y_f=10ft\end{equation*}

(23)   \begin{equation*}x_0=0ft\end{equation*}

(24)   \begin{equation*}x_f=390ft\end{equation*}

(25)   \begin{equation*}\theta=45\end{equation*}

(26)   \begin{equation*}V_in=90\frac{miles}{hour}=132\frac{ft}{sec}\end{eqaution} \begin{equation*}\Delta t=10.0ms\end{eqauation} \begin{equation*}m_b_a_l_l=0.33 lbm\end{equation*}

(27)   \begin{equation*}a_x=0.0\frac{ft}{s^2}\end{equation*}

(28)   \begin{equation*}a_y=g=-32.2\frac{ft}{s^2}\end{equation*}

To find the force required we will use the impulse momentum equation:

(29)   \begin{equation*}F\Delta t=m\Delta v\end{equation*}

The problem statement tells us the change in time, mass, and the velocity in. To be able to solve for the force, the velocity at which the ball leaves the bat must be found. Once this velocity is known the change in velocity will be known and the force attainable.

The first stage of this problem will be concerned with finding what minimum initial velocity is required to get the ball up and over the wall. This involves the position, velocity and acceleration of the ball – so the kinematic equations will be useful.

First we will link points A to B in the y direction using kinematic equations:

(30)   \begin{equation*}V_B_y=V_A_y+g*t_A_B\end{equation*}

(31)   \begin{equation*}0=V_a_y-32.2*t_A_B\end{equation*}

(32)   \begin{equation*}Y_B=Y_A+V_A_Yt_A_B+\frac{1}{2}*g*t^2_A_B\end{equation*}

(33)   \begin{equation*}Y_B=4+V_A_Yt_A_B+\frac{1}{2}(32.2)t^2_A_B\end{equation*}

Next link point B to C in the direction using the same two equations:

(34)   \begin{equation*}V_C_Y=V_B_Y+g*t_B_C\end{equation*}

(35)   \begin{equation*}V_C_Y=0-(32.2)t_B_C\end{equation*}

(36)   \begin{equation*}Y_C=Y_B+V_B_Yt_B_C+\frac{1}{2}gt^2_B_C\end{equation*}

(37)   \begin{equation*}10=Y_B+(0)*t_B_C-\frac{1}{2}(32.2)*t^2_B_C\end{equation*}

We can repeat these four equations in the x direction:

(38)   \begin{equation*}V_B_x=V_A_x+a_x*t_A_B\end{equation*}

(39)   \begin{equation*}V_B_x=V_A_x+(0)t_A_B\end{equation*}

(40)   \begin{equation*}X_B=X_A+V_A_x*t_A_B+\frac{1}{2}*a_x*t^2_A_B\end{equation*}

(41)   \begin{equation*}X_B=0+V_A_x*t_A_B+\frac{1}{2}*(0)*t^2_A_B\end{equation*}

(42)   \begin{equation*}V_C_x=V_B_x+a_x*t_B_C\end{equation*}

(43)   \begin{equation*}V_C_x=V_B_x+(0)*t_B_C\end{equation*}

(44)   \begin{equation*}x_C=x_B+V_B_x*t_B_C+\frac{1}{2}*a_x*t^2_B_C\end{equation*}

(45)   \begin{equation*}390=x_B+V_B_x*t_B_C+\frac{1}{2}*(0)*t^2_B_C\end{equation*}

The final equation comes from the knowledge that the velocity vector at A is at a known angle:

These 9 equations air with the 9 unknowns, solving the system gives the the following results:

(46)   \begin{equation*}V_A_y=79.9\frac{ft}{sec}\end{equation*}

(47)   \begin{equation*}V_A_x=79.9\frac{ft}{sec}\end{equation*}

(48)   \begin{equation*}V_B_x=79.9\frac{ft}{sec}\end{equation*}

(49)   \begin{equation*}V_C_x=79.9\frac{ft}{sec}\end{equation*}

(50)   \begin{equation*}V_C_y=77.4\frac{ft}{sec}\end{equation*}

(51)   \begin{equation*}X_B=198 ft\end{equation*}

(52)   \begin{equation*}Y_B=103 ft\end{equation*}

(53)   \begin{equation*}t_A_B=2.48 sec\end{equation*}

(54)   \begin{equation*}t_B_C=2.40 sec\end{equation*}

With this information, the magnitude of the velocity vector at A is calculated:

Knowing the velocity at A that is required to get the ball over the fence we can solve the impulse momentum equation to find the force required:

(55)   \begin{equation*}F\Delta t=m\Delta v\end{equation*}

(56)   \begin{equation*}F=\frac{m*\Delta v}{\Delta t}=\frac{(0.33lbm)(132+113\frac{ft}{sec})}{0.01sec}=8,090lbf\end{equation*}

Conversation of Linear Momentum (Inelastic Collision):


Momentum (P) is defined as the product of a moving object’s mass (m) and velocity (v). It can be thought of as a measure of an objects ability to resist changes in its motion. Momentum is a conserved quantity, meaning that when objects collide their total momentum before and after the collision is unchanged – though momentum can be transferred between the objects in the system.

There are two general types of collisions. The first is an inelastic collision where the objects collide and stick together, moving as one combined object. The second is an elastic collision where the objects collide and bounce off of one another, moving independently after the impact.

Problem Statement:

A sugar factory transports sugar using carts on rails. A 50 kg cart is rolling along at 2 meters per second. A 10 kg bag of sugar is dropped into the cart, slowing it don. What is the velocity of the cart with the sugar in it? Neglect frictional losses.



The bag of sugar does not bounce off and away the cart – the two objects continue moving as one lumped abject after the collision. This is an inelastic collision, so we will apply conservation of linear momentum. That is to say that the momentum present in the system before and after the collision is the same. Momentum, P, is the product of mass and velocity.

(57)   \begin{equation*}P_c_a_r_t+P_s_u_g_a_r=P_c_a_r_t_+_s_u_g_a_r\end{equation*}

(58)   \begin{equation*}m_c_a_r_tv_c_a_r_t+m_s_u_g_a_r*v_s_u_g_a_r=m_c_a_r_t_+_s_u_g_a_r*v_c_a_r_t_+_s_u_g_a_r\end{equation*}

(59)   \begin{equation*}(50kg)*(2\frac{m}{s})+(10kg)*(0\frac{m}{s})=(50+10kg)*v_c_a_r_t_+_s_u_g_a_r\end{equation*}

(60)   \begin{equation*}v_c_a_r_t_+_s_u_g_a_r=\frac{100}{60}\frac{m}{s}=1.67\frac{m}{s}\end{equation*}

Conservation of Linear Momentum (Elastic Collision):

Problem Statement:

Some coin operated pool tables use cue balls that have a greater mass than the rest of the balls to identify it when it is sunk so that it may be returned while the other balls are kept until another round is paid for. A 0.2 kg cue ball traveling at 1 m/s hits a 0.1 kg eight ball traveling at 3 m/s head on. What are the velocities of each after the collision?



The balls do not stick together, so they have different velocities after the collision. This is an elastic collision, so it is necessary to apply both conservation of linear momentum and conservation of energy. The only type of energy we are concerned with here is kinetic, because the balls are moving on a level surface.

Kinetic energy is quantifies with this equation:

(61)   \begin{equation*}KE=\frac{1}{2}mv^2\end{equation*}

Conservation of energy says that the total kinetic energy before and after the system will be equal:

(62)   \begin{equation*}KE_c_u_e+KE_e_i_g_h_t=KE_c_u_e_,_a_f_t_e_r+KE_e_i_g_h_t_,_a_f_t_e_r\end{equation*}

(63)   \begin{equation*}\frac{1}{2}m_c_u_ev^2_c_u_e+\frac{1}{2}m_e_i_g_h_tv^2_e_i_g_h_t=\frac{1}{2}m_c_u_e_,_a_f_t_e_rv^2_c_u_e_,_a_f_t_e_r+\frac{1}{2}m_e_i_g_h_t_,_a_f_t_e_rv^2_e_i_g_h_t_,_a_f_t_e_r\end{equation*}

Conservation of linear momentum tells us:

(64)   \begin{equation*}m_c_u_ev_c_u_e+m_e_i_g_h_tv_e_i_g_h_t=m_c_u_e_,_a_f_t_e_rv_c_u_e_,_a_f_t_e_r+m_e_i_g_h_t_,_a_f_t_e_rv_e_i_g_h_t_,_a_f_t_e_r\end{equation*}

The mass is unchanged in the collision, so the masses before and after are equal. The unknowns are v_c_u_e_,_a_f_t_e_r and v_{eight,after}, we have two independent equations and two unknowns, so the system will be solvable.

To solve the system of equations, plug in known values and call v_c_u_e_,_a_f_t_e_r “a” and v_e_i_g_h_t_,_a_f_t_e_r, after “b”.

(65)   \begin{equation*}\frac{1}{2}(0.2)(1)^2+\frac{1}{2}(0.1)(3)^2=\frac{1}{2}(0.2)(a)^2+\frac{1}{2}(0.1)(b)^2\end{equation*}

(66)   \begin{equation*}(0.2)(1)+(0.1)(3)=(0.2)(a)+(0.1)(b)\end{equation*}

Solving equation (2) for a gives:

(67)   \begin{equation*}a=\frac{0.5-0.1b}{0.2}=2.5-0.5b\end{equation*}

Plugging in this value of a to equation(67) and simplifying gives:

(68)   \begin{equation*}1.1=0.2(2.5-0.5b)^2+0.1b^2\end{equation*}

Expanding the squared term gives:

(69)   \begin{equation*}1.1=0.1(0.25b^2-2.5b+6.26)+0.1b^2\end{equation*}

After distributing 0.2 through all the terms and collecting like terms, we get the following quadratic expression:

(70)   \begin{equation*}0=0.15b^2-0.5b+0.15\end{equation*}

Using the quadratic formula, this solves to: b={3,\frac{1}{3}}

Plugging back into equation (66) and solving for a gives: a={1,\frac{7}{3}}

We know there was some exchange of energy and momentum between the balls, so they will not bounce back with the same velocities they had before the collision. We dismiss a=1 and b=3 as non-physical solutions and out final solution is:

(71)   \begin{equation*}a=v_{cue,after}=\frac{7}{3}\approx2.33\frac{m}{sec}\end{equation*}

(72)   \begin{equation*}b=v_{eight,after}=\frac{1}{3}\approx0.33\frac{m}{sec}\end{equation*}

Energy Conservation


Energy can be categorized as either kinetic or potential. Kinetic energy is due to the motion, while potential energy is due to position. The kinetic energy of an object in motion is quantified by:

(73)   \begin{equation*}KE=\frac{1}{2}mv^2\end{equation*}

Potential energy due to an object’s increased elevation is quantified by:

(74)   \begin{equation*}PE=mgh\end{equation*}

Potential energy stored in a compressed spring is quantified by:

(75)   \begin{equation*}PE_s=\frac{1}{2}kx^2\end{equation*}

Where k is the spring constant associated with a particular spring and x is the distance the spring is compressed.

Energy is a conserved quantity, meaning that it is not created or destroyed system. Energy can, however, be converted from one form to another – for example when a box slides against the ground some of the kinetic energy it starts with its converted to thermal energy.

Problem Statement:

Many roller coasters get their speed by climbing to the top of a tall hill, then rolling down. Consider a 4kg ball released from rest at the top of a 10 meter tall hill. What will its speed be at the bottom of the hill? Neglect friction.



At point A the ball has potential energy due to its elevation and no kinetic energy because it is initially at rest. At point B the ball has no potential energy because it is at ground level of zero meters elevation, but it has kinetic energy because it is rolling.

Conservation of energy tells us that the total energy at A and B are equal:

(76)   \begin{equation*}E_{A,total}=E_{B,total}\end{equation*}

Point A has all the potential energy, while point B has all the kinetic energy:

(77)   \begin{equation*}PE_A=KE_B\end{equation*}

Substituting the formulas for potential and kinetic energy:

(78)   \begin{equation*}mgh_A=\frac{1}{2}mv^2_B\end{equation*}

Plug in values and solve:

(79)   \begin{equation*}(10kg)(9.81\frac{m}{s^2})(10m)=\frac{1}{2}(10kg)v^2_B\end{equation*}

(80)   \begin{equation*}v_B=14.0\frac{m}{s}\end{equation*}

Notice that the mass of the ball did not actually matter.

Energy Conservation (with friction)


Consider a large block sitting on a table. We apply a force to it, but it does not move. The force counteracting our push that holds the block in place is due to the force of static friction. When we push sufficiently hard to start sliding the block, it stops moving when we stop applying force because there is a force of kinetic friction resisting the motion. The force of friction always acts opposite the direction of motion.

The force of static friction is:

(81)   \begin{equation*}F_{fs}\leq\mu_sN\end{equation*}

Where N is the normal force and \mu is the coefficient of static friction. The inequality is used because as soon as the force exceeds \muN, the block starts to move and static friction is replaced by kinetic friction. The equality holds when the block is just on the verge of moving.

the kinetic friction acting on an object is:

(82)   \begin{equation*}F_{fk}=\mu_kN\end{equation*}

Where \mu_k is the coefficient of kinetic friction. Both coefficients of friction are found experimentally and can be looked up in reference materials.

Problem Statement:

You have a block of trash with a mass of 2 kg that you wish to put in the trash bin by launching it up a ramp with a 30 degree incline using a spring. The spring has a coefficient of 300 N/m and is 100 cm uncompressed. You compress it to 40 cm before releasing it. The coefficient of friction between the trash block and the ramp is 0.5. If the trash bin is directly off the edge of the ramp, what length and height should the ramp be?

(Be careful not to overshoot the trash bin, you want the block to just make it up the ramp and fall off the edge; though an interesting variation on this problem would be to move the trash bin over a meter or two.)



The velocity of the block at the bottom of the ramp can be determined by balancing all of the potential energy stored in the spring with all of the kinetic energy transferred to the block at the bottom of the ramp.

(83)   \begin{equation*}PE_s=KE_b\end{equation*}

(84)   \begin{equation*}\frac{1}{2}kx^2=\frac{1}{2}mv^2\end{equation*}

(85)   \begin{equation*}\frac{1}{2}(300\frac{N}{m})(0.6m)^2=\frac{1}{2}(2kg)v^2\end{equation*}

(86)   \begin{equation*}v=7.35\frac{m}{sec}\end{equation*}

Consider the following free body diagram that shows the force acting on the black as it slides up the ramp:

The x and y axis have been rotated 30 degrees to match the incline of the ramp. The weight of the block has been decomposed into its x and y components: F_{wx} and F_{wy}.

After drawing the free body diagram, it is appropriate to sum forces in the x and y directions. In the y direction there should be zero acceleration because a non-zero acceleration in the y direction would mean the block was jumping up off of the ramp:

(87)   \begin{equation*}\Sigma{F_y}=ma_y=0\end{equation*}

(88)   \begin{equation*}F_N-F_{wy}=0\end{equation*}

(89)   \begin{equation*}F_N=F_{wy}\end{equation*}

By summing the forces in the y direction we have shown that the normal force is equal to the y component of the block’s weight. Summing forces in the x direction will give the acceleration in the x direction:

(90)   \begin{equation*}\Sigma{F_x}=ma_x\end{equation*}

(91)   \begin{equation*}-F_{wx}-F_f=ma_x\end{equation*}

(92)   \begin{equation*}-mg\sin(\theta)-\muF_N=ma_x\end{equation*}

(93)   \begin{equation*}-mg\sin(\theta)-\mumg\cos(\theta)=ma_x\end{equation*}

(94)   \begin{equation*}-g\sin(\theta)-\mumg\cos(\theta)=a_x\end{equation*}

(95)   \begin{equation*}(9.81\frac{m}{s^2})[\sin(30) +(0.5)\cos(30)]=a_x\end{equation*}

(96)   \begin{equation*}a_x=-9.15\frac{m}{s^2}\end{equation*}

Knowing the initial velocity and the acceleration, this is now a kinematics problem. How far will the block go in the x direction before it runs out of velocity? First determine how much time it
will take to come to rest:

(97)   \begin{equation*}v=v_0+at\end{equation*}

(98)   \begin{equation*}0=7.35\frac{m/s}-(9.15\frac{m}{s^2}t\end{equation*}

(99)   \begin{equation*}t=\frac{7.35}{9.15}=0.80sec\end{equation*}

Now use the position equation:

(100)   \begin{equation*}x_f=x_0+v_0t+\frac{1}{2}at^2\end{equation*}

(101)   \begin{equation*}x_f=0+(7.35)(0.80)+\frac{1}{2}(9.15)(0.8)^2\end{equation*}

(102)   \begin{equation*}x_f=2.95m\end{equation*}

Therefore, the following geometry is known:

The length of the other two sides can be determined by simple trigonometry:

With these ramp dimensions the block will stop just at the top. Make the ramp slightly smaller for the block to go over the edge and land in the barrel.

If the block is 50cm long, it would be ideal to shorten the ramp from 2.95 m to about 2.70 m. Leaving half of the block teetering over the edge when it comes to rest, it will likely fall if its density is uniformly distributed or slightly denser towards the front half.

Static Equilibrium (Forces)


Statics is the study of forces on systems in static equilibrium- that is, systems that have no acceleration. Analysis of these systems includes free body diagrams and summing forces in the x, y, z directions and moments about a point, then setting them equal to zero.

(103)   \begin{equation*}\Sigma{F_{x,y,z}}=\Sigma{m}*a=0\end{equation*}

(104)   \begin{equation*}\Sigma{M_0}=\Sigma{F}*l=0\end{equation*}

A free body diagram is representation of the system that is helpful in analyzing the forces acting on a body. A good free body diagram should include well defined axis, walls and supports indicated, and all known and unknown forces as well as the angles they act at.

Problem Statement:

Consider a 5kg stop light suspended by cables above a road as shown below. The cables connect to rigid poles on one end and to the hanging light on the other. Determine the tensions T1 and T2. Assume static equilibrium and gravity to be 9.81 \frac{m}{s^2}.



The first step in this sort of problem should always be to draw a free body diagram like the one above. Having this diagram will make summing the forces about point A much simpler. Your diagram should include all known and unknown forces and angles.

The next step is to sum the forces in the x direction, then sum the forces in the y direction and set them equal to zero. We set the sum of the forces equal to zero because “static equilibrium” means that there is no acceleration, so the “ma” in F=ma is zero. This will produce a system of two equations(one in the x direction, one in the y) and two unknowns(T1 and T2).

Summing the forces in the x direction gives:

(105)   \begin{equation*}\Sigma{F_x}=ma=0\end{equation*}

(106)   \begin{equation*}T_2\cos(60)-T_1\cos(30)=0\end{equation*}

Note that:

(107)   \begin{equation*}W=mg\end{equation*}

Summing the forces in the y direction gives:

(108)   \begin{equation*}\Sigma{F_y}=ma=0\end{equation*}

(109)   \begin{equation*}T_1\sin(30)-T_2\sin(60)-mg=0\end{equation*}

Solving the two equations:

(110)   \begin{equation*}0.5T_2-0.866T_1=0\end{equation*}

(111)   \begin{equation*}0.866t_2+0.5T_1-(5)(9.81)=0\end{equation*}

(112)   \begin{equation*}T_1\approx24.5N\end{equation*}

(113)   \begin{equation*}T_2\approx42.5N\end{equation*}

Static Equilibrium (Moments)


A moment is a force that acts at a perpendicular distance from a reference point, causing the object it is acting upon to experience bending, torsion or rotation. A moment is calculated as the product of the magnitude of the force and the perpendicular distance between the force and the reference point. This means that the same force applied at a farther distance can produce the same moment as a larger force applied closer. This principal is clearly demonstrated when using a wrench to loosen tight bolts- the longer the wrench is, the easier it is to loosen a bolt.

Forces that go directly through the reference point have zero length lever arms, thus they produce no moment.

The sign of the moment depends on what direction the force induces spin. It is best to use visualize the problem and curl your fingers in the direction of rotation, then if your thumb is pointing out of the paper call the sign positive, if it is pointing into the paper call it negative. You can choose which is which- into the paper or out of the paper can both be positive or negative, but it is very important to choose a convention and maintain it throughout the problem.

Problem Statement:

A snowman has climbed onto a massless beam to retrieve his lunch which is teetering at the edge. The snowman weighs 50 kg and is 0.3 meters away from the wall. The lunch weighs 5 kg and is at the end of the 0.75 meter long beam. There is a support holding the beam up pinned at the wall on one end and at the edge of the beam in the vertical middle on the other. Find the beam’s reaction forces at the wall and the force in the support. The system is in static equilibrium.



The first step is to draw the free body diagram shown above. For even more clarity, the vector F_s can be broken into its x and y components.

There are three unknowns in this problem: R_x, R_y, and F_s (the components of F_s are found by decomposing the vector with its known angle of action). Because the problem is static – the “a” in “ma” is zero, begin by summing forces and setting them equal to zero in the x and y direction:

(114)   \begin{equation*}\Sigma{F_x}=R_x+F_{sx}=ma_x=0\end{equation*}

(115)   \begin{equation*}\Sigma{F_y}=R_y+F_{sy}-W_s-W_L=ma_y=0\end{equation*}

There are three unknowns (R_x, R_y, and F_s which shows up in component form), so a third equation is necessary. Sum the moments about the beam’s pin (where the reaction forces are) and set them equal to zero.

(116)   \begin{equation*}\Sigma{M_R}=W_sx_s+W_Lx_L-F_{sy}x_s=0\end{equation*}

Note that the reaction forces and F_{sx} do not appear in this equation. This is because they go directly through the pin, thus their lever arm is zero and they generate no moment about the pin.

Substituting known values:

(117)   \begin{equation*}\Sigma{F_x}=R_x+F_s\cos(30)=0\end{equation*}

(118)   \begin{equation*}\Sigma{F_y}=R_y+F_s\sin(30)-(5okg(9.81\frac{m}{s^2}))-(5kg(9.81\frac{m}{s^2}))=0\end{equation*}

(119)   \begin{equation*}\Sigma{M_R}=(5okg(9.81\frac{m}{s^2}))(0.3m)+(5kg(9.81\frac{m}{s^2}))(0.75m)-F_s\sin(30)(0.75m)=0\end{equation*}

Simplifying and solving gives:

(120)   \begin{equation*}R_x=-425N\end{equation*}

(121)   \begin{equation*}R_y=294N\end{equation*}

(122)   \begin{equation*}F_s=491N\end{equation*}

Centripetal Acceleration


Centripetal acceleration is the rate of change of tangential velocity. Centripetal acceleration is always inwards towards the center, along the radius vector of the circle. The magnitude is given by:

(123)   \begin{equation*}a_c=\frac{v^2_t}{r}\end{equation*}

Tangential velocity relates to angular velocity by:

(124)   \begin{equation*}v_t=r\omega\end{equation*}

Therefore centripetal acceleration can also be expressed as:

(125)   \begin{equation*}a_c=\frac{r^2\omega^2}{r}=r\omega^2\end{equation*}

Problem Statement:

In geosynchronous orbit a satellite always has to be found at the same
location with respect to the surface of the Earth. This means that the satellite must be at a height such that its revolution period is the same as that of the Earth, 24 hours. At what height must a 100 kg satellite be to achieve geosynchronous orbit?


While truly an ellipse, the path of an orbiting satellite is very nearly a circle and will be modeled as such.

A summary of the information given and useful information available in reference materials will help to keep the solution organized:

(126)   \begin{equation*}G=6.67*10^{-11}\frac{Nm^2}{kg^2}\end{equation*}

(127)   \begin{equation*}T=24hr=86400sec\end{equation*}

(128)   \begin{equation*}m_{earth}=5.98*10^{24}kg\end{equation*}

(129)   \begin{equation*}m_{sat}=100kg\end{equation*}

(130)   \begin{equation*}r_{earth}=6.38*10^6m\end{equation*}

The satellite is held in orbit by centripetal acceleration, defined as:

(131)   \begin{equation*}a=\frac{v^2}{r}\end{equation*}

Newton’s second law of motion requires:

(132)   \begin{equation*}F=m_{sat}a\end{equation*}

Newton’s law of universal gravitation states:

(133)   \begin{equation*}F=\frac{Gm_{earth}m_{sat}}{r^2}\end{equation*}

The satellite’s velocity is given by:

(134)   \begin{equation*}v=\frac{distance}{time}=\frac{2\Pi*r}{T}\end{equation*}

Where G is the gravitational constant and r is the average distance from the center of the earth to the satellite. The system of four independent equations has four unknowns(r, a, F, and v) so it is solvable.

Solving the system of equations yields:

(135)   \begin{equation*}r=4.23*10^7m\end{equation*}

Subtracting the radius of the earth gives the height of the satellite from Earth’s surface:

(136)   \begin{equation*}h=4.23*10^7-6.38*10^6=3.59*10^7m\end{equation*}

This converts to about 22,300 miles.

Coulmb’s Law


Coulomb’s law relates the electrostatic force between two electric point charges to the magnitude of the charges and the distance between them. When using Coulomb’s law it’s important to remember that it applies only to point charges, it applies only to electrostatics and the electric forces can be summed to get a net electric force just like other forces.

When determining the direction of the force, remember that opposite charges will attract and like charges will repel one another.

The magnitude of the force is given by:

(137)   \begin{equation*}F_{AonB}=\frac{K|q_A\parallel{q_B}|}{r^2_{AB}}\end{equation*}

Problem statement:

Three charged particles with q_1 = -40 nC, q_2 = +40 nC and
q_3 = +30 nC are placed on the corners of a 4 cm by 8 cm rectangle as shown. What is the net force on q_3 due to the other two charges?



First identify the forces and radii that will be involved in the problem:

q_1 and q_3 are opposite charges, so force vector \vec{F}_{1on3} is an attractive force toward q_1. q_2 and q_3 are like charges, so force vector \vec{F}_{2on3} is a repulsive force away from q_2. \vec{F}_3 is the sum of \vec{F}_{1on3} and \vec{F}_{2on3} and it is the unknown we are trying to find.

The magnitude of \vec{F}_{1on3} is found by applying Coulomb’s law:

(138)   \begin{equation*}F_{1on3}=\frac{K|q_1\parallel{q_3}|}{r^2_{13}}=\frac{(9.0*10^9\frac{Nm^2}{C^2})(40*10^-9C)(30*10^-9C)}{(0.08m)^2}=1.69*10^-3N\end{equation*}

The vector \vec{F}_{1on3} points straight down, so it will be written as:

(139)   \begin{equation*}\vec{F}_{1on3}=-1.69*10^-3\hat{j}N\end{equation*}

Before calculating the magnitude F_{2on3} find the radius r_{23}:

(140)   \begin{equation*}=\sqrt{(4cm)^2+(8cm)^2}\approx8.94cm\end{equation*}

Now calculate the magnitude F_{2on3}:

(141)   \begin{equation*}F_{2on3}=\frac{K|q_2\parallel{q_3}|}{r^2_{23}}=\frac{(9.0*10^9\frac{Nm^2}{C^2})(40*10^-9C)(30*10^-9C)}{(0.0894m)^2}=1.35*10^-3N\end{equation*}

Before calculating the vector \vec{F}_{2on3} find the \theta:

(142)   \begin{equation*}\theta=\tan^{-1}\frac{8cm}{4cm}=63.4\end{equation*}

Now calculate the vector \vec{F}_{2on3}:

(143)   \begin{equation*}\vec{F}_{2on3}=-F_{2on3}\cos(\theta)\hat{i}+F_{2on3}\sin(\theta)\hat{j}\end{equation*}

(144)   \begin{equation*}\vec{F}_{2on3}=-(1.35*10^{-3}N)\cos(63.4)\hat{i}+(1.35*10^{-3}N)\sin(63.4)\hat{j}\end{equation*}

(145)   \begin{equation*}\vec{F}_{2on3}=-6.04*10^{-4}\hat{i}+1.21*10^{-3}\hat{j}\end{equation*}

Now sum to find \vec{F}_3:

(146)   \begin{equation*}F_3=F_{1on3}+F_{2on3}\end{equation*}

(147)   \begin{equation*}\vec{F}_3=-1.69*10^{-3}\hat{j}+(-6.04*10^{-4}\hat{i})+1.21*10^{-3}\hat{j}N\end{equation*}

(148)   \begin{equation*}\vec{F}_3=-6.04*10^{-4}\hat{i}-4.80*10^{-4}\hat{j}N\end{equation*}

This result can be expressed in terms of magnitude and direction:

(149)   \begin{equation*}F_3=\sqrt{F^2_{3x}+F^2_{3y}}=\sqrt{(-6.04*106{-4})^2+(-4.80*10^{-4})^2}=7.72*10^{-4} N\end{equation*}

(150)   \begin{equation*}\phi=\tan^{-1}|\frac{F_{3y}}{F_{3x}}|=tan^{-1}|\frac{4.80}{6.04}|=38.5\end{equation*}

Magnetic Force


The magnetic force on a moving charged particle is perpendicular to both its velocity vector,
v and the magnetic field, B that it is acting on it. The magnetic force on a moving charged particle is given by:

(151)   \begin{equation*}\vec{F}=q\vec{v}\times{\vec{B}}\end{equation*}

The magnitude of the force can be found:

(152)   \begin{equation*}F=qvB\sin(\theta)\end{equation*}

Where \theta is the angle between the velocity and the magnetic filed vectors. The magnetic force on a stationary charge, or on a charge moving parallel to the magnetic field is zero. When using this formula, the direction of the force vector is given by the right hand rule.

Problem Statement:

A long wire carries a 5 A current from left to right. An electron 0.5 cm above the wire is traveling to the right at a speed of 1.5*10^{7}\frac{m}{s}. What are the magnitude and direction of the magnetic force on the electron?



The current and electron are both moving to the right. The right-hand rule states that the wire’s magnetic field above the wire is out of the page, so the electron is moving perpendicular to the field.

The electron charge is negative, so the direction of the force is opposite the direction of \vec{v}\times{\vec{B}}. The right-hand rule says that \vec{v}\times{\vec{B}} points down toward the wire, so \vec{F} points up, away from the wire.

B is given by:

(153)   \begin{equation*}B=\frac{\mu_0I}{2\Pi*d}=\frac{(1.26*10^{-6}\frac{Tm}{A})(5A)}{2(\Pi)(0.005m)}=2.01*10^{-4}T\end{equation*}

The magnitude of the force is given by:

(154)   \begin{equation*}F=|q|vB=evB=(1.60*10^{-16}C)(1.5*10^{7}\frac{m}{s})(2.01*10^{-4}T)=4.82*10^{-13}N\end{equation*}

Making our final solution:

(155)   \begin{equation*}\vec{F}=4.82*10^{-13} N, up\end{equation*}

Resistors in Series


When looking at two terminals in a circuit with multiple resistors elements between them, it is possible to find an equivalent value of resistance and effectively replace the multiple resistors with a single resistor. The equivalent value is the value that an ohmmeter would read if it was applied at terminals. The equivalent resistance of series connected resistors is:

(156)   \begin{equation*}R_{equiv}=R_1+R_2+...+R_{N-1}+R_N\end{equation*}

Series connected elements are ones that have the same current flowing through all of them.

Problem Statement:

Find the equivalent resistance between terminals A and B.



The goal is to replace the four resistors with one equivalent resistor. The current at A has no where to go excpt through all of the resistors and out at B, the same current flows through each of the four resistors.

The resistors are in series, so the resistances add. The equivalent resistance is thus:

(157)   \begin{equation*}R_{equiv}=5k\Omega+15k\Omega+10k\Omega+5k\Omega=53k\Omega\end{equation*}

Resistors in Parallel


When looking at two terminals in a circuit with multiple resistor elements between them, it is possible to find an equivalent value of resistance and effectively replace the multiple resistors with a single resistor. The equivalent value is the value that an ohmmeter would read if it was applied at the terminals. The equivalent resistance of N parallel connected resistors is:

(158)   \begin{equation*}\frac{1}{R_{equiv}}=\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_{N-1}}+\frac{1}{R_N}\end{equation*}

(159)   \begin{equation*}R_{equiv}=\frac{1}{\frac{1}{R_{equiv}}=\frac{1}{R_1}+\frac{1}{R_2}+...+\frac{1}{R_{N-1}}+\frac{1}{R_N}}\end{equation*}

Parallel connected elements are ones that have the same voltage drop across them.

Problem Statement:

Find the equivalent resistance of the given circuit.



The goal is to replace the four given resistors with one equivalent resistor. The current that enters at A will split between all of the resistors, more current will go through lower resistance and less will go through higher resistance – so the the resistors are not in series. The voltage drop across all of the resistors will be the same.

The resistors are in parallel, so use the reciprocal addition rule:

(160)   \begin{equation*}\frac{1}{R_{equiv}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}\end{equation*}

(161)   \begin{equation*}\frac{1}{R_{equiv}}=\frac{1}{15k\Omega}+\frac{1}{10k\Omega}+\frac{1}{15k\Omega}+\frac{1}{5k\Omega}=\frac{13}{30}k\Omega\end{equation*}

Note that:

(162)   \begin{equation*}\frac{1}{a}+\frac{1}{b}\neq\frac{1}{a+b}\end{equation*}

To sum the fractions you must first give them all a common denominator, then add the numerators.

Finally take the reciprocal of \frac{1}{R_{equiv}}=\frac{13}{30}k\Omega to find that:

(163)   \begin{equation*}R_{equiv}=\frac{30}{13}\approx2.3k\Omega\end{equation*}

Capacitors in Series


When considering two terminals in a circuit with multiple capacitor elements between them, it is possible to find an equivalent value of capacitance and effectively replace the multiple capacitors with a single capacitor. The equivalent capacitance of series connected capacitors is:

(164)   \begin{equation*}\frac{1}{C_{equiv}}=\frac{1}{C_1}+\frac{1}{C_2}+...+\frac{1}{C_{N-1}}+\frac{1}{C_N}\end{equation*}

(165)   \begin{equation*}C_{equiv}=\frac{1}{\frac{1}{C_1}+\frac{1}{C_2}+...+\frac{1}{C_{N-1}}+\frac{1}{C_N}}\end{equation*}

Series connected elements are ones that have the same current flowing through all of them.

Problem Statement:

Determine the equivalent capacitance between terminals A and B in the circuit shown.



The current at A has no where to go except through all of the capacitors and out at B, the same current flows through each of the four capacitors. Therefore, they are in series.

Capacitors in series can be combined with the reciprocal addition rule. They do not all need to be combined at once. Consider the right-most capacitors:

(166)   \begin{equation*}C_{new}=\frac{1}{\frac{1}{0.4}+\frac{1}{0.2}}=\frac{(0.4)(0.2)}{(0.4+0.2)}=0.13\mu{F}\end{equation*}

The new circuit:

These two remaining capacitors can be combined in a similar manner:

(167)   \begin{equation*}C_{equiv}=\frac{1}{\frac{1}{0.1}+\frac{1}{0.13}}=\frac{(0.1)(0.13)}{(0.1+0..13)}=0.057\mu{F}\end{equation*}

Making the final solution:

Capacitors in Parallel


When looking at two terminals in a circuit with multiple capacitor elements between them, it is possible to find an equivalent value of capacitance and effectively replace the multiple capacitors with a capacitor. The equivalent resistance of N parallel connected capacitors is:

(168)   \begin{equation*}C_{equiv}=C_1+C_2+...+C_{N-1}+C_N\end{equation*}

Parallel connected elements are ones that have the same voltage drop across them.

Problem Statement:

Find the equivalent capacitance of the given circuit.



The current tat enters at A will split between all of the capacitors, more current will go through lower impedance and less will go through higher impedance – so the capacitors are not in series. The voltage drop across all of the capacitors will be the same. Therefore,they are in parallel.

Capacitors in parallel can be combined by adding their capacitance. Consider the two right-most capacitors:


Combining these two capacitors gives:

The remaining two capacitors can be combined in the same manner:

(169)   \begin{equation*}C_{equiv}=1.2+0.2=1.4\mu{F}\end{equation*}

This gives the final solution:

Identifying Series and Parallel Components


Circuits can contain a combination of both series and parallel elements. The rules for combining
elements still apply, but the application of those rules may not be as clear. Circuits will not
always be drawn in a simple, organized manner. It is very common to not be able to identify if
elements are in series or parallel at a quick glance. When this happens, recall that elements in
series have the same current going through them- if there is no where for current to split between
elements then they are in series. Also recall that elements with the same voltage drop across
them are in parallel. Applying these rules can be simplified by examining how many common
nodes two elements share. If they have one common node, they are in series. If they have two
common nodes they are in parallel. If they have no common nodes, more simplifications are
required before you compare the two elements.

Problem Statement:

Simplify the given circuit by identifying resistors in series and parallel, then replacing them with
equivalent resistances.



Two resistors are in series if the current that flows through them is the same. Every point where
one or more resistors meet is called a node. Elements in series have one common node while
elements in parallel have two common nodes. The following picture identifies and labels all of
the nodes in this circuit:

For this circuit it will be easiest to start at the far right, and work towards the terminals. Consider
the 4k\Omega and 6k\Omega resistors. The 4k\Omega resistor is connected to nodes ii and iii, while the 4k\Omega is connected to iii and iv. Because they only share one common node (iii), these two resistors are in series, so they can be replaced by a 4+6=10k\Omega resistor as shown.

Note that node iii no longer exists, only nodes i, ii, and iv remain.

Next consider the 10k\Omega resistor on the far right and the 2k\Omega resistor to the left of it. The 10k\Omega resistor is connected to nodes ii and iv, the 2k\Omega resistor is connected to nodes ii and iv also. Because both resistors are connected to the same nodes, these two are in parallel, so we calculate their equivalent resistance using the reciprocal addition method:

(170)   \begin{equation*}\frac{1}{R_{equiv}}=\frac{1}{2}+\frac{1}{10}=\frac{3}{5}k\Omega\end{equation*}

(171)   \begin{equation*}R_{equiv}=\frac{5}{3}\approx1.67k\Omega\end{equation*}

As a shortcut, when adding two resistors in parallel simply multiply their values, then divide by the sum of the two. Note that this only works for two resistors (not 3 or more) that are in parallel.

Now consider the 1.67k\Omega and remaining 10k\Omega resistors. The 1.67k\Omega resistor is attached to nodes ii and iv, while the 10k\Omega is attached to nodes i and ii. Because they only have one common node, these two resistors are in series, so they will be replaced with a 10+1.67=11.7k\Omega resistor as shown.

 he final two resistors are both connected to nodes i and iv, so they are in parallel. Using the multiply and divide by the sum shortcut:

(172)   \begin{equation*}R_{equiv}=\frac{(5)(11.7)}{5+11.7}=3.50k\Omega\end{equation*}

The final simplified circuit looks like this:

Hydrostatic Pressure


Hydrostatic problems deal with fluids – which can be liquids or gasses – that are at rest. Hydrostatics is used to analyze the forces acting on floating or submerged objects. When an object is submerged in fluid, the weight of the fluid above it exerts pressure on it. Pressure is defined as force per unit area. The density of the fluid the object is in will effect how much pressure it experiences because the denser the fluid above it is, the more weight will be pushing on it. Density is defined as mass divided by volume. The depth of the object will also impact the pressure because the deeper the object is, the more fluid there is on top of it.

Problem Statement:

You drop your physics book in a tank of water 10 meters deep and it falls to the bottom. The density of water is approximately 1000 \frac{kg}{m^3}. What is the total pressure being exerted on your textbook? (Do not solve experimentally.)



The solution to this problem requires a combination of the definition of pressure and density as well as Newton’s second law:

(173)   \begin{equation*}P=\frac{F}{A}\end{equation*}

(174)   \begin{equation*}\rho=\frac{m}{V}\end{equation*}

(175)   \begin{equation*}F=ma=mg\end{equation*}

Rearranging equation (174) and recognizing that volume is equal to the height of the water column above the book times the area of the book:

(176)   \begin{equation*}m=\rho{V}=\rho{h}A\end{equation*}

Plugging this formula for mass into equation (175) gives:

(177)   \begin{equation*}F=\rho{h}Ag\end{equation*}

Plugging this force into the pressure equation (173):

(178)   \begin{equation*}P=\frac{F}{A}=\frac{\rho{h}Ag}{A}=\rho{g}h\end{equation*}

Plugging in known values to this equation:

(179)   \begin{equation*}P=\rho{g}h=(1000\frac{kg}{m^3})(9.81\frac{m}{s^2})(10m)=98100\frac{kg}{ms^2}=98.1\frac{kN}{m^2}=9.81kPa\end{equation*}



Anyone who has been in a pool has experienced the force of buoyancy. It is what allows people, ships and other objects to float on top of water. The fact that these objects do not sink is a strong indication that the water is exerting an upward force. It is, and that force is called the Buoyant force. The Buoyant force comes directly from the fact that pressure increases with depth in a fluid. Consider a flat plate that is submerged parallel to the fluid surface:

The plate is at a depth d, has height h and area A. There are forces acting on all sides of the plate, but the horizontal forces will offset one another so the Buoyant force is related to only the forces on the top and bottom of the plate. Recall that:

(180)   \begin{equation*}F=ma=mg\end{equation*}

(181)   \begin{equation*}\rho=\frac{m}{V}\end{equation*}

Solving equation (181) for mass and rewriting volume as the product of area and height of the fluid column on top of the object, then substituting into equation (180) gives the following result:

(182)   \begin{equation*}F_{TOP}=\rho{g}dA\end{equation*}

(183)   \begin{equation*}F_{BOTTOM}=\rho{g}(d+h)A\end{equation*}

The Buoyant force is the net force in the Y direction, so it can be defined as:

(184)   \begin{equation*}F_B=F_{BOTTOM}-F_{TOP}=\rho{g}(d+h)A=\rho{g}dA\end{equation*}

(185)   \begin{equation*}F_B=\rho{g}hA=\rho{g}V\end{equation*}

This result has important implications. First, the Buoyant force is equal to the weight of the volume of fluid that a submerged object displaces. Second, the Buoyant force is independent of depth in the fluid, it only depends on the volume of fluid displaced, gravity and the density of the fluid. Finally, consider the effect that the object’s density will have on its behavior once submerged. If the density of the object is less than that of the fluid, it will float on top of the fluid. If the density of the object is equal to that of the fluid, it will suspended in the fluid. If the density of the solid is greater than the density of the fluid, the object will sink.

Problem Statement:

A metal sphere with density 3000 \frac{kg}{m^3} and a radius of 2 m is being lowered into a pool of water with density 1000 \frac{kg}{m^3}
using a crane. Determine the tension in the crane’s cable before and after the sphere is submerged.



First consider the sphere out of water, suspended in the air. The density of air is about 1.5 \frac{kg}{m^3}, the density of water is about 1000 \frac{kg}{m^3}. The density of air is relatively very small compared to that of air, so the Buoyant effect of air will be ignored- though it does exist.

A free body diagram of the sphere out of water looks like this:

Summing the forces in the y direction gives:

(186)   \begin{equation*}\Sigma{F_y}=ma=0\end{equation*}

(187)   \begin{equation*}T-W=0\end{equation*}

(188)   \begin{equation*}T=W=mg\end{equation*}

To find the mass, first find the volume:

(189)   \begin{equation*}V=\frac{4}{3}\Pi{r^3}=\frac{4}{3}\Pi{(2m)^3}=35.3m^3\end{equation*}

Then the mass is:

(190)   \begin{equation*}m=\rho_s{V}=(3000\frac{kg}{m^3})(33.5m^3)=1.01*10^{5}kg\end{equation*}

Making the tension:

(191)   \begin{equation*}T=mg=(1.01*10^{5}kg)(9.81\frac{m}{s^2})=9.86*10^{5}N\end{equation*}

Now consider this free body diagram of the submerged sphere:

Summing the forces in the y direction gives:

(192)   \begin{equation*}\Sigma{F_y}=ma=0\end{equation*}

(193)   \begin{equation*}T+F_B-W=o\end{equation*}

(194)   \begin{equation*}T=W-F_B=mg-\rho_f{V}g\end{equation*}

(195)   \begin{equation*}T=(1.01*10^{5}kg)(9.81\frac{m}{s^2})-(1000\frac{kg}{m^3})(33.5m^3)(9.81\frac{m}{s^2}=6.67*10^{5}N\end{equation*}

Be careful to use the density of the solid when determining the mass of the solid sphere and the density of the fluid when determining the buoyant force from the volume of the displaced fluid.

Bernoulli Equation


The Bernoulli equation relates flow energy, kinetic energy and potential energy in a steady, incompressible flow along a streamline in inviscid regions of flow. Before applying the Bernoulli equation, it is important to check that it is valid for the scenario being analyzed.

Steady flow is flow that is smooth and uniform and has constant velocity with respect to time at any point. A Reynolds number less than 1000 typically indicates steady flow.

In-compressible flow can be thought of as having a working fluid with constant density. More precisely, in-compressible flows are those which have zero divergence of velocity but the constant density definition should be adequate for lower level fluid mechanics studies.

A streamline is a curve that is always tangent to the velocity vector. Analysis with the Bernoulli equation requires all points be on the same streamline. The thin, arrow lines below represent streamlines while the thicker lines are fluid flow boundaries.

Inviscid flow is flow where the viscous or frictional forces are relatively small compared to inertial forces. flow is flow where the viscous or frictional forces are relatively small compared to inertial forces.

Now that the conditions on its use are understood, the Bernoulli equation is presented:

(196)   \begin{equation*}\frac{v^2}{2}+gz+\frac{P}{\rho}=constant\end{equation*}

Problem Statement:

A dunk tank is full of water 1 meter high. After it is used, it is drained by removing a plug from an outlet near the bottom. Find the maximum velocity of the water coming out of the outlet.



Point 1 is at the water surface, exposed to atmospheric pressure at 1m elevation. Point 2 is at the outlet with 0m elevation. Bernoulli’s equation gives:

(197)   \begin{equation*}\frac{v^2_1}{2}+gz_1+\frac{P_1}{\rho}=\frac{v^2_2}{2}+gz_2+\frac{P_2}{\rho}\end{equation*}

This equation can be simplified considerably. Both points 1 and 2 are exposed to the atmosphere, so they are both at atmospheric pressure- the pressure terms will cancel. The height at two is zero, so the z_2 term will go to zero. The velocity at 1 is essentially zero, so the v_1 term will go to zero. This leaves:

(198)   \begin{equation*}gz_1=\frac{v^2_2}{2}\end{equation*}

Solving for the velocity at the outlet:

(199)   \begin{equation*}v_2=\sqrt{2gz_1}=\sqrt{2(9.81\frac{m}{s^2})(1m)}=4.43\frac{m}{s}\end{equation*}

Pascal’s Principal


Pascal’s law says that a change in the pressure of an enclosed, in-compressible fluid will be distributed throughout the entire fluid and to the walls of the container. This simple idea has very useful implications. Hydraulic lifts which allow small forces to lift very heavy objects operate on this principal. By connecting two hydraulic piston-cylinder apparatuses of different diameters, the larger one can exert a larger force than is applied to the smaller one. The following diagram illustrates the concept:

If the heights of the pistons are the same, then P_1=P_2. Therefore:

(200)   \begin{equation*}\frac{F_1}{A_1}=\frac{F_2}{A_2}\end{equation*}

Problem Statement:

Black Walnuts are some of the toughest nuts to crack. You are using a hydraulic press to crush one. If the walnut can withstand 3 kN of force and you are able to exert only 500 N, how large should the area of the large piston be to crack the nut if the smaller piston has an area of 0.1 m^2?


Simply apply the proportion and solve for the larger area:

(201)   \begin{equation*}\frac{F_1}{A_1}=\frac{F_2}{A_2}\end{equation*}

(202)   \begin{equation*}A_2=\frac{A_1F_2}{F_1}=\frac{(0.1m^2)(3000N)}{(500N)}=0.6m^2\end{equation*}