# Forces In Physics

These are my notes on forces in physics.

**Intro**

In the development of mechanics, the first thing to learn is the interrelations

of position, velocity, and acceleration. These interrelations are described with

the four kinematic equations of motion.

In this section, we relate constant force to constant acceleration and then to

all of kinematics. Force can be defined as what makes masses accelerate.

Actually, it is the unbalanced force on a mass that makes it accelerate. So,

unbalanced forces make masses accelerate.

Force, acceleration, and mass are related by Newton's second law as F = ma.**Example 1**

A 40 N force is applied to a 20 kg block resting on a horizontal frictionless

table. Find the acceleration.

Solution:

The acceleration is \(a=\frac{f}{m} = \frac{40N}{20kg} = 2.0m/s^2\)

Once the acceleration is known, the kinematic equations of motion can be applied

to find position, velocity, and acceleration as a function of time.**Example 2**

Now place the 20 kg mass on a frictionless 50 degree inclined plane. What is the

acceleration of the mass?

Solution:

The force acting on the mass is due to gravity, so setup a vector diagram

starting with this 196 N force acting down. The mass is constrained to move down

the plane. The gravitational force is shown with components down the plane and

normal to the plane. Notice the geometry of the situation, where the 50 degree

angle of the plane is the same as the angle between the gravitational force and

the normal force.

So, the force straight down is \(F=20kg(9.8m/s^2)=196N\)

However, the mass is going down an incline plane at 50 degrees.

\(F_{p}=(196N)(sin50)=150N\)

The 150 N force acting down the plane causes the 20 kg mass to accelerate at:

\(a=\frac{150N}{20kg}=7.5m/s^2\)**Example 3**

Place a 10 kg mass on a frictionless 35 degree incline plane, and attach a

second 20 kg mass via a cord to hang vertically. Calculate the acceleration of

the system.

Solution:

\(mg = 98N\)

\(98N*sin35 = 56N\)

\(98N*cos35 = 80N\)

Mass 1 is 10 kg at 35 degree down incline plane.

Mass 2 is 20 kg and hangs straight down. It has a downward force of 196N.

This problem introduces the concept of the tension in the connecting cord. The

most convenient way to visualize this tension is that if the cord were cut and a

force meter inserted, it would read a certain tension(force). This tension acts

on each mass. Notice the 35 degree angle between the normal and the direction of

mg.

To write equations relating the unbalanced force to the acceleration, we need to

make an assumption as to which way the masses move. Assume that the system moves

to the right or clockwise.

Start by writing equations relating the unbalanced force on each mass. On m1,

the unbalanced force is T-56N, and this force makes the 10 kg accelerate.

\(T-56N = 10kg*a\)

Likewise on mass2, the unbalanced force to make the mass accelerate is:

\(196N-T = 20kg*a\)

The assumption that that system accelerates clockwise dictates the directions of

the forces in these two equations. Look again at these equations, and notice

that if we had assumed the acceleration were counterclockwise, then the signs

would be different. In frictionless problems it is alright to have a negative

acceleration. The negative sign means that you guessed wrong in assuming the

acceleration direction. This is not true with friction problems. Add the two

equations, and take the acceleration as the same. The accelerations are the same

because the masses are linked together with the cord.

\(140N = 30kg * a | a = 4.7m/s^2\)

Now the tension in the cord is easily calculated from the equation for mass 1.

\(T = 56N + 10kg * 4.7m/s^2 = 103N\)**Example 4**

The next complication in force problems with inclined planes is a double

inclined plane.

Mass 1 is 12 kg going down to the left of the apex at a 40 degree angle.

Mass 2 is 20 kg going down to the right of the apex at a 25 degree angle.

Solution: Create a vector diagram, it is helpful.

Assume the system moves clockwise.

We will start with mass 1 first.

The force straight down is 118 N, from (12 kg * 9.8 m/s^2).

Now, lets calculate the force down the incline plane to the left.

118 N * sin 40 = 76 N

Now, lets go to mass 2.

The force straight down is 196 N, from (20 kg * 9.8 m/s^2).

Now, lets calculate the force down the incline plane to the right.

196 N * sin 25 = 83 N

The equation for mass 1 is: T-76N = 12 kg * a

The equation for mass 2 is: 83N-T = 20 kg * a

Add the two equations

7.0N = 32 kg * a

a = 0.22 m/s^2

The tension in the cord is:

T = 76N + 12 kg * 0.22 m/s^2 = 79 N**Example 5**

Another popular problem is the Atwood machine. Find the acceleration of the

system.

Solution: Assume that the system moves counterclockwise.

The equation for mass 1 is 98N-T = 10kg * a

The equation for mass 2 is T-147N = 15kg * a

Add the two equations

-49N = 25kg * a

a = -2.0 m/s^2

Force of mass 1 is 10kg * 9.8 m/s^2 = 98N

Force of mass 2 is 15kg * 9.8 m/s^2 = 147N

The system accelerates clockwise. The tension in the cord is from the equation

for mass 1.

T = 98N + 10kg * 2.0m/s^2 = 118N**Friction**

A good first problem in friction is where the problem is first done without

friction and then with friction.

For the purposes of doing problems, there are two important properties of

frictional forces:

They oppose the motion

They are less than or equal to a constant (coefficient of friction), times the

normal force, the force at the frictioning surface.**Example 6**

Two masses are arranged with one, of 50kg, on a frictionless table and the

other, of 30 kg, attached by a cord and hanging, over a frictionless pulley, off

the edge of a table. Find the acceleration of the system.

Solution: Assume that the system moves clockwise.

The equation for mass 1 is T=50kg * a

The equation for mass 2 is 294N-T = 30kg * a

Add the two equations:

294N = 80kg * a

a = 3.7 m/s^2**Example 7**

For the situation in Example 6, add a coefficient of friction \mu=.20 between

the block and the table.

Solution:

As above, horizontal:

\mu N = mu*m*g = 0.20*50kg*9.8m/s^2 = 98N

Vertical:

294N

The equation for mass 1 is T-98N=50kg*a

The equation for mass 2 is 294N-T=30kg*a

Add the equations:

196N=80kg*a

a=2.4m/s^2**Example 8**

For the situation in the previous problem, increase the coefficient of friction

to 0.80. What is the acceleration of the system?

Solution:

\(f=\mu N=\mu*m*g=0.80*50kg*9,8m/s^2=392N\)

\(294N\)

The equation for mass 1 is \(T-392N=50kg*a\)

The equation for mass 2 is \(294-T=30kg*a\)

Add the two equations

\(-98N=80kg*a\)

\(a=-1.2m/s^2\)

Clearly, the system does not accelerate counterclockwise. This is an

illustration of the second property of frictional forces: the frictional force

is less than or equal to \(\mu N\). In this case the frictional force just

balances the tension in the cord. The frictional force can be up to 392N. In

order for mass 2 not to move, the tension in the cord must be 294N. If mass 2 is

not moving, the forces must be in balance. Therefore, the forces on mass 1 must

be 294N due to the tension in the cord and 294N due to friction.

Another important point with frictional forces is the minimal force.**Example 9**

A 50kg sled is pulled along a level surface at constant velocity by a constant

force of 200N at an angle of 30 degrees. What is the coefficient of friction

between the sled and the surface?

Solution: The first step in this problem is to find the components of the

applied force and the mg of the sled. These are shown in the vector diagram.

Notice the normal force, the actual force between the sled and the friction

surface, is mg minus the component of the applied force lifting up the sled.

Sometimes it is easy to miss this vertical component of the applied force. This

is a major mistake in this type of problem. To help visualize this vertical

component of force, try placing the origin for the vectors in the middle of the

sled rather than at one end, where a rope would be attached.

\(f = \mu*N\)

\(mg = 50kg*9.8m/s^2 = 490N\)

\(200N * sin 30 = 100N\)

\(200N * cos 30 = 173N\)

The normal force is mg minus the vertical component of the applied force, 100N,

or 390N total. Since the sled is moving at constant velocity, the forces must be

in equilibrium. The horizontal component of the applied force must equal the

frictional retarding force, the coefficient of friction times the normal force.

\(\mu*390N = 173N\)

\(\mu = 0.44\)**Example 10**

A 65 degree inclined plane has a mass of 20kg on the plane where the coefficient

of friction is 0.40 and a mass of 50kg hanging free. Calculate the acceleration

of the system.

Solution: Setup the vector diagram assuming that the motion is counterclockwise.

In the diagram, the frictional force is calculated as a maximum of 33N.

The equation for mass2 is \(490N-T=50kg*a\)

The equation for mass1 is \(T-178N=50kg*a\)

Add the equations: \(279N=70kg*a\)

\(a=4.0m/s^2\)

\(mg = 50kg*9.8m/s^2 = 490N\)

Mass2 of 50kg is hanging straight down off a pulley.

Mass1 of 20kg is sliding down an incline plane. The plane is a 65 degree angle.

Coefficient of friction is 0.40

\(196N cos65=83N\)

\(196N sin65=178N\)

\(0.40*83N=33N\)

In doing friction problems like this, you have to be careful with the frictional

force. If the forces up and down the plane are nearly balanced, it is important

to remember that the frictional force is not 33N but that it can be up to 33N.

In a problem similar tot his, it could easily happen that the blocks would not

move.**Example 11**

Consider a double inclined plane with angles, masses, and coefficients of

friction as shown. Calculate the acceleration of the system.

You have a sort of triangle with the apex at the top. To the left and right of

the apex are inclined planes. On the left inclined plane, you have mass1 of

6.0kg, coefficient of friction of 0.30, and going down at an angle of 30

degrees.

On the right inclined plane, you have mass2 of 9.0kg, coefficient of friction of

0.16, and sliding down at an angle of 45 degrees.

Solution: Assume that the system moves counterclockwise, and set up the vector

diagram.

The equation for mass1 is \((30-15-T)N=6.0kg*a\)

The equation for mass2 is \((T-62-10)N=9.0kg*a\)

Add the equations \(-57N=15kg*a\)

\(59N sin30=30N\)

\(59N cos30=51N\)

\(f=0.30*51N=15N\)

\(88N sin45=62N\)

\(88N cos45=62N\)

\(f=0.16*62N=10N\)

This is a negative acceleration. We cannot simply reverse the sign of "a" to

find the correct answer because if the system were moving the other way, the

frictional forces would have to be reversed. further analysis requires us to

consider another vector diagram assuming that the system moves in a clockwise

direction.

The new equations are \((T-30-15)N=6.0kg*a\)

\((62-T-10)N=9.0kg*a\)

Add the equations: \(7.0N=15kg*a\)

\(a=0.54m/s^2\)

The tension in the cord is from the first equation:

\(T=45N+6.0kg*0.54m/s^2)=48N\)

If the acceleration of the system were negative for both the clockwise and

counterclockwise calculations, then we would conclude that the system would not

move.

The final complication in these inclined plane problems is the introduction of

three masses and two connecting rods.**Example 12**

Consider a flat surface with a coefficient of friction of 0.20 with a 2.0kg mass

and a 3kg mass connected together with a 6kg mass along a slant as shown. The

coefficient of friction on the slant is 0. The slant is going down at an angle

of 25 degrees. Calculate the acceleration of the system.

Solution:

In this problem, note that the tensions are different, leading to 3 unknowns,

T1, T2, and a. The solution will require three equations from three vector

diagrams. Assume that the acceleration of the system is clockwise.

The frictional retarding force on the 2kg block is:

\(f=\mu N=0.20*19.6N=3.92N\)

The frictional retarding force on the 3kg block is:

\(f=\mu N=0.20*29.4N=5.88N\)

\(58.8N sin 25 =24.8N\)

The equations for the masses are:

\(T2-3.92N=2.0kg*a\)

\(T1-T2-5.88N=3.0kg*a\)

\(24.8N-T1=6.0kg*a\)

The last two equations added together are:

\(18.9N-T2=9.0kg*a\)

Adding this to the first equation is:

\(15.0N=11kg*a\)

\(a=1.36m/s^2\)

The tensions come from the first and third equations for the masses:

\(T2=3.92N+2.0kg*1.36m/s^2=6.64N\)

\(T1=24.8N-6.0kg*1.36m/s^2=16.6N\)**Circular Motion**

A particle moving at constant speed in a circle is moving in uniform circular

motion. There is an acceleration and a force associated with such a particle.

While the length of the velocity vector remains the same, the direction changes

continually. Acceleration points radially inward.

Example 13

A certain car driven in a circle can exert a maximum side force of 0.95g. What

is the maximum speed for this car driven in a circle of 160m radius.

Solution:

A side force of 0.95g means a side force or acceleration directed at right

angles to the direction of travel of 0.95*9.8=9.3m/s^2. Using \(a_{rad}=v^2/r\):

\(v=\sqrt{a_{rad} r} = \sqrt{9.3m/s^2*160m} = 38.6m/s\)

At a speed greater than 38.6 m/s, the car will slide out of the circle.

The acceleration directed toward the center of the circle for a mass in uniform

circular motion is called centripetal acceleration. The force associated with

moving a mass in a circle is \(ma_{rad}\) and is called centripetal force.**Example 14**

What is the centripetal force produced by the tires acting on the pavement for

the car of the previous problem if the car is 1200kg of mass.

Solution:

\(F=ma_{rad}=1200kg*9.3m/s^2=11200N\)**Example 15**

A 0.60kg rubber stopper is whirled in a horizontal circle of 0.80m radius at a

rate of 3.0 revolutions per second. What is the tension in the string?

Solution:

Three revolutions per second means three \(2\pi r\) circumferences per second,

so: \(v=\frac{3.2 \pi *0.80}{1.0s}=15m/s\)

The tension in the string, which is providing the centripetal force, is:

\(F=m\frac{v^2}{r} = 0.60kg*\frac{(15m/s)^2}{0.80m} = 170N\)**Example 16**

A 1 oz gold coin is placed on a turntable at 33 1/3 rpm. What is the coefficient

of friction between the coin and the turntable if the maximum radius, before the

coin slips, is 0.14m?

Solution:

The frictional force between the coin and turntable provides the center directed

force to keep the coin on the turntable. This center directed force must equal

\(mv^2/r\).

\(f=\mu N = \mu mg\)

This is a force balance problem. In equation form: frictional force=centripetal

force, or: \(\mu mg=mv^2/r\)

The velocity is 2\pi r\) times the number of \(2\pi r's\) per minute, 100/3 or:

\( v=2\pi*0.14m*(100/3)*(1/min)*(min/60s)=0.49m/s\)

Therefore: \( \mu = \frac{v^2}{rg} = \frac{(0.49m/s)^2}{0.14m*9.8m/s^2}=0.18\)**Example 17**

A conical pendulum is a mass on the end of a cord, where the mass moves at

constant speed in a circle with the cord tracing out a cone. A conical pendulum

of length 1.2m moves in a circle of radius 0.20m. What is the period of the

pendulum?

Solution:

Note that the mass is not given. Start with a vector diagram of the forces on

the mass.

The mg must equal the vertical component of the tension in the string \(T cos

\theta = mg\). The horizontal component is due to the centripetal force, so \(T

sin \theta = mv^2/r\). Dividing the second equation by the first eliminates T and

m. The angle is from \(sin \theta=r/L\), making \(tan \theta = 0.17\) and:

\(v=\sqrt{rg tan \theta} = \sqrt{.20m*9.8*.17}=.57m/s\)

The period is from \(2 \pi r/T=v\)

\(T=\frac{2\pi r}{v} = \frac{2\pi*0.20m}{0.57m/s}=2.2s\)**Example 18**

What is the speed, at which no side force due to friction is required, for a car

traveling at a radius of 240m on a banked 20 degree road?

Solution:

Start with mg acting down. The normal force(between the car and the road) is

normal to the surface, with the vertical component equal to mg and the

horizontal component equal to \(mv^2/r\).

We have: \(N cos 20 = mg\) and \(N sin 20 = mv^2/r\)

Then: \(\frac{N sin 20}{N cos 20} = tan 20 = \frac{v^2}{rg} \)

So: \(v = \sqrt{rg tan 20} = \sqrt{240m*9.8*tan 20} = 29m/s \)

Notice that in this problem the mass of the car does not enter into the

calculation.

Go back over problems 13-18 and notice how each problem could be written asking

for a different variable. For example, in problem 13, give the side force and

the velocity, and ask for the radius. In problem 16, give the speed and the

coefficient of friction, and ask for the radius. Changing the problems like this

is an excellent way to generate practice problems.**Example 19**

A 2kg ball is whirled in a vertical circle of radius 1m at a constant velocity

of 6.28m/s. Calculate the tension in the cord at the top of the motion, the

tension in the cord at the bottom of the motion, and the minimum velocity to

keep the mass from falling out at the top of the circle.

Solution:

We are considering this problem from the reference frame of the ball, not from

the point of view of the person whirling the ball. From the person's point of

view, the tension is pulling radially outward. From the ball's referencew frame,

the tension is radially inward. If the ball is to move in a circle, this must be

the case. There must be a net center directed force.

Operationally, a good method for solving centripetal force problems is to find

the net center directed force and set this equal to \(mv^2/r\).

\(T+mg=\frac{mv^2}{r}\)

\(T=\frac{mv^2}{r}-mg\)

\(T=\frac{2.0kg*6,28m/s^2}{1.0m} - 2.0kg(9.8m/s^2)=59.4N\)

\(T-mg=\frac{mv^2}{r}\)

\(T=\frac{mv^2}{r}+mg\)

\(T=\frac{2.0kg*6.28m/s^2}{1.0m} + 2.9kg*9.8m/s^2 = 98.6N\)

For part c, the minimum velocity required can be found by considering the

condition where mv^2 is less than or equal mg. In this case, at the top of the

motion, the mass will fall out of the loop. So let's set the two forces equal

and solve for the minimum velocity needed to keep the mass in the loop.

\(\frac{mv^2_{min}}{r} = mg \rightarrow v_{min} = \sqrt{rg} =

\sqrt{1.0m*9.8}=3.13m/s\)**Example 20**

A person on a dirt bike travels over a semicircular shaped bridge with a radius

of curvature of 45m. Calculate the max speed of the bike if its road wheels are

to stay in contact with the bridge.

Solution:

Again, let us find the net center directed force and set this equal to

\(mv^2/r\). So: \(mg-R=\frac{mv^2}{r}\) and: \(R=mg-\frac{mv^2}{r}\).

As v increases, R must decrease because mg is constant. In the limiting case,

when the wheels are just about to leave the ground, R=0, so

\(mg=\frac{mv^2}{r}\).

The mass m cancels out and is not required. So max speed v is given by

\(v^2=rg\).

Plugging in the numbers: \(v=\sqrt{rg} = \sqrt{441} = 21m/s\)

At this point, you might be thinking that from the point of view of the person

on the dirt bike, the \(mv^2/r\) force is radially outward. The rider is

momentarily weightless as the wheels leave the ground. This is correct- from the

reference frame of the person inside the circular motion. Astronauts in the

space station orbiting Earth are weightless. To them, there is no net force, and

the effect of the \(mv^2/r\) force is radially outward to counterbalance

gravity.

If you go around a corner in a car, you seem to feel a force outward from the

circle, not inward. But from the reference frame of a stationary observer, there

must be a net center directed force on both the astronaut(gravity) and the

person toward the center of the circle; otherwise, the person could not be

moving in a circle. Is one force correct and the other incorrect? Is either of

these an illusion? No, they are all real forces.

Take a thoughtful trip in an airplane flying in a horizontal circle with its

wings vertical. The force acting on you is toward the center of the circle, but

you exert an equal and opposite force on the seat that is radially out of the

circle. In some airplanes, if the banking is done too hard, this radially

outward force can be so great as to the blood in the pilot to remain in the

lower part of the body, resulting in a loss of oxygen to the brain and what is

called a blackout. So the blackout is the result of a radially outward force.

Is one reference frame correct and the other incorrect? When you study

Einstein's theory or relativity in physics, you will come across the idea that

there is no one absolute reference frame. You may think that a reference frame

at rest relative to the ground is correct, but the Earth spins on its axis and

rotates around the Sun. The Sun moves around the galactic center. In relativity,

we always have to specify the reference frame we are considering. And the

observed effects of many things change depending on the chosen reference frame.

So, when doing circular motion problems, always think about what reference

frame you are considering. For the method in the preceding problem, our

reference frame is outside the motion, and in this perspective, we take the net

center directed force and set it equal to \(mv^2/r\). This is the default

reference frame that is commonly used to solve circular motion problems.