# High School Physics : Contact Forces

## Example Questions

### Example Question #1 : Contact Forces

Three blocks on a frictionless horizontal surface as in contact with each other.  A force is applied to the first block ().  Determine the net force on block B terms of , , and  and .

Explanation:

To determine the net force on a system, it is important to consider all the forces between each object in the horizontal direction.  The system is considered to be all the blocks together as they will all move with the same acceleration.

According to Newton’s 3rd Law  and  are equal in magnitude and opposite in direction so they will cancel out.  The same goes for  and .

Therefore our equation is reduce to

From Newton’s 2nd law we know that the net force on a system is equal to the mass of the system times the acceleration of the system

Therefore

This means that the acceleration on any one block is

So when solving for the net force on any one block, the net force on the single block will be equal to that block's mass times the acceleration of the system.  Therefore for block B, the net force is

### Example Question #2 : Contact Forces

A child slides down a slide with a 30 degree incline, and at the bottom her speed is precisely half what it would have been  if the slide had been frictionless.  Calculate the coefficient of kinetic friction between the slide and the child.

Explanation:

Consider the net forces acting on the object causing it to accelerate.

To determine the Force of Gravity in the x-direction, we must break the force of gravity into components and examine the side acting in the x-direction.  Using trigonometric functions we get that

We know that the force of gravity is equal to mg

According to Newton’s 2nd law the force is equal to the mass times the acceleration of the object.

The force of friction is directly related to   (the coefficient of friction)  times the normal force.  In this case the normal force is equal to the y component of the force of gravity.

Therefore

If we substitute this in our original net force equation

Notice that mass is in each piece of the equation so we can cancel it out.

We can also use our kinematic equations to determine the speed of the object at the bottom of the incline.  We can represent  to be the length of the slope.

Since the ball is assumed to be at rest at the top of the incline, the initial velocity at the top will be 0.

Therefore

Solving for  on its own we get

In the original situation, the force of gravity is the only force pulling on the object.  Therefore the acceleration is

We can substitute this value into our velocity equation

In the second situation the velocity is one half of the original velocity.  Therefore

We can place this back into our kinematic equation and rearrange it to solve for the acceleration.

We can then substitute that into our net force acceleration equation

Notice that g is in all of these terms so we can cancel it out.

Now we can substitute and solve

### Example Question #3 : Contact Forces

crate slides across a floor for  before coming to rest  from its original position.  What is the coefficient of kinetic friction on the crate? Assume

Explanation:

The equation for the force due to friction is , where μ is the coefficient of kinetic friction. Since there is only one force acting upon the object, the force due to friction, we can find its value using the equation . We can equate these two force equations, meaning that . We can solve for the normal force, but we need to find ma in order to find .

The problem gives us the mass of the crate, but we have to solve for the acceleration.

Start by finding the initial velocity. The problem gives us distance, final velocity, and change in time. We can use these values in the equation below to solve for the initial velocity.

Plug in our given values and solve.

We can use a linear motion equation to solve for the acceleration, using the velocity we just found. We now have the distance, time, and initial velocity.

Plug in the given values to solve for acceleration.

Now that we have the acceleration and the mass, we can return to our first equation for force.

The normal force is the same as the mass times gravity.

In this format, the masses cancel on both sides of the equation/

Now we can plug in our value for acceleration and gravity to solve for the coefficient of friction.

### Example Question #81 : Specific Forces

box is released on a 25 degree incline and accelerates down the ramp at .  What is the coefficient of kinetic friction impeding its motion?

Explanation:

Consider the net forces acting on the object causing it to accelerate.

To determine the Force of Gravity in the x-direction, we must break the force of gravity into components and examine the side acting in the x-direction.  Using trigonometric functions we get that

We know that the force of gravity is equal to mg

According to Newton’s 2nd law the force is equal to the mass times the acceleration of the object.

The force of friction is directly related to  μ (the coefficient of friction)  times the normal force.  In this case the normal force is equal to the y component of the force of gravity.

Therefore

If we substitute this in our original net force equation

Notice that mass is in each piece of the equation so we can cancel it out.

Now we can rearrange and solve for the coefficient of friction

### Example Question #1 : Contact Forces

Two toy cars (15kg and 3kg) are released simultaneously on an inclined plane that makes an angle of 28 degrees with the horizontal.  Which statement best describes their acceleration once they are released.  Assume that the track is frictionless.

None of the above

Both cars accelerate at the same rate

The 3kg car accelerates 5 times faster than the 15kg car

The 15kg car accelerates 5 times faster than the 3kg car

Both cars accelerate at the same rate

Explanation:

Since the only force pulling both cars down the slope is the force of gravity, both cars will accelerate at the same rate.

It is possible to prove this mathematically by examining the x-component of the gravitational pull.

We know that the force of gravity is equal to mg

According to Newton’s 2nd law the force is equal to the mass times the acceleration of the object.

Both sides have mass so this cancels out of the equation.

Since this is independent of mass, this will be true for both objects.

### Example Question #83 : Specific Forces

A horizontal force accelerates a box from rest across a horizontal surface where friction is present at a constant rate.  Then the experiment is repeated.  All conditions remain the same with the exception that the horizontal force is doubled.  What happens to the box’s acceleration.

It increases to less than double its original value

It increases to more than double its original value

It increases to exactly double its original value

It increases somewhat

It increases to more than double its original value

Explanation:

According to Newton’s 2nd Law the net force on the object is equal to the mass times the acceleration.  The net force is also equal to the sum of the forces involved.

In the second experiment the horizontal applied force is doubled.  However, this does not have any effect on the friction on the object.  Therefore the new equation is

From this equation we can see that the new acceleration will be more than double the original value.

To examine this numerically, let’s assume that the mass of the box is , the friction force is  and the applied force originally was .

Then, let’s double the horizontal force, keeping the friction force the same.

This value is more than double the original value, confirming our answer.

### Example Question #84 : Specific Forces

A block is on a smooth horizontal surface (), connected by a tine cord that passes over a pulley to a second block which hangs vertically ().  Find the formula for the acceleration of the system.  Ignore friction and the masses of the pulley and cord.

Cannot determine this with the information provided

Explanation:

To determine the net force on a system, it is important to consider all the forces between each object in the same direction.  The direction to be considered is along the line of the rope.  The system is considered to be all the blocks together as they will all move with the same acceleration.

According to Newton’s 3rd Law, the Tension in the rope caused by A pulling on B and B pulling on A is equal in magnitude and opposite in direction.  Therefore these two Tension forces cancel out.

From Newton’s 2nd law we know that the net force on a system is equal to the mass of the system times the acceleration of the system

Therefore

We can calculate the force of gravity from the equation Fg = mg

Therefore

We can rearrange this equation and solve for a

### Example Question #85 : Specific Forces

Susan is trying to push a  crate across the floor. She observes that the force of friction between the crate and the floor is . What is the coefficient of static friction?  Assume

It cannot be determined

Explanation:

The equation for the force of friction is , where μ is the coefficient of static friction.

The normal force is equal to the mass times acceleration due to gravity, but in the opposite direction (negative of the force of gravity).

Since the problem tells us that the force due to friction is , we can plug these values into our original equation to solve for the coefficient of friction.

The coefficient of friction has no units.

### Example Question #86 : Specific Forces

Which of the following cannot be true of an object on a given surface?

Static friction:  Kinetic friction:

Static friction:  Kinetic friction:

Static friction:  Kinetic friction:

Static friction:  Kinetic friction:

Static friction:  Kinetic friction: