### All AP Physics C: Mechanics Resources

## Example Questions

### Example Question #1 : Understanding Newton's Laws

A paint bucket hangs from the right end of a meter stick, oriented horizontally. The left end of the meter stick rests on a fulcrum so that it may rotate about that point. A rope is to be attached from the left end, so that the system does not rotate. What is the minimum force this rope must support, assuming the meter stick is massless?

**Possible Answers:**

**Correct answer:**

Relevant equations:

Determine the clockwise torque caused by the bucket.

Write an expression for the counterclockwise torque caused by the rope.

Combine the torque of the rope and the torque of the bucket into the net torque equation.

Since the system has no angular acceleration, net torque must be zero, allowing us to solve for the force of the rope.

### Example Question #1 : Newton's Laws And Force Diagrams

An object at rest will remain at rest unless acted on by a(n) __________.

**Possible Answers:**

internal force

external force

positive force

negative force

**Correct answer:**

external force

The correct answer is external force. External forces applied to an object will result in non-zero acceleration, causing the object to move. In contrast, internal forces contribute to the properties of the object and do not result in acceleration of the object.

Either positive or negative forces can result in the acceleration of an object. The sign on the force simply conveys information about its relative direction.

### Example Question #2 : Newton's Laws And Force Diagrams

A 1000kg rocket has an engine capable of producing a force of 30000N. By the third law of motion, when the rocket launches it experiences a reaction force that pushes it upwards of equal magnitude to the force produced by the engine. What is the acceleration of the rocket?

**Possible Answers:**

**Correct answer:**

When the rocket launches it produces a downward force of 30000N. Due to the third law of motion, the rocket experiences a 30000N reaction force that pushes it upwards.

In addition, the rocket experiences the downward force of its own weight. This is given by:

We know that .

We know the mass of the rocket is 1000kg, so we need only to find the net force to solve for acceleration.

We know that (since is directed upwards and is directed downwards).

Finally we solve for acceleration:

### Example Question #3 : Newton's Laws And Force Diagrams

Three boxes tied by two ropes move across a frictionless surface pulled by a force as shown in the diagram. Which of the following is an expression of the acceleration of the system?

**Possible Answers:**

None of these

**Correct answer:**

Since the boxes are all connected by ropes, we know that the acceleration of each box is exactly the same. They all move simultaneously, in tandem, with the same velocity and acceleration.

A quick analysis of each box will produce a very simple system of equations. Each rope experiences some tension, so we will label the tension experienced by the rope between and as . Similarly, we will label the tension experienced by the rope between and as .

For the box on the right (of mass ) we know that the force pulls to the right and the rope pulls it to the left with a tension of .

We get the equation: . is the mass of the box and is its acceleration (which is the same for the other boxes).

For the middle box (of mass ) we have a rope pulling on it on the right with tension and a rope pulling on the left with tension .

So we get the equation: . is the mass of the box and is its acceleration (which is the same for the other boxes).

Finally, for the box on the left (of mass ) we have only one rope pulling it to the right with a tension .

So we get the equation: . is the mass of the box and is its acceleration (which is the same for the other boxes).

Now it is just a matter of simple substitution. We have three equations:

From them we can get an expression for the force. Isolate in the second equation.

Substitute the expression of from the first equation and simplify.

Use this value of in the third equation to get an expression for the force.

So we have:

Solve for acceleration.

### Example Question #4 : Newton's Laws And Force Diagrams

A box is being pushed against a wall by a force as shown in the picture. If the coefficient of static friction between the box and the wall is , which of the following expressions represents what the force must be for the box not to fall?

**Possible Answers:**

None of these

**Correct answer:**

This question requires a 2-dimensional analysis. First identify all the forces acting on the box. Since the box is being pushed against a surface it automatically experiences a normal force . The box experiences the downward force of its own weight given by . Finally, since the box is trying to slide down, friction opposes this motion. The following diagram shows these forces:

The key is that the box should NOT move.

For the horizontal axis, net force must be zero. The two horizontal forces are the applied force and the normal force.

Now, on the vertical axis we have fiction and the object's weight:

Finally, we use the equation for friction force to solve the problem.

Substitute the weight, since it is equal to the force of friction.

Isolate the normal force.

Since the normal force is equal to the applied force, this is our final expression.

### Example Question #5 : Newton's Laws And Force Diagrams

A box of mass 10kg is pulled by a force as shown in the diagram. The surface is frictionless. How much force is the box experiencing along the horizontal axis?

**Possible Answers:**

**Correct answer:**

You need only to obtain the horizontal component of the force. To do this you must use trigonometric properties.

We see that the force makes an angle of 60º with the horizontal axis. This means that the horizontal component of the force is adjacent to this angle. We can view the diagram as a triangle. From trigonometry we know that to calculate the adjacent side of a triangle we need to multiply the hypotenuse by the cosine of the angle.

We can solve for the horizontal component of the force using the applied force as the hypotenuse:

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