### All MCAT Physical Resources

## Example Questions

### Example Question #293 : Mcat Physical Sciences

A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what distance from the left end of the rod should a 0.6kg mass be hung to balance the rod?

**Possible Answers:**

50cm

48cm

45cm

42cm

The rod cannot be balanced with this mass

**Correct answer:**

45cm

The counterclockwise and clockwise torques about the pivot point must be equal for the rod to balance. Taking the fulcrum as the pivot point, the counterclockwise torque is due to the rod’s weight, gravitational force acting downwards at the center of the rod. If we use the pivot as our reference, then the center of the rod is 15cm from the reference.

Set this equal to the clockwise torque due to the additional mass, a distance r to the right of the pivot.

.

Solving for r gives r = 0.05m to the right of the pivot, so 40 + 5 cm from the left end of the rod.

### Example Question #1 : Torque

A 2kg mass is suspended on a rope that wraps around a frictionless pulley attached to the ceiling with a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.

What is the torque on the pulley when the system is motionless?

**Possible Answers:**

9.8N*m

0N*m

10N*m

19.6N*m

**Correct answer:**

0N*m

The net torque on the pulley is zero. Remember that , assuming the force acts perpendicular to the radius. Because the pulley is symmetrical in this problem (meaning the r is the same) and the tension throughout the entire rope is the same (meaning F is the same), we know that the counterclockwise torque cancels out the clockwise torque, thus, the net torque is zero.

In the image below, T_{1} (due to the platform with the 4 0.5kg weights) = T_{2} (the 2kg mass).

### Example Question #2 : Torque

Two students are balancing on a 10m seesaw. The seesaw is designed so that each side of the seesaw is 5m long. The student on the left weighs 60kg and is standing three meters away from the center. The student on the right weighs 45kg. The seesaw is parallel to the ground. Assume the board that makes the seesaw is massless.

What distance from the center should the student on the right be if they want the seesaw to stay parallel to the ground?

**Possible Answers:**

**Correct answer:**

Torque is defined by the equation . Since both students will exert a downward force perpendicular to the length of the seesaw, . In our case, force is the force of gravity, given below, and is the distance from the center of the seesaw.

Since the torque must be zero in order for the seesaw to stay parallel (not move), the lighter student on the right must make his torque on the right equal to the torque of the student on the left. We can determine the required distance by setting their torques equal to each other.

### Example Question #1 : Torque

Two students are balancing on a 10m seesaw. The seesaw is designed so that each side of the seesaw is 5m long. The student on the left weighs 60kg and is standing three meters away from the center. The student on the right weighs 45kg. The seesaw is parallel to the ground. Assume the board that makes the seesaw is massless.

Imagine that the two students are sitting on the seesaw so that the torque is . Which of the following changes will alter the torque of the seesaw?

**Possible Answers:**

Two more students get on the seesaw, each weighing 45kg. They both sit on opposite ends of the seesaw, five meters away from the center.

Both students move toward the center by one meter.

Another student stands perfectly on the center of the seesaw.

The heavier student moves forward 1m, while the lighter student moves forward 1.33m.

**Correct answer:**

Both students move toward the center by one meter.

Torque, in this case, is dependent on both the force exerted by the students as well as their distances from the point of rotation. As a result, both students moving forward by one meter will cause a nonzero torque on the seesaw. This is because the heavier student's ratio of force and distance will result in less torque on his side than the lighter student.

### Example Question #91 : Circular And Rotational Motion

A 3m beam of negligible weight is balancing in equilibrium with a fulcrum placed 1m from it's left end. If a force of 50N is applied on it's right end, how much force would needs to be applied to the left end?

**Possible Answers:**

**Correct answer:**

This a an example of rotational equilibrium involving torque. The formula for torque is , where is the angle that the force vector makes with the object in equilibrium and is the distance from the fulcrum to the point of the force vector. To achieve equilibrium, our torques must be equal.

Since the forces are applied perpendicular to the beam, becomes 1. The distance of the fulcrum from the left end is 1m and its distance from the right end is 2m.

Since the 50N force is twice as far from the fulcrum as the force that must be applied on the left side, it must be half as strong as the force on the left. The force on the left can be found to be 100N.

### Example Question #3 : Torque

One side of a seesaw carries a mass four meters from the fulcrum and a mass two meters from the fulcrum. To balance the seesaw, what mass should be placed nine meters from the fulcrum on the side opposite the first two masses?

**Possible Answers:**

**Correct answer:**

For the seesaw to be balanced, the system must be in rotational equilibrium. For this to occur, the torque the same on both sides.

The total torque must be equal on both sides in order for the net torque to be zero.

Substitute the formula for torque into this equation.

Now we can use the given values to solve for the missing mass.

The acceleration form gravity cancels from each term.

### Example Question #4 : Torque

Two masses hang below a massless meter stick. Mass 1 is located at the 10cm mark with a weight of 15kg, while mass 2 is located at the 60cm mark with a weight of 30kg. At what point in between the two masses must the string be attached in order to balance the system?

**Possible Answers:**

**Correct answer:**

This problem deals with torque and equilibrium. Noting that the string is between the two masses we can use the torque equation of . We can use the equation to find the torque. Since force is perpendicular to the distance we can use the equation (sine of 90^{o} is 1). Force presented in this situation is gravity, therefore F=mg, and using the variable x as a placement for the string we can find r.

x=43, thus the string is placed at the 43cm mark.

### Example Question #1 : Torque

A 2kg mass is suspended on a rope that wraps around a frictionless pulley. The pulley is attached to the ceiling and has a mass of 0.01kg and a radius of 0.25m. The other end of the rope is attached to a massless suspended platform, upon which 0.5kg weights may be placed. While the system is initially at equilibrium, the rope is later cut above the weight, and the platform subsequently raised by pulling on the rope.

The rope is cut just above the 2kg mass, and the platform starts to fall. The rope is still wrapped around the pulley. What is the torque around the pulley?

**Possible Answers:**

**Correct answer:**

In the instant the rope is cut right above the 2kg mass, the pulley now has net torque in the counterclockwise direction because the platform with the masses is still connected to the rope that winds around the pulley. Knowing that , we can determine the torque by knowing that the force (F) on the pulley is due to the tension (T) by the platform. The tension is 19.6N (equal to the weight) and knowing that the radius of the pulley is 0.25m, we can solve for torque.

### Example Question #8 : Torque

Which of these is the correct expression for torque, τ, when θ is the angle at which the force, F, acts around the axis of rotation?

- τ = F cos θ
- τ = F sin θ
- τ = F tan θ
- τ = F cotan θ
- τ = F sec θ

**Possible Answers:**

3

5

2

1

4

**Correct answer:**

2

Choice 2 is correct. Think about trying to change a tire…you apply the force on the tire iron as close to perpendicular as possible in order to generate the most force.

Since the sine of 90 degrees is 1, then you are applying the maximum torque you can generate according to the equation τ = F sin θ.

### Example Question #2 : Torque

An attraction at a science museum helps teach students about the power of torque. There is a long metal beam that has one pivot point. At one end of the bar hangs a full sedan, and on the other end is a rope at which students can pull down, raising the car off the ground.

The beam is 40 meters long and the pivot point is 5 meters from one end. A car of mass 500kg hangs from the short end of the beam. Neglecting the mass of the beam, what is the minimum mass of a student who can hang from the rope and begin to raise the car off the ground?

**Possible Answers:**

More information is needed to answer

**Correct answer:**

We are trying to find what force needs to be applied to the rope to result in a net of zero torque on the beam.

Torque applied by the car:

We can use this to find the mass of a student that will create the same amount of torque while hanging from the rope:

### All MCAT Physical Resources

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