Sixth Grade Assignment: Torque Activity

TORQUE

What is Torque?

Torque is more than just a force which is responsible for an object to rotate, it is more of a twist.

Let us take the simplest example to understand this concept. When you push a door applying force to the edge of the door, it is simpler to push. However, if you try pushing the same door somewhere in between the edge and the axis or closer to the pivot, it becomes that much more difficult. It is because of torque. Torque depends not just on the force but also on the distance of the application of force from the pivot point. This distance is called as the moment arm and is denoted by 'r'.

With this information in hand, let us understand the torque equation.

Torque Equation

Torque is the cross product or the vector product of the distance vector 'r' and the force vector 'F'. Torque is denoted by the Greek letter 'τ' called tau. Therefore,

τ = r × F

The direction of the torque vector can be determined using the right hand rule. When we put our fingers in the direction of r, and curl them to the direction of F, then the thumb points in the direction of the torque vector. From this torque equation, we can identify that if you apply a force at the pivot of the door or in the direction of the distance vector, the torque produced will be zero and hence the door will be unmoved. The SI unit of torque is Newton-meter, which is the same as for energy (though it is called joule).

EXAMPLE PROBLEM ON TORQUE: The Swinging Door

Question
In a hurry to catch a cab, you rush through a frictionless swinging door and onto the sidewalk. The force you extered on the door was 50N, applied perpendicular to the plane of the door. The door is 1.0m wide. Assuming that you pushed the door at its edge, what was the torque on the swinging door (taking the hinge as the pivot point)?

Hints
 
 

1. Where is the pivot point?
2. What was the force applied?
3. How far from the pivot point was the force applied?
4. What was the angle between the door and the direction of force?

Solution

The pivot point is at the hinges of the door, opposite to where you were pushing the door. The force you used was 50N, at a distance 1.0m from the pivot point. You hit the door perpendicular to its plane, so the angle between the door and the direction of force was 90 degrees. Since  = r x F

Figure 1 Diagram of Example Problem 1

then the torque on the door was:

= (1.0m) (50N)
= 50 N m = 50 Joules = 50J

Note that this is only the magnitude of the torque; to complete the answer, we need to find the direction of torque. Using the right hand rule, we see that the direction of torque is out of the screen.

Practice Problems

1.      If you apply 10 Newtons of Force to open a gate that is 5 meters wide, what is the torque that you apply on that gate?

2.      If you apply 20 Newtons of Force to open that same gate, what is the torque that you apply on that gate?

3.      If you apply 10 Newtons of Force but try to open that door only 1 meter away from the pivot point, what is the torque you apply on that gate?

4.      If you apply 40 Newtons to a nut using a wrench, and the wrench is 0.2 meters long, what would be the torque you apply to that nut?

Extra Credit Problem

If you know that the torque you apply to open and big industrial nut is 10,000 Joules and the force that you applied was 500 Newtons, what is the length of the wrench you used?