Torque: Understanding the Complexity Behind the Physics

In summary: In both cases there is a weight in the center, a force is being applied at one end, and the center of gravity is at the other.In summary, a torque is a force that is greater when applied closer to the axis of rotation. This is fallacious because it requires a larger force to apply a torque at a closer distance. However, a torque can be created by applying a force at the end of a long arm, and the magnitude of the torque is based on the distance the arm is from the axis of rotation.
  • #1
Felix83
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I have a few years of physics experience, and as I was thinking about a few things the other day, something about the concept of torque seemed a little weird to me.

Simple Forces are easy to understand - gravity pulls down on an object, the ground pushes back up with an equal and opposite force, etc, etc. Now torque - If you think about it from the perspective of work it makes sense - If you apply a force at the end of a long arm to rotate it, the force is greater closer to the axis of rotation because it moves a shorter distance, but performs the same amount of work.

However, consider a rigid bar in static equilibrium, with a fixed axis of rotation at one end. There is a weight in the middle, say 100 kg. You push up farther away from the axis so you only have to push with a 50kg force to keep it from falling. At that instant, there is no motion, so you cannot say your hand travels a greater distance than the weight so the force needed is less.

With no motion, how does the system "know" (for lack of a better word) to take your 50kg force at the end, and apply 100kg of upward force where the weight is?
 
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  • #2
Felix83 said:
Simple Forces are easy to understand - gravity pulls down on an object, the ground pushes back up with an equal and opposite force, etc, etc. Now torque - If you think about it from the perspective of work it makes sense - If you apply a force at the end of a long arm to rotate it, the force is greater closer to the axis of rotation because it moves a shorter distance, but performs the same amount of work.

This sounds fallacious but I think I know what you mean. To apply a torque of equal magnitude at a closer distance requires a larger force.

Felix83 said:
However, consider a rigid bar in static equilibrium, with a fixed axis of rotation at one end. There is a weight in the middle, say 100 kg. You push up farther away from the axis so you only have to push with a 50kg force to keep it from falling. At that instant, there is no motion, so you cannot say your hand travels a greater distance than the weight so the force needed is less.
I'm assuming the bar is horizontal. Firstly, you are mistaking weights, masses and forces. A weight is a force due to gravity, and a mass is just a mass for the purposes of this argument. Having a 100kg weight at a distance 'r' from the AoR will yield a torque of 100kg*9.8m/s^2*r. To counter this torque, you would need to apply a torque of equal magnitude in the opposite direction. Let's say your arm can apply a force of 100N

[tex] \tau_{net} = 0 = \tau_{cw} + \tau_{ccw} \ and \ so \ \tau_{cw} = -\tau_{ccw}[/tex]

Say the weight provides the clockwise torque, then

[tex] \tau_{cw} = mgr = 980r_1 [/tex]

[tex] \tau_{ccw} \mbox{ will need to be of the same magnitude,} 100r_2 [/tex]

[tex] 980r_1 = 100r_2 \ and so \ r_2 = 9.8r_1 [/tex]. Only with radiuses described by this relationship will the system be in equilibrium.
felix83 said:
With no motion, how does the system "know" (for lack of a better word) to take your 50kg force at the end, and apply 100kg of upward force where the weight is?

The system doesn't 'know' per sai, it will simply follow through any actions that forces acting on it dictate. If your arm wasnt pushing, the entire system would begin rotating. You deliberately position your arm at a certain radius so that the system is not rotating.

I hope this is along the liens of what you were looking for.
 
  • #3
Think about it a different way, using center of gravity. Take a rigid beam and support it on each end. Put a weight at the center - it is distributed evenly between the two supports. Put it 2/3 of the way to one side and one support takes double the force of the other. How does the support "know" how much force with which to push up?

For some reason, people have no problem with that scenario, but they do with a lever. But from a standpoint of statics (the way you analyze the system), this situation and the lever situation you describe are exactly the same (just flipped over).
 

FAQ: Torque: Understanding the Complexity Behind the Physics

1. What is torque?

Torque is a measure of the force that causes an object to rotate around an axis. It is a vector quantity, meaning it has both magnitude and direction.

2. How is torque calculated?

The formula for torque is torque = force x lever arm. The force is the amount of push or pull applied to an object, and the lever arm is the perpendicular distance from the axis of rotation to the line of action of the force.

3. What factors affect torque?

The magnitude of torque is affected by the magnitude and direction of the force, as well as the distance between the force and the axis of rotation. Additionally, the angle at which the force is applied can also affect the torque.

4. What are some real-world applications of torque?

Torque plays a crucial role in many everyday devices, such as door handles, wrenches, and gears. It is also important in larger systems, such as car engines, where torque is necessary to rotate the crankshaft and power the wheels.

5. Why is understanding torque important in physics?

Torque is a fundamental concept in physics and is essential in understanding rotational motion and the behavior of objects in motion. It is also crucial in engineering and design, as it helps in creating efficient and stable systems.

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