Proving that net torque isn't reliant on point of rotation.

In summary, the conversation discusses a horizontal bar with a length of 1m and various forces acting on it at different distances from the point of rotation. The equations for calculating torque are also mentioned. The calculations are shown for two different points of rotation, with the conclusion that the point of rotation does not affect the net torque. A diagram is suggested to better understand the relationship between the distances and angles.
  • #1
P-Fry
1
0

Homework Statement


So we have a horizontal bar. Distance = r Forces = F
All numbers remain constant with the exception of the distance, denoted as r(set)()

Length of bar = 1m

F1 = 10N r1-1 = 0m r2-1 = .25m (behind the point of rotation)
F2 = 5N r1-2 = 0.5m r2-2 = .25m (ahead of the point of rotation)
F3 = 20N r1-3 = 1.0m r2-3 = .75m (ahead)
F4 = 40N r1-4 = 0.75m r2-4 = .50m (ahead)

Theta 1 = 30 degrees clockwise from horizontal (South of east)
Theta 2 = 90 degrees, perpendicular, counterclockwise from horizontal (North)
Theta 3 = 60 degrees counterclockwise from horizontal (North of east)
Theta 4 = 90 degrees perpendicular, clockwise from the horizontal (south)

Essentially, with the first problem, the point of rotation is at the very edge, on the left. The second, it's shifted to the right .25m.

Homework Equations


Torque = r*Force*sin(theta)

The Attempt at a Solution


T1-1 = 0*(10) * sin(30) = 0
T1-2 = .5 *(5) * sin(90) = 2.5 (ccw)
T1-3 = 1*(20) * sin(60) = 17.32 (ccw)
T1-4 = .75 * (40) * sin (90) = 30 (cw)

T1net = 0 - 2.5 - 17.32 + 30 = 10.17

T2-1 = .25 * (10) * sin(30) = 1.25 (ccw)
T2-2 = .25 *(5) * sin(90) = 1.25 (ccw)
T2-3 = .75 * (20) * sin(60) = 12.99 (ccw)
T2-4 = .50 * (40) * sin (90) = 20 (cw)

T2net = -1.25 - 1.25 - 12.99 + 20 = 4.5

I can see logically why point of rotation doesn't have an effect on Net Torque, but I think I'm doing something wrong with the calculations.
 
Physics news on Phys.org
  • #2
You need to explain how all these distances and angles relate to the bar and the force. A diagram would be good.
 

1. What is net torque?

Net torque is the measure of the rotational force applied to an object. It is a vector quantity that takes into account both the magnitude and direction of the force, and is typically measured in units of Newton-meters (Nm).

2. How is net torque calculated?

Net torque is calculated by multiplying the force applied to an object by the distance from the point of rotation (also known as the lever arm). The direction of the torque is determined by the direction of the force and the direction of rotation.

3. Why is the point of rotation important when calculating net torque?

The point of rotation is important because it determines the lever arm, which is a crucial component in calculating net torque. The lever arm is the perpendicular distance from the point of rotation to the line of action of the force, and it affects the magnitude of the torque.

4. How can you prove that net torque isn't reliant on the point of rotation?

One way to prove that net torque isn't reliant on the point of rotation is by using the principle of moments. This principle states that the sum of the clockwise moments is equal to the sum of the counterclockwise moments, regardless of the point of rotation. This proves that the net torque remains the same, regardless of the point of rotation.

5. Are there any real-life examples that demonstrate that net torque isn't reliant on the point of rotation?

Yes, a simple example is a seesaw or a teeter-totter. The net torque on the seesaw remains the same, regardless of the point of rotation (the point at which the seesaw is balanced). This is because the forces acting on each end of the seesaw have equal magnitudes and opposite directions, resulting in a net torque of zero.

Similar threads

  • Introductory Physics Homework Help
Replies
33
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
22
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
4K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
2
Replies
44
Views
5K
  • Introductory Physics Homework Help
Replies
2
Views
2K
Back
Top