# Homework Help: Proving that net torque isn't reliant on point of rotation.

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1. Apr 4, 2016

### P-Fry

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Torque = r*Force*sin(theta)

3. 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.

2. Apr 4, 2016

### haruspex

You need to explain how all these distances and angles relate to the bar and the force. A diagram would be good.