# Multiple torques/forces on a disk, finding net torque

1. Apr 29, 2007

### aliaze1

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

The 20-cm-diameter disk in the figure can rotate on an axle through its center.

What is the net torque about the axle?

image: http://session.masteringphysics.com/problemAsset/1000963/5/knight_Figure_13_14.jpg

2. Relevant equations

τ=Frsinφ

3. The attempt at a solution

τnet= τ1+τ2+τ3+τ4

τ1 = 30N * 0.1 * sin(90)
τ2 = 30N * 0.05 * sin(-45)
τ3 = 20N * 0.05 * sin(-90)
τ4 = 20N * 0.1 * sin(45)
_____________________
τnet =~2.35355 Nm

this was my original solution, which came out incorrect, a friend then told me that τ4 is zero because the force of 20N is not being applied to a point on the disk itself, so then:

τ1 = 30N * 0.1 * sin(90)
τ2 = 30N * 0.05 * sin(-45)
τ3 = 20N * 0.05 * sin(-90)
_____________________
τnet =~0.94 Nm

the issue is that I have only one attempt left at this problem and I want to confirm this before I submit the problem
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Apr 30, 2007

### denverdoc

That looks kike it should be the right answer, but I think it is less confusing and error prone if you keep the signs out of the sin arguments, in other words if a force is tending to cause conterclockwise rotation, make the entire torgue negative,

eg, T2=-30N*.05m*sin(45)
T3=-20N*0.05*sin(90)

Else it looks fine.

3. May 1, 2007

### aliaze1

got it, yea the signs were messing me up, i got 0.94 when its actually -0.94,

thanks