Multiple torques/forces on a disk, finding net torque

  • Thread starter aliaze1
  • Start date
  • #1
174
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Homework Statement



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

Homework Equations



τ=Frsinφ

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

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
960
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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
174
0
got it, yea the signs were messing me up, i got 0.94 when its actually -0.94,

thanks
 

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