Calculate the net torque about the axle of the wheel shown in https://jigsaw.vitalsource.com/books/9781269392464/content/id/ch08fig39 [Broken]. Assume that a friction torque of 0.40 m · N opposes the motion. Where F=18N (in photo).
The picture is here: http://www.webassign.net/giancoli/8-39alt.gif
τ (tau) = rF
The Attempt at a Solution
Direction of the positive torque: counterclockwise direction
T1 = 28N*24cm = 6.72 N-m
T2 = -35N*12cm = -4.2 N-m
T3 = -18N*24cm = -4.32 N-m
Now you add all these torques
6.72 N-m -4.2 N-m -4.32 N-m = -1.8 N-m
Since I know the direction of rotation is clockwise, I added the friction torque to that result in the opposite direction:
-1.8 N-m +0.40 N-m = -1.4 N-m (clockwise)
This is the only way I was able to get the right answer. However, my first try was using -35Nsin135. Why do I not have to take the component of F in this example?
Last edited by a moderator: