# Net torque

## Homework Statement

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?

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Suraj M
Gold Member
Aren't all the forces in the diagram, tangential?
And how did you arrive at -35Sin135 as the component?

That's where I was confused. I assumed it wouldn't be tangential because it was on the inside of the wheel.

Suraj M
Gold Member
Inside of the wheel? I don't really think that matters, it still has to be tangential, even if it's not, again, how did you arrive at -35 sin 135?

I was thinking that the F of 35N would have to be resolved into components and the only one being effective in producing rotational motion would be F sin θ. I'm pretty sure I'm just mixing up the theory behind torque.

Suraj M
Gold Member
That ##\theta## in ## \sin (\theta) ## is the angle which the force makes with the radius, in this case I don't thing 135 is that angle, hence I feel we should take it as a tangential force.

I definitely agree with that. I just have to revisit some of the theory so I'm not confusing concepts.