Calculating Net Torque on Wheel Axle

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Homework Help Overview

The discussion revolves around calculating the net torque on a wheel axle, considering various forces and their directions. The problem involves understanding torque in the context of rotational motion and the effects of opposing friction torque.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of torque using different forces and their components, questioning the necessity of resolving forces into tangential components. There is also confusion regarding the angle used in calculations and its relevance to torque.

Discussion Status

Participants are actively engaging with each other's reasoning, exploring the concepts of torque and the effects of force components. Some guidance has been offered regarding the tangential nature of forces, but there is no explicit consensus on the correct approach to the problem.

Contextual Notes

There is mention of a specific diagram and the forces acting on the wheel, but the exact details of the diagram are not provided in the discussion. Participants express uncertainty about the angles and components involved in the torque calculations.

Jam51
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Homework Statement


Calculate the net torque about the axle of the wheel shown in https://jigsaw.vitalsource.com/books/9781269392464/content/id/ch08fig39 . 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

Homework Equations


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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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.
 
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.
 
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.
 

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