Calculate the net torque about the axle of the wheel

In summary, the problem is asking to calculate the net torque about the axle of a wheel with three forces acting on it, one of which is a friction torque opposing the motion. After calculating the individual torques for each force, the net torque is found to be -4.15, but this is incorrect due to a mistake in the direction of the friction torque. The correct net torque is -3.29, considering the friction torque as a positive value.
  • #1
confusedbyphysics
62
0
This problem is due on my webassign homework in 3 hours and I can't get it right!

Calculate the net torque about the axle of the wheel shown in Fig. 8-39. Assume that a friction torque of 0.43 mN opposes the motion and that F = 26.

The picture is here: http://www.webassign.net/giancoli/8-39alt.gif

What I did:

The 35 N force will move the wheel clockwise so the torque will be -

28 N force will move it counterclockwise so torque will be +

26 N force will move it clockwise so force will be -

They forces are all perpendicular to the radius so its just T = r X F

35 N force T = 35 N X .12 m = -4.2 (moves cw)

28 N force T = 28 X .24 m = 6.72 (moves ccw)

26 N force T = 26 X .24 m = -6.24 (moves cw)

So then including the friction torque which opposes the motion of .43 I put

6.72 - 4.2 - 6.24 - .43 = -4.15 is the net torque and because its - it movies in the clockwise direction

Web assign says -4.15 is WRONG! What am I doing wrong, please help!
 
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  • #2
You haven't said yet which direction (cw , ccw) the disk is rotating...
if it was already rotating ccw, the friction torque would be cw.
have you tried that?
 
  • #3
It says 0.43 mN, which isn't the same as 0.43N
 
  • #4
Ohhhh...so if the motion is going clockwise than the friction will be counter clockwise so it would be +.43 instead of -.43?

6.72 - 4.2 - 6.24 + .43 = -3.29

is that right?? ( I only have one guess left on my webassign and I don't want to screw it up, lol)
 
  • #5
Well, I don't know the whole scenario of the problem, but ...
the 3 forces in the diagram MIGHT be slowing the spin of the disk.
There should be some indication of the spin, in the problem text.

I expect your .43 mN means meter x Newton (not milliNewton)
since it is a torque - unless it is 43 milliN applied at the .12m hub!
 
  • #6
Seems right.
 
  • #7
Ok I just guessed and -3.29 is right. That was stupid of me not to catch that the friction is positive. Thanks for the help!
 

FAQ: Calculate the net torque about the axle of the wheel

What is the definition of torque?

Torque is a measure of the rotational force applied to an object. It is calculated by multiplying the magnitude of the force by the distance from the axis of rotation to the point where the force is applied.

How is net torque calculated?

Net torque is calculated by adding together all the individual torques acting on an object. This can be done by using the formula: net torque = force * distance, where the force and distance are perpendicular to each other.

What is the role of the axle in calculating net torque?

The axle is the axis of rotation for the wheel. It is the point where the net torque is calculated and is crucial in determining the direction and magnitude of the rotation.

How does the direction of the force affect the net torque?

The direction of the force is crucial in determining the direction of the net torque. If the force is applied in the same direction as the rotation, it will increase the net torque. If the force is applied in the opposite direction, it will decrease the net torque.

Does the weight of the wheel affect the net torque calculation?

Yes, the weight of the wheel can affect the net torque calculation. A heavier wheel will have a greater moment of inertia and therefore require more torque to rotate. This can impact the overall net torque on the axle.

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