Calculate the net torque about the axle of a wheel

[ssjdbz101]

Homework Statement

Could someone explain this problem a little better to me.

Problem:
Calculate the net torque about the axle of the wheel shown in Fig. 8-39. Assume that a friction torque of 0.45 m·N opposes the motion and that F = 18.

*see attachment for picture.


I'm confused as how a wheel can have to forces (28N and 18N) on it in different directions on it at the same time.
Also, the 35N force looks like it is tangential to the smaller wheel, so do I have to use the 135 degrees in finding it's torque?

Torque equation: T=rFsin()
r=lever arm
F=force

Thanks in advance.

 

[Feldoh]

Torque is actually a cross product of r x F. To find the magnitude we've got:

$$\tau = rFsin(\theta)$$ Where theta is the angle between the moment arm (r) and the force.

So will the 135 degrees be an issue on the 35N force? It looks like the 35N force is tangential from the smaller radius.

 

[ogb p]

I can provide some clarification on this problem for you. First, let's define some terms. Torque is the measure of a force's ability to rotate an object around an axis or pivot point. In this case, the axle of the wheel is the axis or pivot point. The net torque is the sum of all the individual torques acting on the object.

In this problem, there are three forces acting on the wheel: 28N, 18N, and 35N. The 28N and 18N forces are acting in opposite directions, so they will create a net torque that is the difference between the two forces. The 35N force is acting tangentially to the smaller wheel, so it will create a torque that is perpendicular to the radius of the smaller wheel.

To calculate the net torque about the axle, you will need to use the torque equation you provided: T = rFsin(). The lever arm, or distance from the axle to where the force is applied, will be different for each force. For the 28N and 18N forces, the lever arm will be the radius of the larger wheel. For the 35N force, the lever arm will be the radius of the smaller wheel.

To find the angle in the sin() function, you will need to use the 135 degrees given in the problem. This angle represents the angle between the force and the radius of the smaller wheel.

Once you have calculated the individual torques for each force, you can add them together to find the net torque about the axle. Remember to take into account the friction torque of 0.45 m·N, which will act in the opposite direction.

I hope this helps clarify the problem for you. If you have any further questions, please don't hesitate to ask. Good luck with your calculations!