Rotational Motion: Understanding Net Force and Torque

[GreenPrint]

Homework Statement

See attachment

Homework Equations

The Attempt at a Solution

I'm confused as to why it's

Ʃ F = m a = 20 + f

as opposed to

Ʃ F = m a = 20 - f

Also when it states the net torque

Ʃ torque = I alpha = ( 20 - f ) ( 0.10 ) = 0.02 ( a / 0.10 )

Why is it 20 - f (as opposed to the original 20 + f ) ?

Thanks

 

[vela]

Homework Statement

See attachment

Homework Equations

The Attempt at a Solution

I'm confused as to why it's

Ʃ F = m a = 20 + f

as opposed to

Ʃ F = m a = 20 - f

They're assuming friction points to the left, which probably seems counter-intuitive to you. The friction will point in whatever direction is needed to keep the cylinder from sliding across the floor. If the torque due to the 20-N force will cause the cylinder to spin too quickly, the frictional force will be directed such that its torque will oppose the torque of the applied force, which happens to be the case this time.

You can assume the force points to the right, so f gets replaced by -f in the equations, and you'll find that f=-6.7 N. The negative sign simply means you guessed wrong about the direction of f.

Also when it states the net torque

Ʃ torque = I alpha = ( 20 - f ) ( 0.10 ) = 0.02 ( a / 0.10 )

Why is it 20 - f (as opposed to the original 20 + f ) ?

Thanks

What does the right hand rule tell you about the direction of the torque for each force? Assume f points to the left as they did in the solution.

 

[GreenPrint]

Interesting. I'll keep that in mind for friction.

So if both the forces point in the same direction (towards the left) then by the right hand rule wouldn't the torques be in the same direction?

 

[vela]

Remember torques are calculated relative to some axis. In this problem, the axis goes through the center of the cylinder.

 

[GreenPrint]

So by the right hand rule

If I hold my right hand out and point my thumb into the axis of rotation (into the paper) and curl my fingers towards the force by the string I get that the torque goes inwards (into the paper) and I also get the same for friction sense I curl my fingers towards the left and my thumb still points into the paper

 

[vela]

So by the right hand rule

If I hold my right hand out and point my thumb into the axis of rotation (into the paper) and curl my fingers towards the force by the string I get that the torque goes inwards (into the paper)

No, the lever arm r goes from the center of the cylinder to the top. That's the way your fingers point. Then you curl the fingers to point in the direction of the force, and your thumb will point in the direction of the torque, which will be out of the page.

and I also get the same for friction sense I curl my fingers towards the left and my thumb still points into the paper

This time the lever arm points down, but the force points in the same direction, so the torque is in the opposite direction.

A simpler way to look at it is to look at which way a force will cause the cylinder to spin. The top force will make it spin counterclockwise, so by the sign convention chosen for this problem, it would have a positive torque. The friction force, on the other hand, would cause the cylinder to spin in the clockwise direction, so its torque is negative.

 

[GreenPrint]

Thanks that makes much more sense.