Pulleys with friction , FORCES

1. Jul 30, 2017

deuce123

1. The problem statement, all variables and given/known data
A cord connected at one end to a block which can slide on an inclined plane has its other end wrapped around a cylinder resting in a depression at the top of the plane as shown in (Figure 1) .

Determine the speed of the block after it has traveled 1.60 malong the plane, starting from rest. Assume the coefficient of friction between all surfaces is μ= 0.0350. Since the block is much lighter than the cylinder, ignore tension in the string when calculating the normal force on the cylinder. Do not ignore tension in the string when calculating the net torque (including friction) on the cylinder.

2. Relevant equations

F=ma
torque=Iα
I=1/2mr^2

3. The attempt at a solution
There was initially two parts to this problem, one without the presence of friction and now this one with. The problem I encounter is how to deal with the friction in regards to the pulley, and the forces at work. I know the net torque is (T-Ff)r=torque where Ff is the friction force. But I fail to know how to calculate the normal force with regards to the pulley. I've tried to just use the normal force (just mxg) but it failed. I got the answer to try and use different methods too see where I went wrong and what I found out was that an angle is present which decreases the normal force of the pulley. But I fail too see where the angle is? The normal force too me seems perfectly perpendicular to the horizontal but this obviously can't be the case. Please someone explain too me how to go about this. The answer was v=.43 m/s.

2. Jul 30, 2017

haruspex

I'm not convinced there is a way.
We can consider two extremes:
• The normal force supporting the weight is concentrated at the lowest point of the arc. This is what you tried.
• The normal force is concentrated in two regions, one at each end of the arc. Since these will oppose each other, the sum of their magnitudes will exceed mg, and the frictional torque would be greater.
In between, we might try supposing that the normal force per unit length is a constant around the arc, but that would lead to a net leftward force on the cylinder, so some adjustment would need to be made.
Looks to me like a flawed question.

3. Jul 30, 2017

scottdave

Since you know the expected value of v, you can calculate what number they werewanting for Ff. From there you can see what Normal force they wanted and see if it was the combination of regions., I am guessing they may have been thinking that.

4. Jul 30, 2017

deuce123

Initially we weren't given the value, I just wanted the answer too see if I can figure out why it wasn't working. I did find Ff, and in turn N, but it doesn't make sense.

5. Jul 30, 2017

haruspex

I just tried it using the first of my options, all the normal force concentrated at the lowest point. I thought you had tried that and got a wrong answer. I get 0.412 m/s. Any other interpretation will give a lower number.

6. Jul 31, 2017

deuce123

I end up getting .577 m/s. How do you set up the force equation for the pulley? Is it not mg=n, for the normal force (vertical components) and for the torque I get torque=(T-Ff)R----- where R is the radius of the pulley

7. Jul 31, 2017

deuce123

Nevermind haha I just got it , I made an algebra mistake. Thank you.