Finding the Normal Force on the Rod of a Paper Roll

[student34]

Homework Statement

A large 16.0 kg roll of paper with radius R=18.0 cm rests against the wall and is held in place by a bracket attached to a rod through the center of the roll. The rod turns without friction in the bracket, and the moment of inertia of the paper and rod about the axis is 0.260 kg m^2. The other end of the bracket is attached by a frictionless hinge to the wall such that the bracket makes an angle of 30.0° with the wall. The weight of the bracket is negligible. The coefficient of kinetic friction between the paper and the wall is μ= 0.25. A constant vertical force F= 60.0 N is applied to the paper, and the paper unrolls.

(a) What is the magnitude of the force that the rod exerts on the paper as it unrolls?

(b) What is the magnitude of the angular acceleration of the roll?

Here is a picture of that someone else found for the same question, https://www.physicsforums.com/showthread.php?t=544377

Homework Equations

∑Fy = 0

∑Fx = 0

F(reaction) - Tension = 0, so F(rod) = T

F(friction) = μfx(normal)

The Attempt at a Solution

I will try part (a), but this attempt gives me the wrong answer.

∑Fy = T*cos30° - (mg + 60N) = 0

T = (mg + 60N)/cos30° = (16*9.8 + 60N)/cos°30 = 250.3391N

T = F(rod) = 250.3391N.

But the answer is 293N.

I probably need to use the friction, but I don't understand why.

 

[SteamKing]

If you will draw a proper free body diagram for the paper roll, it should be apparent what forces are used to solve this problem.

 

[student34]

If you will draw a proper free body diagram for the paper roll, it should be apparent what forces are used to solve this problem.

My work that I have shown above comes from free body diagrams that I made.

I have drawn 3 different free body diagrams and spent about 5 hours on this question in total. Any help would be greatly appreciated.

 

[SteamKing]

Can you post your FBDs?

 

[student34]

Can you post your FBDs?

Here is an attachment.

 

[SteamKing]

There is going to be a frictional force developed between the wall and the paper roll, which will retard the unrolling of the paper. This frictional force should be included in your FBD.

 

[student34]

There is going to be a frictional force developed between the wall and the paper roll, which will retard the unrolling of the paper. This frictional force should be included in your FBD.

I did have a strong feeling that the Ff would have an effect, but I still don't know why. I thought of a case where μ = 1 so that the roll wouldn't even spin versus the case where μ = 0. Both cases seem to have the same effect on the center of mass where the rod is.

It's wrong, but I will show what I did with friction.

μ*T*sin30° = 0.25*250.339N*sin30° = 31.292N. I will assume that this adds to the ∑Fy. So now,

∑Fy = T*cos30° + (-31.292N) - (mg + 60N) = 0

T = 286.47N, which is close but not quite enough.

 

[student34]

Can anyone see anything wrong with my last post?

 

[haruspex]

μ*T*sin30° = 0.25*250.339N*sin30° = 31.292N. I will assume that this adds to the ∑Fy. So now,

You can't plug in a value for T here. T is what you're trying to calculate. Work it all the way through symbolically. Don't plug in any numbers until you have an equation in the form T = ...

 

[student34]

You can't plug in a value for T here. T is what you're trying to calculate. Work it all the way through symbolically. Don't plug in any numbers until you have an equation in the form T = ...

Oh thanks, I got it!