What Is the Coefficient of Static Friction Between the Ring and the Shaft?

[bigu01]

Homework Statement

The 5-kg cylinder is suspended from two equal length cords. The end of each cord is attached to a ring of negligible mass that passes along a horizontal shaft. If the rings can be separated by the greatest distance d= 400 mm and still support the cylinder, determine the the coefficient of static friction between each ring and the shaft.

Homework Equations

F=μ*N; ∑Fx= 0; ∑Fy= 0

The Attempt at a Solution

Firstly, I am finding the tension in the cords. By taking the FBD of the point that it is connecting all the three cords I am getting T= 73.57 N and the angle 19.47 (angle between the shaft and the cords). Then drawing the FBD of the shaft, the friction forces are turning to be in opposite directions, thus canceling each other.I know something is wrong, as the book result of the solution suggests it too.Can you please give me hints where to look for my mistake?

 

[tiny-tim]

hi bigu01!

Firstly, I am finding the tension in the cords. By taking the FBD of the point that it is connecting all the three cords I am getting T= 73.57 N and the angle 19.47 (angle between the shaft and the cords). Then drawing the FBD of the shaft, the friction forces are turning to be in opposite directions, thus canceling each other.I know something is wrong, as the book result of the solution suggests it too.Can you please give me hints where to look for my mistake?

can you show us your calculations?

(particularly, how do you get 73.57 ?)

 

[nvn]

bigu01: 19.47 deg is not the angle between a cord and the shaft. Try again. The tensile force in each cord, T = 73.57 N, currently looks incorrect. Try again. Also, remember to leave one space character between sentences.

 

[bigu01]

hi bigu01!


can you show us your calculations?

(particularly, how do you get 73.57 ?)

I uploaded two pictures of my work, I hope it is clear.

 

[nvn]

bigu01: 19.47 deg is incorrect. Try again.

 

[bigu01]

bigu01: 19.47 deg is not the angle between a cord and the shaft. Try again. The tensile force in each cord, T = 73.57 N, currently looks incorrect. Try again. Also, remember to leave one space character between sentences.

Okay, I think I got the correct result now, my angle is 70.5 deg and the tensile force in each cord is, T = 26 N. I think this is correct. But, still in the FBD of shaft my forces are cancelling each other. About leaving a space character between sentences I am trying to do it every time, but I have not got used to it. So until, I make it as a habit I will be forgetting from time to time. However, thank you for reminding it to me.

 

[nvn]

bigu01: Good, you got theta correct ... except, from item 1, below, it should be 70.53 deg. And T is correct, except it should be T = 26.01 N.

Now compute the horizontal component of T; i.e., force Fs. Also compute the normal (vertical) component of T.

What is the relationship between friction force (Fs) and normal force? Hint 1: Use mu, static coefficient of friction (COF), to write the relationship (equation), then solve for mu.

(1) Generally always maintain at least four significant digits throughout all your intermediate calculations, then round only the final answer to three significant digits.

 

[bigu01]

Okay, I got it. Fs = mu * N, and also Fs T * cos theta , I found N and then put it into the equations got the correct result for mu. So, in this question we did not need to write FBD for the shaft, because I could not solve it in that way, using components of the force was much easier. Thank you very much. About that significant digits throughout calculations, I do that only in exam time. Which is very bad, but I am trying to improve myself every time, as in the case of writing to physicsforums.com :). By the way, mu = 0.354 (rounded-up :) )

 

[nvn]

Excellent work, bigu01. Your answer is correct.