# Dimension of the unidirectional mechanical clutch

This is a problem from the book Engineering Mechanics Statics vol-1 by Meriam Kraige.

The elements of a unidirectional mechanical clutch are shown in the figure.A torque $M$ applied to the outer ring is transmitted to the output shaft through frictional interaction between the outer ring and the balls and between the balls and the inner driven member.If the direction of rotation of the outer ring is reversed,the wedging action of the balls is absent,and no torque can be transmitted to the output shaft.For given values of $r,r_o$ and coefficient of friction $\mu$ which is applicable for both pairs of interacting surfaces,the question is to find the minimum dimension $b$ of inner member which will permit the transmission of torque without slipping.

I couldn't get how to approach this problem.i tried drawing FBD of the balls but could get nothing out of it.Any help shall be highly appreciated.Thanks.

• Look up "sprag". Sep 24 '17 at 17:46
• I will try to answer this tomorrow. Looks like you have to improve your imagination skills too. :) Think and imagine the things and forces that will happen. Do not forget the basic formula that $F_f=\mu N$. Sep 26 '17 at 11:28
• @JemEripol yes sir I can imagine the situation what is going on.But one thing I couldn't get.When we reverse the direction of rotation of outer ring the balls loses contact with the inner member but we are not sure if it loses contact with the outer ring.if that's not then the balls might come in contact with the inner member again and thus the torque can be transmitted.so in practical view is there a critical value of $\mu$ for which this isn't possible. Sep 26 '17 at 11:54
• That statement is only saying that if you reverse the input shaft, it will not have contact with the output shaft through the ball, that's why it is unidirectional. The thing is, you need only to compute the case shown in the drawing. Sep 26 '17 at 23:55

Let me give you a headstart.

$T=F*d$, therefore, $F_{f1}=(T/2)/r$. The torque is divided by two because there are two balls.

From there, you can compute $N_1$. Sum up moments at the center of the ball, you will get:

$F_{f1}=F_{f2}$

Now that you have $F_{f2}$, you will get $N_2$.

Looking at the output shaft now, you will get the position of the force as shown in the right figure. You will need to sum up moment at the center of the output shaft. The resulting moment needs to be equal to $T/2$ as we need to transmit the torque to the output shaft.

This will require hard mathematics but I think this will do it. I know there is a simpler solution so you we will have to find out.

• I got it.This however doesn't involve hard mathematics.a very straightforward approach.thanks. Sep 27 '17 at 3:53
• Yeah, I noticed just now. The "hard" terms cancels out anyway. Sep 27 '17 at 5:17