The 20 -kg motor has a center of gravity at G and is pin...

The 20 -kg motor has a center of gravity at $G$ and is pin connected at $C$ to maintain a tension in the drive belt. Determine the smallest counterclockwise twist or torque $\mathbf{M}$ that must be supplied by the motor to turn the disk $B$ if wheel $A$ locks and causes the belt to slip over the disk. No slipping occurs at $A$. The coefficient of static friction between the belt and the disk is $\mu_{x}=0.3$.

Transcript

00:01 Here in this question we are given here 20 kilogram water so we are given a mass m which is equal to 20 kilogram water so that motor we are given here and the has a center of gravity at g as a center of gravity center of gravity at g and is this pin is connected at c and maintain a tension in the drive build it only the smallest counterclockwise twist or we can say torque here we want to find torque m that must be supplied by the motor turn to dix b if the wheel a is lock and close the belt to sleep over the dix so in this question we are given a mu x is equal to 0 .3 and the coefficient of strategic friction we are given here is this so this is our coefficient of state restriction let's find out the answer so first we can write here the value of capital we are given a 50 millimeter here we are given a value of capital is 50 millimeter so here first we can write the capital r so capital which is equal to 50 millimeter now here taking the moment at see so here we can write moment at c so we can write the capital m g m d r maldivore v s 0 .1 which is equal we can write t 1 mulliver we can write 0 .2 mulliver we can write t 2 1 1 0 .1 so the this value is given here now this is our equation number one from the belt here we can write for the build we can write the value of theta which is equal to pi and here we can write t1 divided by t2 which is equal to we can write it is to mu theta and now here you can write t2 divided by t1 which is equal we can write answer is so answer here is 2 .819 so this is our equation number 2 now, here, we can write this small m which is equal to t2 minus t1, mlaba capital which is equal we can write t2 minus t1, moliabwe, we can write 0 .05.

02:26 And this is the equation number.

02:29 Solving this equation and we can write here answer is t1 which is equal to 40 .713 -08 newton.

02:39 And here we can write t2 which is equal to 114 .772 newton...

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