A uniform sphere of weight W and radius R = 3 | vedclass

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(A) Let the center of the sphere be $O$ and the point where the string is attached to the wall be $A$. The point where the sphere touches the wall is

Vedclass · Accepted Answer

$5\,W/4$

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(A) Let the center of the sphere be $O$ and the point where the string is attached to the wall be $A$. The point where the sphere touches the wall is $B$. The distance $OB$ is the radius $R = 3\,m$. The length of the string $OA$ is $L = 2\,m$. The distance $AB$ is the radius $R = 3\,m$.In the right-angled triangle formed by the center of the sphere,the point of attachment on the wall,and the point of contact with the wall,the hypotenuse is the string length $L = 2\,m$ plus the radius $R = 3\,m$,so the hypotenuse is $5\,m$.The base of this triangle is the radius $R = 3\,m$.Let $	heta$ be the angle the string makes with the wall. Then $\sin 	heta = R / (R + L) = 3 / (3 + 2) = 3/5$.Alternatively,if $	heta$ is the angle with the horizontal,$\cos 	heta = 3/5$,which means $\sin 	heta = 4/5$.Looking at the forces on the sphere: The vertical component of tension $T \sin 	heta$ balances the weight $W$,so $T \sin 	heta = W$.Using $\sin 	heta = 4/5$,we get $T (4/5) = W$,so $T = 5\,W/4$.

$A$ uniform sphere of weight $W$ and radius $R = 3\,m$ is held by a string of length $L = 2\,m$ attached to a frictionless wall as shown in the figure. The tension in the string is:

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