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(D) For the sphere to be in equilibrium,the net force acting on it must be zero.Resolving the tension $T$ into horizontal and vertical components:$\Si

Vedclass · Accepted Answer

$T=P+W$

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(D) For the sphere to be in equilibrium,the net force acting on it must be zero.Resolving the tension $T$ into horizontal and vertical components:$\Sigma F_x = P - T \sin 	heta = 0 \implies P = T \sin 	heta$ ...$(i)$$\Sigma F_y = T \cos 	heta - W = 0 \implies W = T \cos 	heta$ ...(ii)Dividing $(i)$ by (ii),we get $\frac{P}{W} = \frac{T \sin 	heta}{T \cos 	heta} = 	an 	heta$,so $P = W 	an 	heta$. This is correct.The vector sum of all forces is zero,so $\vec{T} + \vec{P} + \vec{W} = 0$. This is correct.From $(i)$ and (ii),$T^2 \sin^2 	heta + T^2 \cos^2 	heta = P^2 + W^2$,which gives $T^2 = P^2 + W^2$. This is correct.The expression $T = P + W$ is incorrect because forces are vectors and cannot be added algebraically unless they are in the same direction.

$A$ small sphere is hung by a string fixed to a wall as shown in the figure. The sphere is pushed from the wall by a stick. The forces acting on the sphere are as shown. Then,identify the incorrect expression.

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