Question

A metal sphere is hanging by a light string fixed to a smooth wall. The forces acting on the sphere are as shown. Which of the following statements are correct?

• A. $\overline{R}+\overline{T}+\overline{W}=0$
• B. $|\overline{R}+\overline{W}|=|\overline{T}|$
• C. $R^2+T^2-2TRcos\theta =W^2$
• D. Tension is greater than normal reaction as well as weight.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\overline{R}+\overline{T}+\overline{W}=0$
B $|\overline{R}+\overline{W}|=|\overline{T}|$
C $R^2+T^2-2TRcos\theta =W^2$
D Tension is greater than normal reaction as well as weight.

All the forces i.e. R,T and W are vectors and are in equilibrium. Thus they produce a zero resultant and there summation would be zero. Hence, option A is correct.
Resultant of all the vector forces acting on the sphere is zero and therefore the body is in equilibrium. Hence option B is correct.
Angle between T and R is
$\pi -\theta$
From cosine triangle law,
$W^2=R^2+T^2-2RTCos\theta$
Hence option C is correct.
From the FBD of sphere, we have
$R=TCos\theta$
$W=TSin\theta$
Squaring and adding the above two equations, we get
$R^2+W^2=T^2$
Therefore Tension T is greater than R and W. Hence option D is correct.