Question

A sphere of mass m is held between two smooth inclined walls. For sin 37° = 3/5,
the normal reaction of the wall (2) is equal to :
​

Answer 1

Given that,

Value of angle $\sin 37^{\circ}=\dfrac{3}{5}$

We need to calculate the normal reaction of the wall

Using balance equation

$\sum y=0$

$N_{2}\sin16^{\circ}+mg=N_{2}\sin37^{\circ}$...(I)

$\sum x=0$

$N_{2}\cos16^{\circ}=N_{1}\cos37^{\circ}$....(II)

We need to calculate the value of N₁

Using equation (II)

$N_{1}=\dfrac{N_{2}\cos16^{\circ}}{\cos37^{\circ}}$

Put the value of N₁ in equation (II)

$N_{2}\cos16^{\circ}+mg=\dfrac{N_{2}\cos16^{\circ}}{\cos37^{\circ}}\sin37^{\circ}$

Put the value in the equation

$N_{2}\times0.96+mg=N_{2}\times0.96\times0.75$

$N_{2}=\dfrac{mg}{0.45}$

$N_{2}=\dfrac{20mg}{9}$

Hence, The value of N₁ is $\dfrac{20mg}{9}$