Question

COB
foced
small
surface
All surface are smooth. Find the force applied by
vertical wall on sphere.​

Answer 1

Given that,

A sphere of mass m, is held between two smooth inclined walls. the normal reactions of the wall (2) is equal to

Suppose, $\theta=37^{\circ}$

According to figure,

We need to calculate the vertical force

Using balance equation

$\Sum F_{y}=0$

$N_{2}\sin\alpha+mg=N_{1}\sin\theta$

Put the value into the formula

$N_{2}\sin16+mg=N_{1}\sin37$

We need to calculate the horizontal force

Using balance equation

$\Sum F_{x}=0$

$N_{2}\cos\theta=N_{1}\cos\theta$

Put the value into the formula

$N_{2}\cos16=N_{1}\cos37$

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

Put the value of N₁ in equation (I)

$N_{2}\sin16+mg=\dfrac{N_{2}\cos16}{\cos37}\sin37$.....(III)

We know that,

$\sin16=0.27$

$\cos16=0.96$

$\tan37=\dfrac{3}{4}=0.75$

Put the value in the equation (III)

$N_{2}\times0.27+mg=N_{2}\times0.96\times0.75$

$mg=N_{2}\times0.96\times0.75-N_{2}\times0.27$

$mg=N_{2}(0.96\times0.75-0.27)$

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

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

Hence, The force applied by  vertical wall on sphere is $\dfrac{20 mg}{9}\ N$