Question

a spherical ball of mass m= 5 kg rests between two planes which make angles of 30° and 45° respectively with the horizontal. the system is in equilibrium . find the normal forces exerted on the ball by each of the planes . the planes are smooth.

Answer 1

The normal forces are 43.9 N and 35.9 N.

The ball of weight W rests on two planes which are at angles 30° and 45° with the horizontal. The ball has a mass of 5 kg, its weight is given by,

$W=(5 kg)*(9.8 m/s^2)\\ =49 N$

The planes exert normal forces N₁ and N₂ on the ball. Since the ball is in equilibrium, the normal forces can be determined by drawing a force diagram for the ball and applying the conditions of equilibrium.

Refer to the attached diagram.

The normal forces are resolved into horizontal and vertical components . The horizontal component of N₁ is N₁ cos45 and the vertical component is N₁ sin 45. The horizontal component of N₂ is N₂ cos 30 and the vertical component is N₂sin 30.

Apply the conditions of equilibrium from the force diagram shown in the figure.

For equilibrium along the horizontal direction,

$N_1 cos45 = N_2 cos 30\\ \frac{N_1}{\sqrt{2}} =\frac{N_2*\sqrt{3}}{2} \\ N_1 =N_2*\sqrt{\frac{3}{2}}$

For equilibrium along the vertical direction,

$N_1 sin 45+N_2 sin 45 = 49\\ \frac{N_1}{\sqrt{2}} +\frac{N_2}{2} =49$

Substitute the expression for N₁.

$\frac{N_2}{\sqrt{2}} \sqrt{\frac{3}{2}} +\frac{N_2}{2} =49\\ N_2=35.9 N$

Solve for N₁.

$N_1 =N_2\sqrt{\frac{3}{2}} \\ =(35.9 N)*(\frac{3}{2})\\ =43.9 N$

The normal forces are 43.9 N and 35.9 N.

Answer 2

N(30•)=100/√3-1

N(45•)=50√2/√3-1

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