Question

A 30 kg smooth, solid sphere rests on two frictionless inclines as shown in the figure. The magnitude of contact force in N acting at the point A is (take acceleration due to gravity \( g = 9.81 \, \text{m/s}^2 \) and consider both sphere and inclines to be rigid) _________. 
[round off to 2 decimal places]

Hint:

To calculate the contact force at the point of contact for a rolling sphere, decompose the weight into components along the inclines and use equilibrium conditions.

Answer 1

Correct Answer: 147

Given:
- Mass of the sphere, \( m = 30 \, \text{kg} \),
- Acceleration due to gravity, \( g = 9.81 \, \text{m/s}^2 \),
- The inclines make angles \( 60^\circ \) and \( 30^\circ \) with the horizontal.
First, the weight of the sphere is: \[ W = mg = 30 \times 9.81 = 294.3 \, \text{N}. \] For a smooth, solid sphere, the contact force at point A is the normal force \( N_A \). The sphere experiences both translational and rotational motion, so we use the equations of motion for rolling without slipping. The component of the weight acting along the incline with angle \( \theta = 60^\circ \) is: \[ W_{\parallel 60} = W \sin(60^\circ) = 294.3 \times \frac{\sqrt{3}}{2} = 254.4 \, \text{N}. \] The other component along the incline with angle \( 30^\circ \) is: \[ W_{\parallel 30} = W \sin(30^\circ) = 294.3 \times \frac{1}{2} = 147.15 \, \text{N}. \] Since the system is in equilibrium, the contact force at point A balances these components. Thus, the magnitude of the contact force is: \[ N_A = 147.15 \, \text{N}. \] Thus, the magnitude of the contact force at point A is approximately \( 147.00 \, \text{N} \).