Question

a block of mass ( m_0 ) is at rest on a ramp inclined at an angle of ( \theta_0 ) with the horizontal. the coefficient of static friction between the ramp and block is ( mu_s ). the angle of the ramp must be increased by ( 10^{circ} ) before the block starts to slide. what is the magnitude of the force of friction exerted on the block when ( \theta < 10^{circ} )?

a ( m_0 g sin \theta_0 )
b ( m_0 g cos \theta_0 )
c ( mu_s m_0 g sin \theta_0 )
d ( mu_s m_0 g cos \theta_0 )

Explanation:

Step1: Analyze the forces on the block

When the block is at rest on the inclined - plane, we resolve the gravitational force \(F = m_0g\) into two components. The component of the gravitational force along the incline is \(F_{\parallel}=m_0g\sin\theta\), and the component perpendicular to the incline is \(F_{\perp}=m_0g\cos\theta\).

Step2: Apply Newton's second law in the direction along the incline

Since the block is at rest (\(a = 0\)) when \(\theta<10^{\circ}\), according to Newton's second law \(F_{net}=ma\). In the direction along the incline, \(F_{f}-m_0g\sin\theta = 0\) (where \(F_{f}\) is the force of static friction).

Answer:

A. \(m_{0}g\sin\theta_{0}\)