Question

a box at rest is in a state of equilibrium half way up on a ramp. the ramp has an incline of 42°. what is the force of static friction acting on the box if the box has a gravitational force of 112.1 n? 80 n 70 n 85 n 75 n

Explanation:

Step1: Recall the equilibrium condition on an inclined plane

When an object is in equilibrium on an inclined plane, the static friction force \( f_s \) balances the component of the gravitational force parallel to the ramp. The formula for the component of gravitational force parallel to the ramp is \( F_{\parallel}=mg\sin\theta \), where \( mg \) is the gravitational force (weight) of the object and \( \theta \) is the angle of inclination. Here, \( mg = 112.1\space N \) and \( \theta = 42^{\circ} \).

Step2: Calculate the parallel component of gravitational force

We need to find \( F_{\parallel}=112.1\times\sin(42^{\circ}) \). First, calculate \( \sin(42^{\circ})\approx0.6691 \). Then, \( F_{\parallel}=112.1\times0.6691\approx 112.1\times0.67\approx75.107\space N \approx 75\space N \). Since the object is in equilibrium, the static friction force \( f_s = F_{\parallel} \).

Answer:

75 N (corresponding to the option with 75 N)