Question

a box at rest on a ramp is in equilibrium, as shown. what is the force of static friction acting on the box? round your answer to the nearest whole number n. what is the normal force acting on the box? round your answer to the nearest whole number n.

Explanation:

Step1: Find Static Friction Force

The force of static friction (\(F_{fr}\)) balances the component of the gravitational force down the ramp. The gravitational force \(F_g = 735\) N, and the angle of the ramp is \(30^\circ\). The component of gravity down the ramp is \(F_g \sin(30^\circ)\).
\(F_{fr} = F_g \sin(30^\circ) = 735 \times \sin(30^\circ)\)
Since \(\sin(30^\circ) = 0.5\), we have \(F_{fr} = 735 \times 0.5 = 367.5\) N, which rounds to 368 N.

Step2: Find Normal Force

The normal force (\(F_N\)) balances the component of the gravitational force perpendicular to the ramp. This component is \(F_g \cos(30^\circ)\).
\(F_N = F_g \cos(30^\circ) = 735 \times \cos(30^\circ)\)
Since \(\cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.8660\), we have \(F_N = 735 \times 0.8660 \approx 636.51\) N, which rounds to 637 N.

Answer:

Static friction force: \(\boxed{368}\) N
Normal force: \(\boxed{637}\) N