Question

7 formula 1 point a box with a mass of 6 kilograms is placed on a ramp. the ramp is elevated at an angle of 30 degrees from the horizontal. this is the maximum angle the ramp can be placed at before box starts to move. what must be the coefficient of static friction for the box? answer previous next

Explanation:

Step1: Analyze Forces on Box

The box is in equilibrium (just about to move, so static friction is at maximum). Forces: gravitational force \( mg \), normal force \( N \), static friction \( f_s \). Resolve \( mg \) into components: parallel (\( mg\sin\theta \)) and perpendicular (\( mg\cos\theta \)) to ramp.

Step2: Equilibrium Conditions

Perpendicular direction: \( N = mg\cos\theta \) (no acceleration perpendicular). Parallel direction: \( f_s = mg\sin\theta \) (no acceleration parallel, since it's just about to move, \( f_s = \mu_s N \)).

Step3: Substitute \( N \) into Friction Formula

\( \mu_s mg\cos\theta = mg\sin\theta \). Mass \( m \) and \( g \) cancel out. So \( \mu_s = \tan\theta \).

Step4: Plug in \( \theta = 30^\circ \)

\( \tan(30^\circ) = \frac{1}{\sqrt{3}} \approx 0.577 \) (or \( \frac{\sqrt{3}}{3} \)).

Answer:

\( \frac{\sqrt{3}}{3} \) (or approximately \( 0.58 \))