Question

suppose a block with a mass of 2.92 kg is resting on a ramp. if the coefficient of static friction between the block and ramp is 0.327, what maximum angle can the ramp make with the horizontal before the block starts to slip down? round your answer to the nearest tenth.

Explanation:

Step1: Equate forces at impending slip

At maximum angle, the static friction force balances the component of gravity down the ramp: $f_s = mg\sin\theta$. The maximum static friction is $f_s = \mu_s n$, and the normal force $n = mg\cos\theta$. Substitute to get $\mu_s mg\cos\theta = mg\sin\theta$.

Step2: Simplify to find tangent of angle

Cancel $mg$ from both sides: $\mu_s = \tan\theta$.

Step3: Calculate angle using arctangent

$\theta = \arctan(\mu_s) = \arctan(0.327)$.

Step4: Compute and round the value

Using a calculator, $\arctan(0.327) \approx 18.1^\circ$.

Answer:

$18.1^\circ$