Question

question 5
1 pts
a wooden block of mass 94 g is placed on a horizontal board lying on a table. the coefficient of static friction between the block and the board is 0.33. one side of the board is slowly lifted up so that it creates an angle with the table. at what angle will the wooden block begin to move?
round your answer to the nearest tenths place.

Explanation:

Step1: Analyze forces

When the block is on the inclined board, the gravitational force \(mg\) can be decomposed into two components: \(mg\sin\theta\) (parallel to the incline, causing the block to slide) and \(mg\cos\theta\) (perpendicular to the incline). The maximum static - friction force \(f_s=\mu_sN\), and \(N = mg\cos\theta\). At the moment the block begins to move, \(mg\sin\theta=\mu_smg\cos\theta\).

Step2: Simplify the equation

Divide both sides of \(mg\sin\theta=\mu_smg\cos\theta\) by \(mg\cos\theta\) (since \(m
eq0\) and \(\cos\theta
eq0\) for \(\theta
eq90^{\circ}\)). We get \(\tan\theta=\mu_s\).

Step3: Calculate the angle

Given \(\mu_s = 0.33\), then \(\theta=\arctan(\mu_s)\). Using a calculator, \(\theta=\arctan(0.33)\approx18.3^{\circ}\)

Answer:

\(18.3^{\circ}\)