The system in Fig. 13-29 is in equilibrium, but it begins to slip if any additional mass is added to the 5.0 kg object. What is the coefficient of static friction between the 10 kg block and the plane on which it rests?

0.289

Explanation:

Since the body is in equilibrium (no net movement), then the total upward force must equal the total downward force and also total forward force must equal total backward force

Tension in the rope inclined at 30⁰ to the vertical hhas vertical and horizontal components

vertical component = Tcos 30⁰ and horizontal component = T sin30⁰

T cos 30⁰ = the downward force = mg where m is the mass of 5kg and g is 9.8 m/s² acceleration due to gravity

T cos 30⁰ = 49

T in the inclined rope to the vertical = 49 / cos 30⁰ = 56.58 N

Horizontal component of the tension in the rope = T sin30⁰ = 56.58 sin 30⁰ = 28.29 N

since the 10kg was not moving the frictional force impeding the movement = the horizontal component of the T along the positive x axis since the frictional force will act along the negative x axis

Fr = 28.29 N

coefficient of static friction = Fr / normal (mg) = 28.29 / (10 * 9.8) = 0.289