Question

consider the figure below. (let (w_1 = 150 n) and (w_2=36.5 n)). (a) what is the minimum force of friction required to hold the system of the figure above in equilibrium? (b) what coefficient of static friction between the 150 - n block and the table ensures equilibrium? (enter the minimum acceptable coefficient of friction.) (c) if the coefficient of kinetic friction between the 150 - n block and the table is 0.122, what hanging weight should replace the 36.5 - n weight to allow the system to move at a constant speed once it is set in motion?

Explanation:

Step1: Analyze forces for equilibrium in part (a)

For the system to be in equilibrium, the frictional force \(f\) on the block on the table must balance the weight of the hanging - block. So, \(f = w_2\). Given \(w_2=36.5\ N\), the minimum force of friction required for equilibrium is \(f = 36.5\ N\).

Step2: Calculate the coefficient of static friction in part (b)

The formula for the maximum static - friction force is \(f_s=\mu_sN\). The normal force \(N\) on the \(150 - N\) block on the table is equal to its weight, \(N = 150\ N\). Since the minimum static - friction force required for equilibrium is \(f_s = 36.5\ N\), and \(f_s=\mu_sN\), we can solve for \(\mu_s\): \(\mu_s=\frac{f_s}{N}\). Substituting \(f_s = 36.5\ N\) and \(N = 150\ N\), we get \(\mu_s=\frac{36.5}{150}\approx0.243\).

Step3: Analyze forces for constant - speed motion in part (c)

When the system is moving at a constant speed, the net force on the system is zero. The kinetic - friction force is \(f_k=\mu_kN\). Given \(N = 150\ N\) and \(\mu_k = 0.122\), then \(f_k=\mu_kN=0.122\times150 = 18.3\ N\). For the system to move at a constant speed, the hanging weight \(w\) must be equal to the kinetic - friction force, so \(w = 18.3\ N\).

Answer:

(a) \(36.5\)
(b) \(0.243\)
(c) \(18.3\)