Answer to Question #124636 in Physics for Dr. Horus

Question

A mass m1=6.5kg is attached to a mass m2=3.75kg with a light inextensible string passing over a frictionless pulley at the end of the horizontal plane.
(i) what is the minimum coefficient of static friction that will keep the system of masses stationary?
(ii) if the coefficient is reduced to 0.25, what would the acceleration of m2 be?

Answer 1

Consider the first mass and apply Newton's first law to this mass. Assume that the m2 hangs to the right of m1 and the motion will be to the right for m1 and down for m2. Point Y-axis upward and X-axis to the right.

$$-\mu N_1+T=m_1a,\\N_1=m_1g.\\ \mu=\frac{T-m_1a}{m_1g}.$$

For the second body:

$$T-m_2g=-m_2a,\\ T=m_2(g-a).$$

Substitute to find the coefficient of friction:

$$\mu=\frac{m_2(g-a)-m_1a}{m_1g}.$$

(i) If the coefficient of friction is greater than some value μ, the system will be in equilibrium (a=0):

$$\mu=\frac{m_2}{m_1}=\frac{3.75}{6.5}=0.577.$$

(ii) Express the equation for acceleration from the equation for the coefficient of frictiuon:

$$a=\frac{g(m_2-\mu m_1)}{m_1+m_2}=\frac{9.8(3.75-0.25\cdot 6.5)}{6.5+3.75}=2.03\text{ m/s}^2.$$