Question

Block 1 of mass m1 = 2 kg is placed on block 2 of mass m2 = 3 kg which is then placed on a table. A string connecting block 2 to a hanging mass M passes over a pulley attached to one end of the table, as shown in Fig. 1. The mass and friction of the pulley are negligible. The coefficients of friction between block 1 and block 2 are μ1s = 0.2, μ1k = 0.1 and between block 2 and the tabletop air μ2s = 0.3, μ2k = 0.15. Draw the complete free body diagram of the system and determine the largest value of M for which the blocks can remain at rest.

Answer 1

Explanations & Calculations

• The free-body diagram,

• The second block has all the forces that influence the static equilibrium of this system.
• The static friction forces are larger than the kinetic friction forces, hence to remain stationary this condition should be satisfied

$\qquad\qquad \begin{aligned} \small T&\leq \small f_{1-max}+f_{2-max}\\ \small Mg&\leq\small \mu_sR_{1s}+\mu_{s2}R_2\\ &\leq\small 0.2(m_1g)+0.3(R_1+m_2g)\\&\leq\small 0.2(m_1g)+0.3(m_1g+m_2g)\\ &\leq\small (0.5m_1+0.3m_2)g\\ \small M&\leq\small0.5m_1+0.3m_2\\ \small M_{max}&=\small 0.5m_1+0.3m_2\\ \small\bold{M_{max}}&=\small \bold{1.9\,kg} \end{aligned}$