Question

M=0.500kg on level table, Fs= 0.600. Three strings tide together in a knot, 1 attach to mass, 2 30°above horizontal attached to wall, 3 hangs M2 vertically what is the Maximum weight of M2 for M1 & 2 to remain in equilibrium

Answer 1

For the equilibrium:

$$m_2g=T$$

$$N=m_1g-T\sin 30=m_1g-0.5T$$

$$T=T\cos{30}+\mu_sN=T\cos{30}+\mu_s(m_1g-0.5T)$$

$$T=T\cos{30}+\mu_s(m_1g-0.5T)$$

$$T=T\cos{30}+0.6((0.5)(9.8)-0.5T)$$

Maximum weight of M2 for M1 & 2 to remain in equilibrium

$$T=6.77\ N$$