Question #103754

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

Expert's answer

For the equilibrium:

"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

Learn more about our help with Assignments: MechanicsRelativity

## Comments

## Leave a comment