Two masses m and M are attached with strings as shown For the system to be in equilibrium we - Physics - Laws Of Motion

Question

Two masses m and M are attached with strings as shown For the
system to be in equilibrium we have

(A) tanθ = 1 + 2 M m (B) tanθ = 1 + 2 m M (C) tanθ =
1 + M 2 m (D) tanθ = 1 + m 2 M

Ajay Patil · Accepted Answer

Dear Student,

Firstly find the all the forces acts on a first junction,
resolve all the inclined force, mg act downward and T1
and T2 resolved
we can write,
 ∑Fy=0-mg+sin45o(T1+T2)=0T1+T2=mg2-----i∑Fx=0T1cos45o-T2cos45o=0T1=T2--------iiT1=T2=mg2

Free body diagram for mass M
∑Fx=0T3cosθ=T1cos45o-----iii∑Fy=0T3sinθ=T1sin45o+Mg----ivdivide equation iv by iiitanθ=T1sin45o+MgT1cos45o=mg/2+Mgmg/2= 1+2Mm
so, option (a) is correct

Two masses m and M are attached with strings as shown. For the system to be in equilibrium we have (A) tanθ = 1 + 2 M m (B) tanθ = 1 + 2 m M (C) tanθ = 1 + M 2 m (D) tanθ = 1 + m 2 M

Two masses m and M are attached with strings as shown. For the system to be in equilibrium we have

(A) $tanθ = 1 + \frac{2 M}{m}$ (B) $tanθ = 1 + \frac{2 m}{M}$ (C) $tanθ = 1 + \frac{M}{2 m}$ (D) $tanθ = 1 + \frac{m}{2 M}$