Maximum force of friction between M1 and inclined plane
f1 = m1M1g cos θ = (0. 75)(4)(9. 8)(0.8) = 23.52 N
M1g sin θ = (4)(9.8)(0.6) = 23.52 N = F1 (say)
Maximum force of friction between M2 and inclined plane
f2 = m2M2g cos θ = (0. 25)(2)(9.8)(0.8) = 3.92 N
M2g sin θ = (2)(9.8)(0.6) = 11.76 N = F2 (say)
Both the blocks will be moving downwards with same acceleration a. Different forces acting on two blocks are as shown in figures.
Equation of motion of M1
T + F1 – f1 = M1a ... (i)
or T = 4a
Equation of motion M2
F2 – T – f2 = M2a
or 7.84 – T = 2a ... (ii)
Solving eqs. (i) and (ii), we get
a = 1.3 m/s2 and T = 5.2 N