Answer:
a) F = μ g (M_a + M_b), b) T = μ g M_b
Explanation:
a) as the box goes at constant speed, the relation is zero and we can use the translational equilibrium relation
∑ F = 0
Y axis
N₁ - W₁ = 0
N₂-W₂ = 0
N₁ = W₁
N₂ = W₂
X axis
F -fr₁ -fr₂ = 0
F = fr₁ + fr₂
the friction force has the expression
fr = μ N
indicates that the friction coefficient is the same for the two boxes
fr₁ = μ M_a g
Fr₂ = μ M_b g
we substitute
F = μ g (M_a + M_b)
b) let's apply the equilibrium for box A
F -fr₁ - T = 0
T = F -fr1
boc B
T - fr2 =0
T = fr2
our system equation
T = F -fr1
T = fr2
solved
2 T = F - fr₁ + fr₂
2T = [μ g (M_a + M_b) - μ g M_a + μ g M_b]
2T = 2 μ g M_b
T = μ g M_b
