Answer:
6 N
Explanation:
Let's start with the small block m on top. There are four forces:
Weight force mg pulling down, normal force N₁ pushing up, tension force T pulling right, and friction force N₁μ pushing left (opposing the direction of motion).
Now let's look at the large block M on bottom. There are seven forces:
Normal force N₁ pushing down (opposite and equal from block m),
Friction force N₁μ pushing right (opposite and equal from block m),
Weight force Mg pulling down,
Tension force T pulling right,
Applied force F pulling left,
Normal force N₂ pushing up,
and friction force N₂μ pushing right (opposing the direction of motion).
So you've correctly identified the free body diagrams.
Now apply Newton's second law. Sum of forces in the y direction for block m:
∑F = ma
N₁ − mg = 0
N₁ = mg
Sum of forces in the x direction:
∑F = ma
T − N₁μ = 0
T = N₁μ
T = mgμ
Sum of forces in the y direction for block M:
∑F = ma
-N₁ − Mg + N₂ = 0
N₂ = N₁ + Mg
N₂ = mg + Mg
Sum of forces in the x direction:
∑F = ma
N₁μ + T − F + N₂μ = 0
F = N₁μ + T + N₂μ
F = mgμ + mgμ + (mg + Mg)μ
F = gμ(3m + M)
Since M = 2m:
F = 5gμm
Plug in values:
F = 5 (10 m/s²) (0.400) (0.300 kg)
F = 6 N
