Answer:
(a) μ ≈ 0.57
(b) not enough information
Explanation:
Draw a free body diagram.
There are four forces acting on block B:
- Weight force Mb g pulling down
- Normal force N pushing up
- Friction force Nμ pulling left
- Tension force T₁ pulling right
There are three forces acting on the node:
- Tension force T₁ pulling left
- Tension force T₂ pulling 35° above the horizontal
- Weight force Ma g pulling down
(a)
Sum of forces on block B in the y direction:
∑F = ma
N − Mb g = 0
N = Mb g
Sum of forces on block B in the x direction:
∑F = ma
T₁ − Nμ = 0
T₁ = Nμ
T₁ = Mb g μ
Sum of forces on the node in the y direction:
∑F = ma
T₂ sin 35° − Ma g = 0
T₂ sin 35° = Ma g
Sum of forces on the node in the x direction:
∑F = ma
T₂ cos 35° − T₁ = 0
T₂ cos 35° = T₁
T₂ cos 35° = Mb g μ
Divide the previous equation by this equation, eliminating T₂.
tan 35° = Ma g / (Mb g μ)
μ = Ma / (Mb tan 35°)
μ = 0.4 / tan 35°
μ ≈ 0.57
(b) T₂ sin 35° = Ma g = 0.4 Mb g
Without knowing the value of Mb, we cannot find the value of the tension force T₂.
