Question

Consider the situation illustrated in the figure below. If θ = 35° and the masses of blocks B and A are, respectively, Mb and Ma= 0.4Mb, (a) what should be the minimum value of the static coefficient of friction μs between the table and block B so that the system remains in balance? (b) In this case, what will be the tension in the rope segment connected to (node) and the vertical wall?

Answer 1

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.