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1. A 65-kg painter stands 1.5 m from the left end of a 4-m long uniform

plank that is supported by a cable at each end. If the plank weighs
50 kg, what is the tension in each cable?

User Luan
by
3.0k points

1 Answer

4 votes

Answer:

Step-by-step explanation:

Given:

m₁ = 65 kg

m₂ = 50 kg

L = 4 m

L₁ = 1.5 m

L₂ = L / 2 = 2 m

g = 10 m/s²

____________

T₁ - ?

T₂ - ?

Let's make a drawing.

Forces:

F₁ = m₁·g = 65·10 = 650 N

F₂ = m₂·g = 50·10 = 500 N

By the rule of moments (with respect to point A):

F₁·L₁ + F₂·L₂ = T₂·L

T₂ = (F₁·L₁ + F₂·L₂ ) / L

The tension in each cable:

T₂ = (650·1.5 + 500·2) / 4 ≈ 490 N

T₁ = (F₁ + F₂) - T₂ = (650 + 500) - 490 = 660 N

1. A 65-kg painter stands 1.5 m from the left end of a 4-m long uniform plank that-example-1
User Gianluca Bargelli
by
3.2k points