Question

20. A bag of cement of weight 325 N hangs in equilibrium from three wires as shown in Figure P5.20. Two of the wires make angles 0, = 60.0° and 0, - 25.0° with the hor izontal. Assuming the system is in equilibrium, find the tensions 7,, Ty, and T, in the wires.

Answer 1

Answer:

T₁ = 296 N

T₂ = 163 N

T₃ = 325 N

Explanation:

Draw a free body diagram of the bag. There are two forces:

Weight force mg pulling down,

Tension force T₃ pulling up.

Sum of forces in the y direction:

∑F = ma

T₃ − mg = 0

T₃ = mg

T₃ = 325 N

Draw a free body diagram of the connection between the wires. There are three forces, T₁, T₂, and T₃.

Sum of forces in the y direction:

∑F = ma

T₁ sin θ₁ + T₂ sin θ₂ − T₃ = 0

Sum of forces in the x direction:

∑F = ma

T₂ cos θ₂ − T₁ cos θ₁ = 0

T₂ cos θ₂ = T₁ cos θ₁

T₂ = T₁ cos θ₁ / cos θ₂

Substitute:

T₁ sin θ₁ + T₁ cos θ₁ tan θ₂ − T₃ = 0

T₁ (sin θ₁ + cos θ₁ tan θ₂) = T₃

T₁ = T₃ / (sin θ₁ + cos θ₁ tan θ₂)

T₁ = 325 N / (sin 60° + cos 60° tan 25°)

T₁ = 296 N

T₂ = T₁ cos θ₁ / cos θ₂

T₂ = 296 N cos 60° / cos 25°

T₂ = 163 N