An object weighing 400 N is hanging from two ropes that make angles of 40° and 50° with the ceiling. Calculate the tension (magnitude only) in the two ropes.

Answer 1

Please draw a picture with all three forces.

Let

F = 400 N be the weight of the object,

θ₁ = 40° be the angle of one rope,

θ₂ = 50° be the angle of the other rope,

T₁ be the tension in the first rope (to be calculated),

T₂ be the tension in the other rope (to be caclulated).

As the object is in equilibrium, the vector sum of all forces must be 0.

That gives two equations concerning the magnitudes of forces:

1) The downward components of all forces must equal the upward components of all forces.

F = T₁sin(θ₁) + T₂sin(θ₂) (1)

2) The leftward components of all forces must equal the rightward components of all forces.

T₁cos(θ₁) = T₂cos(θ₂) (2)

From (2)

T₂ = T₁cos(θ₁)/cos(θ₂) (3)

Substitute (3) into (1)

F = T₁sin(θ₁) + T₁sin(θ₂)cos(θ₁)/cos(θ₂)

= T₁(sin(θ₁)cos(θ₂) + sin(θ₂)cos(θ₁))/cos(θ₂)

= T₁sin(θ₁+θ₂)/cos(θ₂)

From that express

T₁ = Fcos(θ₂)/sin(θ₁+θ₂) (4)

Substitute (4) into (3)

T₂ = Fcos(θ₁)/sin(θ₁+θ₂) (5)

Substituting actual numbers into (4) and (5)

T₁= 400cos(50°)/sin(40°+50°) = 400cos(50°) = 257.1 N

T₂ = 400cos(40°)/sin(40°+50°) = 400cos(40°) = 306.4 N