Question

The figure shows a structure subject to a force of 60 kN and a torque of 30 kN.m. Determine the horizontal (left-to-right oriented) and vertical (bottom-up oriented) force modulus acting on point A and also the tension developed on the BC rope. The blue arrow indicates the "direction of rotation" caused in the structure by the action of the torque in relation to point A.

Answer 1

Answer:

Fₓ = 21.9 kN

Fᵧ = 84.3 kN

T = 32.7 kN

Explanation:

Draw a free body diagram (assuming the weight of the structure is included in the 60 kN force).

There are vertical and horizontal reaction forces at A (Fᵧ and Fₓ), and a tension force T at B pulling down along the rope.

The length of BC is √(2.7² + 3²) = √16.29.  Using similar triangles, the vertical and horizontal components of the tension force are:

Tᵧ = 3 T / √16.29 ≈ 0.743 T

Tₓ = 2.7 T / √16.29 ≈ 0.669 T

Sum of moments about A in the counterclockwise direction:

∑τ = Iα

Tᵧ (1 m) + Tₓ (3 m) − 60 kN (1 m) − 30 kNm = 0

Tᵧ + 3 Tₓ = 90 kN

0.743 T + 3 (0.669 T) = 90 kN

2.750 T = 90 kN

T = 32.7 kN

Sum of forces in the +x direction:

∑F = ma

Fₓ − Tₓ = 0

Fₓ = Tₓ

Fₓ = 0.669 T

Fₓ = 21.9 kN

Sum of forces in the +y direction:

∑F = ma

Fᵧ − Tᵧ − 60 kN= 0

Fᵧ = Tᵧ + 60 kN

Fᵧ = 0.743 T + 60 kN

Fᵧ = 84.3 kN