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
Answer:0.669
Explanation:
Given
mass of clock 93 kg
Initial force required to move it 610 N
After clock sets in motion it requires a force of 514 N to keep moving it with a constant velocity
Initially static friction is acting which is more than kinetic friction
thus 613 force is required to overcome static friction