A block-and-tackle pulley hoist is suspended in a warehouse by ropes of lengths 2 m and 3 m. the hoist weighs 430 n. the ropes,
fastened at different heights, make angles of 50° and 38° with the horizontal. find the tension in each rope and the magnitude of each tension. (let t2 and t3, represent the tension vectors corresponding to the ropes of length 2 m and 3 m respectively. round all numerical values to two decimal places.)
1 answer:
Refer to the diagram shown below.
For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂ (1)
For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392 (2)
Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N
Answer:
T₂ = 339.06 N
T₃ = 276.57 N
For horizontal equilibrium,
T₃ cos38 = T₂ cos 50
0.788 T₃ = 0.6428 T₂
T₃ = 0.8157 T₂ (1)
For vertical equilibrium,
T₂ sin 50 + T₃ sin 38 = 430
0.766 T₂ + 0.6157 T₃ = 430
1.2441 T₂ + T₃ = 698.392 (2)
Substitute (1) into (2).
(1.2441 + 0.8157) T₂ = 698.392
T₂ = 339.058 N
T₃ = 0.8157(399.058) = 276.571 N
Answer:
T₂ = 339.06 N
T₃ = 276.57 N
You might be interested in
Answer:
Part a)
Part b)
t = 12 s
Explanation:
Part a)
Tension in the rope at a distance x from the lower end is given as
so the speed of the wave at that position is given as
here we know that
now we have
Part b)
time taken by the wave to reach the top is given as
Answer:
b) 5 J
Explanation:
Work is the energy transferred by an object when acted by a force along a displacement. Work is the product of force and displacement. The SI unit of work is the joules (J)
To calculate the work done by the force, we have to first get the displacement (D) of the object. Hence:
Displacement (D) = Q(3, 8) - P(1, 3) = (3 - 1, 8 - 3) = (2, 5) = 2i + 5j
The work done is the dot product of the force and the displacement. Force = 5i - j. Hence:
Work done = (5i - j)(2i + 5j) = 10 - 5 = 5 J
