Question

A uniform rope of cross-sectional area 0.95 cm² breaks when the tensile stress in it reaches 5.50 ? 10⁶ N/m².

(a)

What is the maximum load (in N) that can be lifted slowly at a constant speed by the rope?

___________ N

(b)

What is the maximum load (in N) that can be lifted by the rope with an acceleration of 5.00 m/s²?

___________ N

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Cross sectional area of the rope = A = 0.95 cm² = 0.95 x 10^-4 m²

Maximum tensile stress of the rope = = 5.5 x 10⁶ N/m²

Tension in the rope = T

T = A

T = (5.5x10⁶)(0.95x10^-4)

T = 522.5 N

Maximum load that can be lifted slowly at constant speed = W_a

The load is being lifted at constant speed therefore there is no acceleration, hence the forces are balanced.

T = W_a

W_a = 522.5 N

Maximum load that can be lifted with an acceleration of 5 m/s² = W_b

Acceleration of the load = a = 5 m/s²

Mass of the load = W_b/g

(W_b/g)a = T - W_b

(W_b/9.81)(5) = 522.5 - W_b

1.51W_b = 522.5

W_b = 346.03 N

a) Maximum load that can be lifted slowly at a constant speed by the rope = 522.5 N

b) Maximum load that can be lifted by the rope with an acceleration of 5 m/s² = 346.03 N