Question

At a construction site, a helicopter is lifting two steel beams, each with a mass of 290 kg. One steel beam is attached directly to the helicopter by a cable. The second steel beam is not attached to the helicopter directly, but it is instead hanging directly below the first steal beam and attached to the first beam by another cable. Assume the cables have no mass.

(a) If the steel beams are at rest, what is the tension in each cord?

(lower cable) _______ N

(upper cable) _______ N

(b) If the steel beams are pulled upward with an acceleration of 1.70 m/s² calculate the tension is each cable.

(lower cable) ______ N

(upper cable) _______ N

Answer 1

a) when the steel beams are at rest, the only force acting on the cords are the weight of the beams.
In the lower cable, only one steel beam is exerting its weight on the cord.

Hence, using W=mg,

W = 9.81N/kg * 290 Kg = 2844.9 N acting on the lower cable.

In the upper cable, 2 steel beams are exerting its weight on the cord. Hence the tension experience in the upper cord is twice the tension experience in the lower cable, so

2844.9N * 2 = 5689.8 N.

b) In the second case, there will be more tension in the cords since the steel beams are accelerating upwards.

F(Resultant)=Total force used - the weight of the beams

Note: the Total force used will be the tension in the rope since it is exerted on the rope.

In the lower cord, we have one steel beam exerting its weight.
Using the formula, you get
(1.70m/s^-2)(290kg)=Total force used-(290kg)(9.81m/s^-2)
Total force used= 2844.9 + 493 = 3337.9
Hence the tension in the lower cord is 3337.9 N

As for the upper cord, we have 2 steel beam exerting its weight on it.
Using the formula again, you get
(1.70m/s^-2)(290kg)*2 = Total force used - 2*(290kg)(9.81m/s^-2)
Total force used = 5689.8 + 986 = 6675.8

Hence tension in upper cord = 6675.8N

Comment below if you have further doubt. Happy to help.