# Physics Tension Question?

Can someone help me with this question? The steel I-beam in the drawing has a weight of 8.76 × 103 N and is being lifted at a constant velocity. What is the tension in each cable attached to its ends? Relevance

Since it is being lifted at a constant velocity, the force acting on the I-beam is that of Gravity (and of course the two cables).

Note that we expect the tension on the two cables to be equal.

Consider ALL force in the y-direction only:

Weight of beam = vertical force on LHS cable + vertical force on RHS cable

eqn 1: 8.76 * 10^3 = Flhs * sin(70) + Frhs * sin (70), where Flhs and Frhs are tension in the cable respectively

Consider ALL forces in the x-direction only:

eqn 2: 0 = Flhs * cos(70) - Frhs * cos(70), (i take +'ve force in the positive x-direction, ie from left to right, that's why I subtracted the Frhs x component because the force is going from rhs to lhs or in the -ve x-direction)

from eqn 2: Flhs = Frhs substitute this result in eqn 1 and solve.

I hope this helps

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• Since the angle is the same for each cable, the magnitude of tension is the same for each cable. Since the I beam is being lifted at a constant velocity, the net vertical force on the I beam is 0 N. This means the sum of the vertical components of the tensions is equal to the weight of the beam.

Vertical component = T * sin 70

2 * T * sin 70 = 8.76 * 10^3

T = 8.76 * 10^3 ÷ 140

The tension is approximately 62.57 N.

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