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?
- GFLv 66 years agoFavorite Answer
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
- electron1Lv 76 years ago
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.