The steel I-beam in the drawing has a weight of 8-13 103 N and is bei.docx

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The steel I-beam has a weight of 8.13 kN that is being lifted at a constant velocity by two cables attached at its ends. As the situation is symmetrical, each cable must provide half the lifting force. The tension in each cable is calculated to be 4.065 kN by dividing the weight between the two cables. Some of the tension is used to apply compression along the beam rather than lift it, but this effect is neglected. Using trigonometry, the lifting component of the tension is calculated to be 4.33 kN, so the total tension in each cable is 4.33 kN.

$The steel I-beam in the drawing has a weight of 8.13 Ã— 103 N and is being lifted at a constant velocity. What is the tension in each cable attached to its ends? Units Solution First, each cable is identical, and we're probably to assume that the I-beam is of constant density all along its length, so that if you flip the I-beam around, the situation is exactly the same. That means that whatever the tension we find in one cable, the tension in the other is the same. That means that 8.13 kN /2 = 4.065 kN is held by each cable. The cables aren't hanging straight down. This means that some fraction of the tension doesn't go into lifting the I-beam, but (in this case) into applying a compressional force along the I- beam. This can be neglected (presumably the beam isn't so weak that it's going to crumple). So, what is the component of the tension that actually lifts the beam? It's T*sin(70deg) ~ 0.94T. Each cable provides this lift, so we have 2*0.94*T = 8.13 kN ==> T = 4.33 kN = 4.33 x 10^3 N$

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1. The steel I-beam in the drawing has a weight of 8.13 Ã— 103 N and is being lifted at a constant velocity. What is the tension in each cable attached to its ends? Units Solution First, each cable is identical, and we're probably to assume that the I-beam is of constant density all along its length, so that if you flip the I-beam around, the situation is exactly the same. That means that whatever the tension we find in one cable, the tension in the other is the same. That means that 8.13 kN /2 = 4.065 kN is held by each cable. The cables aren't hanging straight down. This means that some fraction of the tension doesn't go into lifting the I-beam, but (in this case) into applying a compressional force along the I- beam. This can be neglected (presumably the beam isn't so weak that it's going to crumple). So, what is the component of the tension that actually lifts the beam? It's T*sin(70deg) ~ 0.94T. Each cable provides this lift, so we have 2*0.94*T = 8.13 kN ==> T = 4.33 kN = 4.33 x 10^3 N