Calculating Tension in a Nonmassless Rope with Unequal Vertical Forces

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To calculate the tension in a nonmassless rope with unequal vertical forces, first determine the net acceleration by analyzing the forces acting on the rope. The tension acts internally within the rope, and when cut, it exerts an outward force at both ends. Using a mass of 1 kg for the rope, the weight is calculated as 10 N, which acts at the center of mass. By applying the equation Fnet = ma, the acceleration is found to be 10 m/s². Finally, analyzing either half of the rope leads to a tension of 20 N, confirming the consistency of the calculations.
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When 2 unequal vertical forces act on the 2 ends of a vertical rope, how do I calculate the tension at the midpoint?
First I calculated the net acceleration of the rope, then I choose any 1 half of the rope for fbd of it and then where and how does the tension act?
 
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The tension of the rope acts inside the rope. If you cut the rope, you will find the tension, acting outwards in either of the cut ends, as seen in the picture. Then, once you know that you can calculate the tension using anyone half of the ropes. In the example, let's say the rope has mass 1 kg. Then, the weight of the role is 1 (10) = 10 N, acting at the center of mass of the rope, that is halfway up or down, i.e. in the middle, not shown on the diagram. Then, for acceleration a, the equation is, Fnet = ma. ∴ 30-10-10(weight of the rope) = 1 (a). This gives a = 10ms-2. Now, taking the upper half, which has half the mass, and again applying the equation, we have 30-T-5 (weight of the upper half now acting downwards at the center of the half) = 0.5 (10). Thus, T = 20 N. You will arrive at the same answer, if you work on the bottom half of the rope. Hope it helped. :smile:
 

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