Calculating Tension in a Nonmassless Rope with Unequal Vertical Forces

  • Thread starter Thread starter agam goel
  • Start date Start date
  • Tags Tags
    Rope Tension
AI Thread Summary
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.
agam goel
Messages
1
Reaction score
0
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?
 
Physics news on Phys.org
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:
 

Attachments

  • Rope.png
    Rope.png
    6.8 KB · Views: 827
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...

Similar threads

Back
Top