Calculating Tension in a Nonmassless Rope with Unequal Vertical Forces

  • Context: Undergrad 
  • Thread starter Thread starter agam goel
  • Start date Start date
  • Tags Tags
    Rope Tension
Click For Summary
SUMMARY

The discussion focuses on calculating the tension in a nonmassless rope subjected to unequal vertical forces. The method involves determining the net acceleration of the rope and analyzing one half of the rope using free body diagrams (FBD). For a rope with a mass of 1 kg, the weight is calculated as 10 N, leading to an acceleration of 10 m/s². By applying the equation Fnet = ma, the tension at the midpoint is determined to be 20 N, consistent across both halves of the rope.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with free body diagrams (FBD)
  • Basic knowledge of tension in ropes
  • Ability to perform algebraic calculations involving forces and acceleration
NEXT STEPS
  • Study advanced applications of Newton's laws in multi-body systems
  • Learn about the dynamics of nonmassless ropes in different configurations
  • Explore the effects of varying mass distributions on tension calculations
  • Investigate real-world applications of tension in engineering and physics
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in understanding the dynamics of tension in ropes under unequal forces.

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: 842

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
732
  • · Replies 20 ·
Replies
20
Views
890
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 40 ·
2
Replies
40
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K