Finding Tension force in a SHM Pendulum

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Homework Help Overview

The problem involves calculating the tension force in a swing at a specific angle during simple harmonic motion (SHM). The scenario includes a panda of mass 110 kg on a swing of length 3.0 m, reaching a maximum height at a 20° angle with the vertical.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the use of force diagrams and the relationship between tension and gravitational force. There is a focus on the components of forces acting on the panda and the correct mathematical manipulation of these forces.

Discussion Status

Some participants have pointed out potential errors in the original poster's reasoning regarding the manipulation of the force equations. Guidance has been offered on considering different axes for analyzing the forces, suggesting a more nuanced approach to the problem.

Contextual Notes

Participants are discussing the assumptions made about vertical acceleration and the implications for the force equations. There are also notes on the formatting of mathematical expressions within the forum.

JessicaHelena
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Homework Statement


A 110 kg panda is riding on a 3.0 m long swing whose mass can be considered negligible. The highest point of its arc occurs when the swing makes a 20° angle with the vertical. What the magnitude of the total tension in the ropes of the swing at that point?

m (mass of panda) = 110 kg
##\theta = 20 ^\circ##
L = 3.0 meters

Homework Equations



Force diagrams (?) — I am not really sure if any specific equation is needed here.

The Attempt at a Solution



Drawing a force diagram of the panda at that point, there is a tension force ##F_t## pointing towards the upper left along the rope, and a gravitational force ##F_g = mg## that points directly downwards. These forces do not balance each other out horizontally because of the horizontal component of ##F_t## but the ##F_{t(y)}## and ##F_g## do. Then we have ##F_t \cos \theta = F_g##. Rearranging, we have ##F_t = F_g / \cos \theta##. Plugging in the right values, ##F_t = (110)(9.8)/(\cos 20°)##. This gives approximately 1147.18 N.
 
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For some reason, you divided by cosθ instead of multiplying.
 
@ Doc Al
Then is my derivation wrong? From $F_t \cos \theta = F_g$, to isolate F_t I think the only way is to divide by $\cos \theta$...
 
JessicaHelena said:
@ Doc Al
Then is my derivation wrong? From $F_t \cos \theta = F_g$, to isolate F_t I think the only way is to divide by $\cos \theta$...
Yes, your derivation is wrong.

Try this: Consider axes parallel and perpendicular to the rope. In particular, consider force components parallel to the rope. What must they add to? (What's the acceleration that direction?) Set up a force equation and you can solve for the tension.
 
Doc Al has told you the right way to analyse this, but I will point out what went wrong with your way:
JessicaHelena said:
##F_t \cos \theta = F_g##
If you are summing forces in the vertical direction then the resultant should correspond to any vertical acceleration. You seem to be assuming that is zero.

Re the LaTeX, in these forums you need to bracket it with a double hash (#) or double dollar. If you use the double dollar it will force a newline before and after.
 

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