Finding Tension force in a SHM Pendulum

AI Thread Summary
The discussion focuses on calculating the tension force in a pendulum swing with a panda weighing 110 kg at a 20° angle from the vertical. The initial approach involved using the equation F_t cos(θ) = F_g, but this was identified as incorrect due to a misunderstanding of force components and acceleration. The correct method involves analyzing forces parallel and perpendicular to the rope to establish a proper equation for tension. The user is advised to reconsider their assumptions about vertical acceleration when summing forces. The conversation emphasizes the importance of accurate force diagrams and component analysis in solving physics problems.
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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