Find the Maximum Mass for Riding a Swing

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Homework Help Overview

The problem involves a swing made from a rope with a maximum tension limit, where the swing is initially pulled back and released. The objective is to determine the maximum mass of a person who can ride the swing without exceeding the tension limit of the rope.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the dynamics of the swing, including centripetal force and tension at different points in the swing's motion. There is mention of using conservation of energy to relate potential and kinetic energy. Some participants question the point at which tension is maximized and suggest drawing free body diagrams to clarify forces involved.

Discussion Status

There is an ongoing exploration of different approaches to the problem, including energy methods and force analysis. Some participants have made progress in their calculations and are seeking validation of their results, while others are still clarifying concepts and relationships between forces.

Contextual Notes

Participants note the importance of understanding the conditions under which tension is maximized and the relevance of free body diagrams in analyzing the forces acting on the swing. There is also a reference to the specific angle at which the swing is released and its implications for the calculations.

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Homework Statement



A swing is made from a rope that will tolerate a maximum
tension of 800N without breaking. Initially, the swing
hangs vertically. The swing is then pulled back at an
angle of 60.0 degree with respect to the vertical and
released from rest. What is the mass of the heaviest
person who can ride the swing?



Homework Equations


Fc = mv^2/r
trig ratios


The Attempt at a Solution



I initially just thought of doing 800N / 9.81m/s^2 which give 81.5kg or so but i thought that was too easy and can't be right so I did some more complex stuff:

using unit circle:
r = sin60r = 0.5r


Fc = mv^2/r
I forgot how but i somehow ended up in the step below, where 0.5r cancels out and in the end and the end result goes back to m = 800/9.81
800 = (9.81m(0.5r))/0.5r

still pretty sure that is wrong and I am really lost, so help please o;
 
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The swing is basically, undergoing motion in a vertical circle. So for this to happen, at every instant the swing needs a radially inward component of the net force(centripetal) which is equal to mv^2/R.

Now it is crucial in this problem that you understand, the tension in the string will be maximum, when it is at, the lowest point in it's motion, I.e when weight and tension are anti parallel.

So at this instant,

T - mg= mv^2/RNow you want to know the maximum mass of the person, so put the maximum value in for the tension.

Use conservation of energy to find v^2 and solve the above eqn for m
 
At what point after release will the tension in the rope be the greatest? Once you decide on that, then draw a free body diagram of the weight on the swing noting all forces. Relate those forces to the tension in the rope. From the sound of the problem statement, I assume this is a one rope swing.
 
The problem can be worked from an energy standpoint or from the acceleration of a mass as it is affected by gravity. I don't know which subject you are studying.
 
Ok so I made some progress from all the help you guys give and now I am at :

T - mg = mv^2/r

mgh = 1/2mv^2
v^2 = 2gh

and since i figured h was 0.5r so:

v^2 = 9.81r

replacing that in initial yields:
800 - 9.81m = 9.81rm/r
800 = 19.62m
m = 40.8kg

DID I DO IT RIGHT?
 
Looks good to me.
 

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