Finding Tension in Suspended Spheres with Given Mass and String Length

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Homework Help Overview

The discussion revolves around finding the tension in strings supporting two suspended spheres, each with a mass of 15 kg. The spheres are in contact and suspended from a common point on the ceiling, with the problem involving geometric considerations and forces acting on the spheres.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the forces acting on the spheres, including gravity and tension, and question how the tension can exceed the weight of the spheres. They discuss the geometry of the setup and the implications of the angle formed by the strings.

Discussion Status

Participants are actively engaging with the problem, attempting to clarify the relationships between the forces and the geometry involved. Some have suggested drawing free body diagrams and checking the equations used to analyze the forces. There is a recognition that tension may be greater than the weight of the spheres, prompting further exploration of the mathematical relationships.

Contextual Notes

There is a mention of differing values for gravitational acceleration (g), with one participant using 10 m/s² and another referencing 9.8 m/s², which may affect the calculations. The discussion also highlights the assumption that the spheres are in contact and the implications of this on the forces at play.

iloveannaw
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This isn't really homework, just doing some revision.

Homework Statement


Find the tension in each string. Two spheres both of mass, m = 15kg. Diameter of each is 25cm. Each string 35cm long. Spheres are suspended on strings with other end attached to same point on ceiling. Therefore spheres are touching. Hopefully, that is not too difficult to understand, sorry no picture!

Taking g as 10 m/s/s

2. Homework Equations plus attempt
A vertical line drawn from point where string attached to ceiling makes a triangle with string + radius (hypotenuse) and radius of ball (opposite side).
hypotenuse = 0.35 + 0.125 = 0.475 meters
therefore angle alpha = arcsin(opp/ hyp) = arcsin(0.125 / 0.475)

Tension in each secondary string = T2 = mgcos(alpha) = 145 Newtons

However answer at back of book gives 152N - how can this be? g cannot be greater than 10! Really would like some help
 
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How many forces act on each sphere?
 
well, i'd say for each sphere gravity (mg), which pulls directly downward, causing a tension in the string, which acts along string at an angle alpha from the vertical. also the spheres are in contact, so there is a contact force in the x-direction.

still i don't understand. if g = 10 m/s/s, then weight of each sphere is 150N. how can the tension in each supporting string be greater than weight of the thing it supports. I would love an example.
 
You have an example. Draw a free body diagram for one of the spheres and put in all the forces. In what direction is the contact force? Say that the sum of all vertical components is zero, solve for the tension and see what you get.
 
thank you!

sum Fy = 0 = Tsin(a) - mg >>> Tsin(a) = mg

sum Fx = mgsin(a)cos(a)

T = sqrt(Fx^2 + Fy^2) = 152 , when g = 9.8

is that what you were hinting at?
 
Not exactly. Look at your first equation. It says

Tsin(a) = mg

Can you solve it to find T?
Given that the angle is not zero, is T greater than, equal to or less than mg?
 
T >mg but I've worked the angle, a, to be 15.3 deg.

T = mg/ sin(a) = 15*10 / sin(15.3) = 568N

this is far too large.
if length of string = 0.35m and radius of sphere = 0.125m, we have a triangle of:
opposite side = radius = 0.125m
hypotenuse = string + radius = 0.475m
therefore sin(a) = opp/hyp = 0.125/ 0.475
and angle a = arcsin(0.125/ 0.475) = 15.3 deg

but i do see how T can be greater than mg now (at least mathematically), otherwise it still seems like a rather weird idea!
 

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