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3 balls connected by massless strings

by akorn3000
Tags: balls, connected, massless, strings
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akorn3000
#1
Mar26-12, 09:30 PM
P: 7
Suppose you have 3 balls, all of equal mass, M. They are connected to each other by equal-length (each length L) strings of negligible mass, such that one ball is suspended in the middle and the two balls at the end are themselves suspended from the ceiling at two points, a distance X from one another. Using M, X, and L, how would I go about finding the exact angles the balls make with the horizontal? I tried summing the forces and the moments, but I think I must be doing something wrong. Any help would be appreciated.
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K^2
#2
Mar26-12, 09:35 PM
Sci Advisor
P: 2,470
1) Use symmetry to assume that middle mass is equidistant from the other two. (Seems you've done that.)

2) Solve the problem for just one mass hanging from the other two as a function of distance between the later two. Call that distance y.

3) Solve for force between ceiling and the mass attached to it using the solution from above as input. You should treat y as one of the variables you are solving for.

Edit: Just realized that there is a much easier way. What can you say about horizontal component of T1 vs horizontal component of T2?
akorn3000
#3
Mar26-12, 10:34 PM
P: 7
Quote Quote by K^2 View Post
Edit: Just realized that there is a much easier way. What can you say about horizontal component of T1 vs horizontal component of T2?
They must be balanced, so T1*cos(theta1) = T2*cos(theta2)

akorn3000
#4
Mar26-12, 10:37 PM
P: 7
3 balls connected by massless strings

The way I see it, each string wants to reduce its tension by hanging straight down. The strings on top are pulled inward by the middle mass they have to support, and so they pull outward, while the strings on the bottom want to reduce their tension and pull inward.
K^2
#5
Mar26-12, 10:45 PM
Sci Advisor
P: 2,470
Quote Quote by akorn3000 View Post
They must be balanced, so T1*cos(theta1) = T2*cos(theta2)
And you know what the vertical components are. So given theta1, you should have no trouble computing theta2, right?
akorn3000
#6
Mar26-12, 10:59 PM
P: 7
I got as far as writing out the force diagrams for each mass: Taking the left mass for instance, balancing the forces in the y direction gives T2*sin(theta2) = mg + T1*sin(theta1), and for the bottom mass, 2*T1*sin(theta1) = mg. Substituting gives 3*T1*sin(theta1) = T2*sin(theta2), and if you divide that by T1*cos(theta1) = T2*cos(theta2) you get 3*tan(theta1) = tan(theta2). This relationship seems right at a cursory glance, since the slope of the upper strings must be steeper than the bottom ones, although I still do not know what the angles are independently and based on X, L, and M. It's quite possible that I'm missing something very obvious.

I also know that X = L*cos(theta1) + L*cos(theta2). I forgot to mention before but I do not want either of the tensions in the final answer, which seems to make this problem tricky for me.
K^2
#7
Mar26-12, 11:04 PM
Sci Advisor
P: 2,470
That's the relationship. So theta2=arctan(3*tan(theta1)). Not the most pleasant expression, but there you go. Now, can you express X in terms of theta1, theta2, and L?
akorn3000
#8
Mar26-12, 11:12 PM
P: 7
Quote Quote by K^2 View Post
That's the relationship. So theta2=arctan(3*tan(theta1)). Not the most pleasant expression, but there you go. Now, can you express X in terms of theta1, theta2, and L?
The simplest I can see is X = 2*L*[cos(theta1) + cos(theta2)] So I suppose if you substitute theta2 = arctan(3*tan(theta1)) you'd have (X/2L) = cos(theta1) + cos(arctan(3*tan(theta1))... yeesh.
K^2
#9
Mar26-12, 11:22 PM
Sci Advisor
P: 2,470
Yeah. And as far as I can tell, that's about as good as it gets. You can also build a 4th order polynomial equation for T_x = Tcos(theta), which obviously has a general solution, but it's a bit lengthy to say the least.

Point is, I don't think you can get a clean algebraic solution here. But given specific number for X/L, you can easily solve it numerically.
akorn3000
#10
Mar26-12, 11:22 PM
P: 7
And if you isolate theta1 instead, I suppose you would have theta1 = arctan((1/3)*tan(theta2)), so in that case you'd have (X/2L) = cos(arctan((1/3)*tan(theta2)) + cos(theta2). Heh, I guess nobody said the answer had to be pretty.
akorn3000
#11
Mar26-12, 11:23 PM
P: 7
Thank you for all your help, I appreciate it.


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