# Seemingly Simple Statics Problem

by lugita15
Tags: seemingly, simple, statics
 P: 1,583 1. The problem statement, all variables and given/known data A uniform rigid rod of mass M and length L is suspended by three massless strings, as shown in the following picture: Two of the strings are at either ends of the rod. The third string is a length x from the left end. Find the tension in these three strings. 2. Relevant equations F=ma Torque=I(alpha) 3. The attempt at a solution Label the three tensions from left to right as T1, T2, and T3. Since the net force is 0, T1+T2+T3=Mg Since the net torque about the left end of the rod is 0, T2*x+T3*L=Mg*L/2 I'm stuck from here; I need another equation in order to solve for the three tensions. I can't take torque about any other point because that will just give me an equivalent equation. What do I do?
 Mentor P: 41,579 Looks like you've not been given enough information to solve the problem.
P: 1,583
 Quote by Doc Al Looks like you've not been given enough information to solve the problem.
There certainly seems to be enough information from a physical standpoint. What else could the tensions possibly depend on other than M, L, and x?

HW Helper
P: 3,025
Seemingly Simple Statics Problem

 Quote by lugita15 1. The problem statement, all variables and given/known data A uniform rigid rod of mass M and length L is suspended by three massless strings, as shown in the following picture: Attachment 9960 Two of the strings are at either ends of the rod. The third string is a length x from the left end. Find the tension in these three strings. 2. Relevant equations F=ma Torque=I(alpha) 3. The attempt at a solution Label the three tensions from left to right as T1, T2, and T3. Since the net force is 0, T1+T2+T3=Mg Since the net torque about the left end of the rod is 0, T2*x+T3*L=Mg*L/2 I'm stuck from here; I need another equation in order to solve for the three tensions. I can't take torque about any other point because that will just give me an equivalent equation. What do I do?
No, it won't give you an equivalent equation! You can get a third, independent equation by considering another point to calculate the net torque. For example, use the point at the far right for your axis and impose that the net torque with respect to that point is zero.
 P: 34 you have 1 equation relating T1 T2 AND T3 you have a second equation relating T2 to T3, so you can substitute into the first for only 2 unknowns. Now you can get torque around another point to relate T1 to T3, or T1 to T2, and you should be all set. I recommend getting the torque around the center of mass, then substituting T3 with T2 for this and the first equation. Then you will have 2 equations and 2 unknowns and it's easy.
Mentor
P: 41,579
 Quote by lugita15 There certainly seems to be enough information from a physical standpoint. What else could the tensions possibly depend on other than M, L, and x?
To see that you don't have enough information, consider this: Imagine that you only had the end strings and thus T1 and T3, but no T2. In that case, you can easily figure out the needed T1 and T3 to support the rod. Now add the middle string, but don't pull it too taut; it removes some of the load from T1 and T3, and now T2 is nonzero. Now pull the middle string a bit more firmly; T1 and T3 are reduced even more while T2 increases. This gives three perfectly valid solutions, all different! You need an additional constraint to fix the values of all three.

 Quote by nrqed No, it won't give you an equivalent equation! You can get a third, independent equation by considering another point to calculate the net torque. For example, use the point at the far right for your axis and impose that the net torque with respect to that point is zero.
Try it and see!
P: 1,583
 Quote by SEG9585 Now you can get torque around another point to relate T1 to T3, or T1 to T2, and you should be all set.
No you won't. If you get a torque equation about another point, and then you try solving the equations, you'll find something along the lines of T1=T1, and that obviously gets you nowhere. Try taking the torques about another point, like the center of mass, like you suggested. I already tried it and it didn't work.
P: 1,583
 Quote by Doc Al To see that you don't have enough information, consider this: Imagine that you only had the end strings and thus T1 and T3, but no T2. In that case, you can easily figure out the needed T1 and T3 to support the rod. Now add the middle string, but don't pull it too taut; it removes some of the load from T1 and T3, and now T2 is nonzero. Now pull the middle string a bit more firmly; T1 and T3 are reduced even more while T2 increases. This gives three perfectly valid solutions, all different! You need an additional constraint to fix the values of all three. Try it and see!
I'm not sure you how you can pull the middle string "a bit more firmly." It is pulled naturally when you attach one end to the rod and the other end to the ceiling (not drawn). I think that we should put one "additional constraint," although I really don't think we'll need it: assume that all three strings are of equal length y>>L.
HW Helper
P: 3,025
 Quote by Doc Al To see that you don't have enough information, consider this: Imagine that you only had the end strings and thus T1 and T3, but no T2. In that case, you can easily figure out the needed T1 and T3 to support the rod. Now add the middle string, but don't pull it too taut; it removes some of the load from T1 and T3, and now T2 is nonzero. Now pull the middle string a bit more firmly; T1 and T3 are reduced even more while T2 increases. This gives three perfectly valid solutions, all different! You need an additional constraint to fix the values of all three. Try it and see!
You are absolutely right. My mistake and my apologies!

It's indeed obvious that there is no unique solution, As Doc Al said. There is a solution with only two strings. Therefore, there is an infinite number of solutions with a third string.

For example, the possible values of T_1 range from a minimum value of 0 (in which case $T_3=Mg \frac{L/2-x}{L-x}}$ and $T_2=\frac{MgL}{2(L-x)}$up to a maximum value of Mg/2 (in which case $T_2 =0, T_3 = Mg/2$).
Mentor
P: 41,579
 Quote by nrqed It's indeed obvious that there is no unique solution, As Doc Al said. There is a solution with only two strings. Therefore, there is an infinite number of solutions with a third string.
Right. This is an example of a statically indeterminate system.

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