Seemingly Simple Statics Problem

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In summary, a uniform rigid rod of mass M and length L is suspended by three massless strings, with two strings at either ends of the rod and the third string a length x from the left end. To find the tension in these three strings, the net force must be zero, resulting in the equation T1+T2+T3=Mg. Additionally, the net torque about the left end of the rod must also be zero, resulting in the equation T2*x+T3*L=Mg*L/2. However, these two equations are not sufficient to solve for the three tensions, as there is not enough information provided. Another equation can be obtained by considering the net torque about another point, such as the far right
  • #1
lugita15
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Homework Statement


A uniform rigid rod of mass M and length L is suspended by three massless strings, as shown in the following picture:
Tension Problem.JPG

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.

Homework Equations


F=ma
Torque=I(alpha)

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?
 
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  • #2
Looks like you've not been given enough information to solve the problem. :frown:
 
  • #3
Doc Al said:
Looks like you've not been given enough information to solve the problem. :frown:
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?
 
  • #4
lugita15 said:

Homework Statement


A uniform rigid rod of mass M and length L is suspended by three massless strings, as shown in the following picture:
View 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.

Homework Equations


F=ma
Torque=I(alpha)

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.
 
  • #5
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.
 
  • #6
lugita15 said:
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.

nrqed said:
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!
 
  • #7
SEG9585 said:
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.
 
  • #8
Doc Al said:
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.
 
  • #9
Doc Al said:
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!:frown:

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 [itex] T_3=Mg \frac{L/2-x}{L-x}} [/itex] and [itex] T_2=\frac{MgL}{2(L-x)}[/itex]up to a maximum value of Mg/2 (in which case [itex]T_2 =0, T_3 = Mg/2[/itex]).
 
Last edited:
  • #10
nrqed said:
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.
 

1. What is a "Seemingly Simple Statics Problem"?

A "Seemingly Simple Statics Problem" refers to a problem in the field of statics, which deals with the study of objects at rest or in constant motion. These problems may seem straightforward at first, but can often involve complex calculations and considerations.

2. How do you solve a "Seemingly Simple Statics Problem"?

The first step in solving a "Seemingly Simple Statics Problem" is to clearly define the problem and identify all the forces acting on the object. Then, using principles of equilibrium and Newton's laws of motion, a free body diagram can be drawn and equations can be set up to solve for unknown forces or variables.

3. What are some common mistakes made in solving "Seemingly Simple Statics Problems"?

One common mistake is forgetting to consider all forces acting on the object, including those caused by friction or tension in ropes or cables. Another mistake is using incorrect or incomplete equations, resulting in incorrect solutions.

4. Can "Seemingly Simple Statics Problems" be solved using computer software?

Yes, there are various computer programs and software that can assist in solving "Seemingly Simple Statics Problems". However, it is important for the user to have a thorough understanding of the principles and concepts involved in order to properly interpret and apply the results.

5. How are "Seemingly Simple Statics Problems" relevant in real-life applications?

"Seemingly Simple Statics Problems" are relevant in many real-life applications, such as in engineering and construction, where the stability and balance of structures and objects must be carefully considered. They are also important in understanding the mechanics of everyday objects and systems, such as bridges, buildings, and machines.

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