Two Unknowns 1 Equation

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Homework Statement:

For the systems in equilibrium in the figure below, find the unknown tensions and masses

Relevant Equations:

ΣF = ma = 0
Here are two questions for my online homework. In part (a) we have to find T₁, T₂, and m.
For part a, I drew a free body diagram and used that ΣFₓ = -30N +T₁cos(60) = 0 To find T₁ = 60N. The software accepted my answer. Now we have to find T₂ and m. We have ΣFᵧ = T₂ + T₁sin(60) + mg = 0. We do not know T₂ and we do not know m. So how can we solve for both?


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  • #2
haruspex
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You need to be clear about what your force balance equation represents. ΣFy on what, exactly? Since you have the three tensions in there, it must be the force balance on the knot. The knot doesn't "know" anything about m - that's somewhere else.
To get mg into it you need to consider the force balance on the vertical string.
 
  • #3
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Homework Statement:: For the systems in equilibrium in the figure below, find the unknown tensions and masses
Relevant Equations:: ΣF = ma = 0

Here are two questions for my online homework. In part (a) we have to find T₁, T₂, and m.
For part a, I drew a free body diagram and used that ΣFₓ = -30N +T₁cos(60) = 0 To find T₁ = 60N. The software accepted my answer. Now we have to find T₂ and m. We have ΣFᵧ = T₂ + T₁sin(60) + mg = 0. We do not know T₂ and we do not know m. So how can we solve for both?


View attachment 257619
mg should not be in the vertical force balance on the junction.
 
  • #4
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mg should not be in the vertical force balance on the junction.
I didnt know we had to consider the junction seperately from the mass. Now I can probably do it myself. Thank you
 
  • #5
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You need to be clear about what your force balance equation represents. ΣFy on what, exactly? Since you have the three tensions in there, it must be the force balance on the knot. The knot doesn't "know" anything about m - that's somewhere else.
To get mg into it you need to consider the force balance on the vertical string.
Thank you. I now know to consider the knot separately from the hanging mass.
 
  • #6
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I didnt know we had to consider the junction seperately from the mass. Now I can probably do it myself. Thank you
The question is, do you know why?
 
  • #7
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The question is, do you know why?
I think I do. It is because all of the tensions are actually acting on the junction, not the mass itselfwhile only one of those tensions are acting on the mass. Am I right?
 
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  • #8
haruspex
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I think I do. It is because all of the tensions are actually acting on the junction, not the mass itselfwhile only one of those tensions are acting on the mass. Am I right?
Yes.
 
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