Solving Newton's Law Problems: Tension & Mass

[prace]

Hello, I have a question with a simple Tension - Mass problem. Here is the question:

For the systems in Equilibrium, find the unknown tensions and masses.
a) http://answerboard.cramster.com/answer-board/image/ed663c5fcb01e8fb6c7522cf963b49f1.jpg

b) http://answerboard.cramster.com/answer-board/image/dc409314d85cd577ab9279e81977fd75.jpg

c) http://answerboard.cramster.com/answer-board/image/d24349d36430cfb757bde12ffdbca0af.jpg

For a), I found T1 = 60N, but I can't seem to find T2 or the mass.

For c), I have found T1 = T3, but that is as far as I have got.

Any help on the other parts?

Thanks!

 

[radou]

Drawing free body diagrams should help.

 

[prace]

Yeah, I drew the free body diagrams, and that is how I figured out the parts that I did, but I am still stuck on the others.

 

[radou]

Yeah, I drew the free body diagrams, and that is how I figured out the parts that I did, but I am still stuck on the others.

Well, let's start with a). Draw a free body diagram for the mass, since I assume you didn't do that, otherwise the problem would be solved.

 

[prace]

Well, let's start with a). Draw a free body diagram for the mass, since I assume you didn't do that, otherwise the problem would be solved.

Actually, I did draw the FBD. It is just a little hard to show on the forum so I did not show it. But just so you believe me, here it is, or what I think it should be.

http://answerboard.cramster.com/answer-board/image/f82c0c65fe75d2ff30bd3844a0167c55.jpg

Here is my work:

For the x direction:

T1cos(\theta)-Fx = 0

0 because the problem states that the hanging mass is in a state of equilibrium.

So then this givesme T1cos(\theta) = F
or
T1 = F/(T1cos(60°)) where F = 30N

This gives me, as I stated in my original post, T1 = 60N.

So I then examined the y direction:

T1sin(\theta)-T2-Fg = ma, where again, ma = 0 because the mass is at its equilibrium position and there is no acceleration.

This then gives me 60Nsin(60°)+T2 = mg. So now I am left with 1 equation and 2 unknowns. This is where my question is coming from.

Thank you.

 

[PhanthomJay]

Actually, I did draw the FBD. It is just a little hard to show on the forum so I did not show it. But just so you believe me, here it is, or what I think it should be.

http://answerboard.cramster.com/answer-board/image/f82c0c65fe75d2ff30bd3844a0167c55.jpg

Here is my work:

For the x direction:

T1cos(\theta)-Fx = 0

0 because the problem states that the hanging mass is in a state of equilibrium.

So then this givesme T1cos(\theta) = F
or
T1 = F/(T1cos(60°)) where F = 30N

This gives me, as I stated in my original post, T1 = 60N.

So I then examined the y direction:

T1sin(\theta)-T2-Fg = ma, where again, ma = 0 because the mass is at its equilibrium position and there is no acceleration.

This then gives me 60Nsin(60°)+T2 = mg. So now I am left with 1 equation and 2 unknowns. This is where my question is coming from.

Thank you.

You've got an extra unknown in your FBD, which is in error. When you take the FBD at the joint, you have T1sin60 =T2. Solve for T2. Then take the FBD of the mass, to get T2 = mg, solve for m. Alternatively, when you look at the FBD of the system, the tension force T2 is internal, so it doesn't enter into the equation. You just have T1sin60 =mg.

In part b, you left out the magnitde of the tension force T3, otherwise, the problem is numerically unsolvable.

In part C, the tension forces around the pulley are equal in magnitude.

 

[prace]

Oh ok, I see. One question, in problem a) why does T1sin60 = T2? Does this always hold true?

 

[radou]

Oh ok, I see. One question, in problem a) why does T1sin60 = T2? Does this always hold true?

It is just the equation of equilibrium for the y-direction. You can imagine as if you 'cut out' the node at which the strings meet and anaylsed it by itself, if that makes things easier.