Tension Forces at Equilibrium Problem

[delecticious]

Homework Statement

Three strings are tied together at one end. The other ends are attached to a rectangular frame as shown below:

The tensions in the strings are labeled Tx, Ty, and Tz. The tension in the vertical string is Tx=47.0N. Angle A= 58.5 deg, and angle B= 30.5 deg. Calculate the magnitude of the tension Tz.

Homework Equations

Tz + Ty + Tx= 0

my TA gave me this one: Ty/sin(A+B) = Tz/sin(90-A)

The Attempt at a Solution

I tried a number of things, thought maybe the tension in Tx would carry through in the y-direction so I could use that as the y-component to make a triangle with Ty, so I could find Ty and use that in the equation my TA gave to me, but that didn't get me the right answer and I got it wrong on the homework assignment. Can anyone help me on finding Ty or give me some sort of step by step process to follow to get the right answer?

 

[Doc Al]

Since you have equilibrium, the sum of the forces must be zero. Write two equations, one for the sum of the vertical components, another for the sum of the horizontal components. Since you are given Tx, you can solve those equations together for the other tensions.

 

[delecticious]

so for Tx, it only has a y-component right? So the sum of the x-components should only include those of Ty and Tz right? What I'm understanding is that the sum of the vector in both the x and y directions should equal zero. So Ty + Tz = 0 in the x direction, so then does Ty + Tz= -Tx in the y direction? Or is that wrong? I'm try to see a way of using the magnitude of Tx to eventually find Tz.

 

[Doc Al]

So Ty + Tz = 0 in the x direction, so then does Ty + Tz= -Tx in the y direction?

Perfectly correct. Now rewrite those two equations while specifying the x- or y-components of the forces as needed. (Use a little trig to find the x & y components of Ty and Tz.)

 

[delecticious]

Since you have equilibrium, the sum of the forces must be zero. Write two equations, one for the sum of the vertical components, another for the sum of the horizontal components. Since you are given Tx, you can solve those equations together for the other tensions.

I think that's the part I'm having trouble with. Just how am I supposed to put the two equations together? I can't see them equal to each other right? And I can't solve for them individually without having 2 unknowns, so what am I supposed to do?

 

[Doc Al]

You have two equations and two unknowns. One way to solve such equations is to use one equation to solve for Ty in terms of Tz (for example), then plug that into the second equation. Then you'll have one equation and one unknown.

First thing to do is to actually write the correct equations using sines and cosines of the given angles to represent the components properly.

 

[TVP45]

It might help if you rename the tensions to T1, T2, T3. Otherwise you have the x component of Ty and so on. Potentially confusing.

 

[fubag]

try:

x = Tucos58.5 - Tzcos30.5 = 0

and in the y direction:

y = Tusin58.5 + Tzsin30.5 + Tx = 0

Now use system of equations, and solve for Tz.

 

[delecticious]

try:

x = Tucos58.5 - Tzcos30.5 = 0

and in the y direction:

y = Tusin58.5 + Tzsin30.5 + Tx = 0

Now use system of equations, and solve for Tz.

when you say "u" do you mean "x" and "y" respectively?

 

[fubag]

no sorry I miss read the diagram i mean it has it labeled as Ty is the one of the tensions rope,

and Tz is the other tensions rope..

so

use the two equations to solve to like Tu in terms of Tz and solve for Tz.