Calculating Tension in Multiple Cords for Suspended Object - 250N Weight

[aligass2004]

Homework Statement

http://i241.photobucket.com/albums/ff4/alg5045/image546.jpg

Find the tension in each cord in the figure if the weight of the suspended object is 250N.

Part A) Find the tension in the cord A for system (a)
Part B) Find the tension in the cord B for system (a)
Part C) Find the tension in the cord C for system (a)
Part D) Find the tension in the cord A for system (b)
Part E) Find the tension in the cord B for system (b)
Part F) Find the tension in the cord B for system (b)

Homework Equations

F=ma

The Attempt at a Solution

I found the tension in both Cs by drawing a free body diagram and setting F=mg = 250N for both.

I've also found the tension in cord A for system b. Since there are two unknowns, I solved the horizontal equation Ax=Bx for B, and then I substituted B into the vertical equation. So the end equation would be A(sin210sin45-cos210sin45) = 250cos45. I got A = 685N.

Then I tried substituting A into the Asin(thetaA) = Bcos(thetaB). I got B = 484N, which was wrong.

I've also tried using the same principle to find the tensions in A and B for system a, but I can't get it.

 

[Q_Goest]

Are you familiar with how to sum forces in the various directions? Let's make the Y direction verticle and the X direction horizontal. Do you know how to set up the equations? Note that the direction of the force each string acts in is along the axis of the string.

 

[aligass2004]

For system a, the horizontal equation is Ax = Bx or Acos(thetaA) = Bcos(thetaB), and the vertical equation is Ay + By = C or Asin(thetaA) + Bsin(thetab) = C. For system b, the horizontal equation is again Ax = Bx or Asin(thetaA) = Bcos(thetaB) because A is below the horizon in this case. The vertical equation By = Ay + C or Bsin(thetaB) = Acos(thetaA) + C.

 

[Q_Goest]

For system a, the horizontal equation is Ax = Bx or Acos(thetaA) = Bcos(thetaB), and the vertical equation is Ay + By = C or Asin(thetaA) + Bsin(thetab) = C.

Good! You now have 2 equations with 2 unknowns.
Acos(thetaA) = Bcos(thetaB)
or
Acos(thetaA) - Bcos(thetaB) = 0

and

Asin(thetaA) + Bsin(thetab) = C

The two unknows are A and B. Note that you have all the angles and you also have C which you indicated in the OP.

Now do you know how to solve the two equations for the 2 unknowns?

The second problem is basically the same.

 

[aligass2004]

Well I already found A for system b = 685N, which was correct. However, when I plugged 685 into the horizontal equation, I got B = 484N, which was incorrect. I've also tried solving for A for system a. For that I got 685N again, but it wasn't right. I know how to solve them. We've done about a million problems in class. I just can't get the right answer.

 

[Q_Goest]

I've also tried solving for A for system a. For that I got 685N again, but it wasn't right. I know how to solve them. We've done about a million problems in class. I just can't get the right answer.

Ok, had a look but I don't get 685 for A on problem a. I got 183 N. Try again and just write out each step. The equations you gave are correct for the first problem:
Acos(thetaA) - Bcos(thetaB) = 0
Asin(thetaA) + Bsin(thetaB) = C

 

[aligass2004]

I meant 685 for A for system b. My mistake.

 

[Q_Goest]

For b, try writing out the two equations as was done for a. The first should be:
A sin(60) = B cos(45)
or
A sin(60) - B cos(45) = 0

Can you get that?
What's the second equation?

 

[aligass2004]

The second equation would be B sin(45) = A cos(210) +250

 

[Q_Goest]

B sin(45) = A cos(210) +250

Close... the force produced by B is vertically up and the force created by A and the 250N load is down as you show, so you have it set up right. But... why A cos(210)? How about A cos(60)?

Think of it this way, cos(60) = Ay / A
(cos of an angle = adjacent/hypotaneus)

so

Ay (ie: verticle force created by string A) = A cos(60)

Now take the two equations and see if you can solve. Start with:

A sin(60) - B cos(45) = 0
and
- A cos(60) + B sin(45) = 250

 

[aligass2004]

Ok. I got everything for system b, and I thank you for your help with that. Now I need help with system a. Would the equation be to solve for A for system a:

A[sin(150)cos(45) + cos(150)sin(45)] = 250cos(45)

 

[Q_Goest]

A[sin(150)cos(45) + cos(150)sin(45)] = 250cos(45)

Where do the 150 degree angles come from?

 

[aligass2004]

I figured it out. I don't know why I keep using the wrong angles. That's what was screwing me up I think. Thanks for your help.