Tension Forces Question

  • #1
myelevatorbeat
55
0

Homework Statement


A bag of cement hangs from three wires as shown. Two of the wires make angles theta1 and theta2, respectively, with the horizontal.
a)Show that, if the system is in equilibrium, then:
T1=Fgcostheta2/sin(theta1+theta2)

b) Given that Fg=325 N, theta1=10.0degrees, and theta2=25.0degrees, find the tensions T1, T2, and T3 in the wires.



The Attempt at a Solution



Part A:

(Fnet)x=max (a=0)
T2cosx2+T1cosx1=0
T2=-T1cosx1/cosx2

(Fnet)y=may (a=0)
T1sinx1+T2sinx2-Fg=0
I substituted in the equation I got for T2 which gave me:

T1sinx1+T1cosx1sinx2/cosx2-Fg=0

I rearranged the above equation to solve for T1 and got:

Fgcostheta1/sinx1cosx2+sinx2

I assume I need a cosx1 in the denominator so it can be changed to sin (x1+x2) but I'm not sure where I get the cosx1 from? Did I skip it in one of the equations?

Now for Part B:

Fg=325 N (cause it's the weight of the cement bag)
x1=10.0 degrees
x2=25.0 degrees

I plugged them into the above equations that I had found:

T1=Fgcosx2/sin(x1+x2)
T1=325 N x cos 25.0 degrees/sin (10.0degrees + 25.0degrees)
T1=514 N

T2= -T1cosx1/cosx2
T2= -514 x cos10.0degrees/cos25.0degrees
T2=-558

I don't understand why tension 2 would be negative though. I would asume they would all be upwards vertical forces. Maybe I solved one of the equations wrong?

I feel like I should just drop the negative sign, but I don't know if that would be correct.
 

Answers and Replies

  • #2
myelevatorbeat
55
0
Ok, I think I see what I've done wrong already. Perhaps T1cosx2 should be negative to begin with and then when I rearrange the equation it will become positive?

I still don't see where the second cosx2 came from though.
 
  • #3
myelevatorbeat
55
0
Maybe more details will help?

I went back over my equation and got the same thing:

T1sinx1+T2sinx2-Fg=0
Substituting in the first equation:
T1sinx1+(T1cosx1/cosx2)sinx2-Fg=0
So now I bring over my numbers:

T1sinx1+T1cosx1/cosx2)sinx2=Fg
Then I divide the cosx2 in the denominator to bring it to the other side:

T1sinx1+T1cosx1sinx2=Fgcosx2

Then I divide over the sinx1, cosx1, and sinx2, which leaves me with:

T1+T1=wcosx2/sinx1+cosx1sinx2

How do I get rid of the second T1 and where doest he cosx2 in the denominator come from? Did I make a mistake in my calculations?
 
  • #4
Doc Al
Mentor
45,447
1,907
A bag of cement hangs from three wires as shown. Two of the wires make angles theta1 and theta2, respectively, with the horizontal.
Can you give a diagram? I don't understand how the wires are arranged. What's the angle of the third wire?
 
  • #5
myelevatorbeat
55
0
Here's a diagram:

http://a373.ac-images.myspacecdn.com/images01/113/l_db80f0d296b290d3e3b1ef076f5fe74c.jpg

Sorry about that.
 
  • #6
Doc Al
Mentor
45,447
1,907
T1sinx1+T2sinx2-Fg=0
Substituting in the first equation:
T1sinx1+(T1cosx1/cosx2)sinx2-Fg=0
OK.
T1sinx1+T1cosx1/cosx2)sinx2=Fg
OK.
Then I divide the cosx2 in the denominator to bring it to the other side:

T1sinx1+T1cosx1sinx2=Fgcosx2
Error: You must multiply all terms by cosx2.
 
  • #7
myelevatorbeat
55
0
So, should I have:

T1sinx1cosx2+T1cosx1sinx2cosx2=Fgcosx2?

I still don't understand what happens to the second T1
 
  • #8
Doc Al
Mentor
45,447
1,907
So, should I have:

T1sinx1cosx2+T1cosx1sinx2cosx2=Fgcosx2?
Almost, but you have an extra cosx2 factor in the second term.

I still don't understand what happens to the second T1
Not sure what you mean by "second" T1. Realize that: A*X + A*Y = A*(X + Y).
 
  • #9
myelevatorbeat
55
0
Let me try to explain where I'm not getting it:

I understand now where the second cosx2 came from, but I wind up with this equation:

T1+T1=Fgcosx2/sinx1cosx1sinx2cosx2

Doesn't that mean it's really
2(T1)=Fgcosx2/sinx1cosx1sinx2cosx2

I want to know how to get just:

T1=Fgcosx2/sinx1cosx1sinx2cosx2
 
  • #10
learningphysics
Homework Helper
4,099
6
So, should I have:

T1sinx1cosx2+T1cosx1sinx2cosx2=Fgcosx2?

I still don't understand what happens to the second T1

This should be T1sinx1cosx2 + T1cosx1sinx2 = Fgcosx2

So T1(sinx1cosx2 + cosx1sinx2) = Fgcosx2

Divide both sides by (sinx1cosx2+cosx1sinx2)

T1 = Fgcosx2/(sinx1cosx2+cosx1sinx2).

there isn't a factor of 2.
 

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