Equilibrium of Hanging Bag of Cement: Tension Forces Question Solution

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Homework Help Overview

The problem involves analyzing the equilibrium of a bag of cement suspended from three wires, with two wires making specific angles with the horizontal. The original poster presents equations related to the forces acting on the system and attempts to derive expressions for the tensions in the wires.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to derive the tension equations based on equilibrium conditions, questioning the arrangement of terms and the presence of cosine factors. Participants explore the implications of negative tension values and seek clarification on the derivation of specific terms in the equations.

Discussion Status

Participants are actively engaging with the equations, providing feedback on each other's reasoning and calculations. There is a focus on clarifying misunderstandings regarding the algebraic manipulation of the tension equations. Some guidance has been offered regarding the correct arrangement of terms, but no consensus has been reached on the final expressions for the tensions.

Contextual Notes

Participants are working under the constraints of the problem statement, including specific angles and the weight of the bag. There is an ongoing discussion about the assumptions made regarding the direction of forces and the arrangement of the wires.

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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.
 
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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.
 
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?
 
myelevatorbeat said:
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?
 
Here's a diagram:

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

Sorry about that.
 
myelevatorbeat said:
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.
 
So, should I have:

T1sinx1cosx2+T1cosx1sinx2cosx2=Fgcosx2?

I still don't understand what happens to the second T1
 
myelevatorbeat said:
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).
 
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
myelevatorbeat said:
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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