Solving Tension Bag Problem: Find T1

[ek]

I'm having trouble with this problem...

A bag of cement of weight Fg hangs from three wires as shown in the attached diagram. Two of the wires make angles theta1 and theta2 with the horizontal. If the system were in equilibrium, show that the tension in the left hand wire is:

T1 = Fgcostheta2/sin(theta 1 + theta2)

Ok, I get that T3 = Fg. I drew a triangle of all the tension forces but that didn't help. Just seems like an odd type of question.

Any help, clues, or suggestions?

 

[Leong]

consider the knot where all the wires meet as a particle. resolve all the tensions into their x and y components. the sum of all the x components must equal to zero. so do the y components because of static equilibrium.

 

[ek]

Ok, so that gives T1sinx1 + T2sinx2 + T3 = 0 (x = >theta)

and

T1cosx1 + T2cosx2 = 0

Nothing wrong so far I hope.

Then, solve the second equation for T2 and substitute in eq1. Then solve for T1. Right?

Maybe my algebra is off, but using this method I'm not coming to the correct answer.

 

[Leong]

upward direction is taken as y-axis ie positive. right direction is taken as x-axis ie positive.
y component :
$$T_{1}sin\theta_{1}+T_{2}sin\theta_{2}+(-Fg)=0$$
x component :
$$T_{2}cos\theta_{2}+(-T_{1}cos\theta_{1})=0$$
Solve it.

 

[ek]

There's this stupid cosine floating around. It won't go away.

T1sinx1 + T1(cosx1)(sinx2)/(cosx2) = Fg

T1 = (Fgcosx2)/(sinx1 + cosx1sinx2)

Why is that cos in the denominator still there?

Or does that denominator somehow become sin (x1+x2)?

 

[Doc Al]

trig identity needed

There's this stupid cosine floating around. It won't go away.

T1sinx1 + T1(cosx1)(sinx2)/(cosx2) = Fg

T1 = (Fgcosx2)/(sinx1 + cosx1sinx2)

You missed a cosx2, that last line should read:
T1 = (Fgcosx2)/(sinx1cosx2 + cosx1sinx2)

Now apply a common trig identity to simplify the denominator.