# Tension problem- why doesn't this solution work?

#### Lola Luck

1. Find the tension in each cord if the weight of the suspended object is w. (I am only asking about part a) 2. Net force on an object in equilibrium = 0

3. part a) I know this isn't the typical solution for this type of problem but I don't understand why it's not working.

I used the sine law: (Ta)/(sin45) = (Tb)/(sin30)
Then cosine law: w2 = (Ta)2 + (Tb)2 - 2(Ta)(Tb)cos75

I combined the equations to solve for Tb in terms of w. The answer I get doesn't match the given answer, so I'm wondering if there is something wrong with how I set up the problem.

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#### jbriggs444

Homework Helper
The law of sines could help you to determine the length of cords A and B. But that is not what you are after. You want to know the tension in cords A and B.

What horizontal force components exist at the junction point between A, B and C?
What vertical force components exist at the junction point?

• Lola Luck

#### Lola Luck

The law of sines could help you to determine the length of cords A and B. But that is not what you are after. You want to know the tension in cords A and B.

What horizontal force components exist at the junction point between A, B and C?
What vertical force components exist at the junction point?
I'm not sure I understand why the law of sines can't be used on forces (tension in this case).

I believe I understand how to do it with components: (Tay) + (Tby) = -w and (Tax) + (Tbx) = 0

#### jbriggs444

Homework Helper
Right. $T_{ax} + T_{bx} = 0$. But that does not involve the law of sines.

• Lola Luck

#### Lola Luck

Yeah. I can solve it that way, but I don't understand why it would be incorrect to use the law of sines.

#### jbriggs444

Homework Helper
See the first response above. The law of sines tells you something about the length of the cords, not their tensions.

But ignore that. Let us assume that the law of sines worked as you state, i.e

$\frac{t_a}{sin 30} = \frac{t_b}{sin 45}$ Which means that $\frac{t_a}{t_b} = \frac{sin 30}{sin 45}$ That's approximately 0.35.

But we also know that

$t_a cos 30 = t_b cos 45$ Which means that $\frac{t_a}{t_b} = \frac{cos 45}{cos 30}$ That's approximately 0.82

They cannot both be right.

• Lola Luck

#### Lola Luck

I understand now. The vectors weren't in an additive form (tip to tail) so it was wrong to apply triangular rules to their magnitudes. Thank you for helping me!