Solve Static Equilibrium for Tac + Tbc

[fordrew]

Hi everyone, first time post at physics forums, I hope to help out some people with calc and I hope to discuss some physics topics too! But however I have a homework question in MECH 234.

Homework Statement

Find Tac + Tbc from the given picture:

Homework Equations

ΣFx = 0
ΣFy = 0

The Attempt at a Solution

Well first, since 500N is being pulled on the x axis-

ΣFx = 0 = T_ac*sin(50)+T_bc*sin(30) - 500

500 = T_ac*sin(50)+T_bc*sin(30)

The right side is the sum of all the Fx components and they must be equal to zero.

ΣFy = 0 = T_ac*cos(50)+T_bc*cos(30)

The right side is the sum of all the Fy components and they must be equal to zero.

When I try to solve the two equations (2 unkowns 2 equations) I just get crazy numbers. I do not know what mistake I could be making otherwise except the angles might be wrong. Or perhaps I should be using the law of cosines or sines?

 

[w3390]

You may be making the mistake in the equation you have for the y dimension. You have that the sum of all y forces is zero, which is correct. However, you may have made your mistake by not setting Tac*cos(50) equal to Tbc*cos(30). Did you mistakingly say Tac*cos(50) equals -Tbc*cos(30)? Also, put up the numbers you got so it is easier to find where you went wrong.

 

[fordrew]

Well I solved for the equations:

0 = Tac*cos(50)+Tbc*cos(30)
500 = Tac*sin(50)+Tbc*sin(30)

and got

Tac ~ 1.98
Tbc ~ -199I did not know that you need to set Tac*cos(50) equal to Tbc*cos(30). Why do they have to be equal? Is it because there is no y component in the 500lb force; therefore, both components need to be equal to each other for equilibrium?My answer for solving the system of equations-

500 = Tac*sin(50)+Tbc*sin(30)
Tac*cos(50) = Tbc*cos(30)

Tbc= 326.3518 N
Tac= 439.6926 N

It is correct. And I would like to thank you in advance.