Tension in cables supporting two hanging signs (13 kg and 4 kg)

[awesome_irl]

Homework Statement

The big sign below (see attached image) is 13 kg and the small sign is 4 kg. Find the tension in all 4 cables.

Homework Equations

Fnet= Fa +Fb

The Attempt at a Solution

ƩFx= T2cos38 - T1cos51=0
T2=T1cos51/cos38
ƩFy=T1sin51 + (T1cos51/cos38)sin38 - 127.4 -39.2 = 0
T1=131.303

 

[ofeyrpf]

Your T₂ is correct.

T₃ = T₄ = 2g

F_y: T₁sin51 + T₂sin38 = 17g

this is because the only forces opposing the vertical components of the tensions are the weight of the 13kg and 4kg masses. So the only thing opposing the vertical components of the tension is a combined weight of 17kg which is 17g.

 

[awesome_irl]

Thanks!
Oh! And merry christmas

 

[ofeyrpf]

Hi, I've updated the post above, which might me helpful to you.

 

[PhizKid]

But the forces are in equilibrium and there is no acceleration so how can there be 17g? And won't the opposing forces be the weights, not the masses?

 

[ofeyrpf]

I think this is correct, I get the same:)

 

[awesome_irl]

Just wondering...was my first attempt correct (my physics teacher posted the answer and her t1 was 131.18 N)
I know I sound paranoid but still... -_-

 

[awesome_irl]

@ phizkid the "13 g" doesn't denote the mass, I think you mistook the g as a gram rather than the acceleration of gravity. Yoy're correct in mentioning that the system is motionless so that's why the net force is 0 for both the x and y components.

 

[PhizKid]

Oh ok then it's all the same

 

[ofeyrpf]

I get T₁ = 131.3N

We agree!

 

[ofeyrpf]

"g" is not grams in this case, it is 9.8 m/s². So by 17g I meant 17 x 9.8m/s².