Force: max tension before rope breaks

[mybrohshi5]

Homework Statement

If the maximum tension either rope can sustain without breaking is 4000 N, determine the maximum value of the hanging weight that these ropes can safely support. You can ignore the weight of the ropes and the steel cable.

The Attempt at a Solution

I figured out that the rope with the 60 degree angle will have the greater tension therefore being able to hold a less amount of weight.

so i did 4000*sin(60)=W

W=3464 Newtons

so the rope that can hold the least amount of tension will be able to hold 3464 N.

i then doubled this to get 6928 N and that is the total weight both ropes can hold if the max tension of the weakest rope can support 4000 N.

Does this sound correct?

Thank you

 

[korican04]

You should draw a free body diagram and get the forces acting on the middle point where the metal cable and ropes meet.
Fy: T1sin(40) + T2sin(60) - mg = 0
Fx: T1cos(40) - T2cos(60)= 0

 

[Chewy0087]

Hmm I'm not so sure it's that easy, I assume that the block must remain in equilibrium and therefore the horizontal forces must be equal aswell, i'd make sure they are. The tensions in each rope may not be the same!

damn that guy above me editing >.<

 

[mybrohshi5]

You should draw a free body diagram and get the forces acting on the middle point where the metal cable and ropes meet.
Fy: T1sin(40) + T2sin(60) - mg = 0
Fx: T1cos(40) - T2cos(60)= 0

Thank you. that worked great and makes sense. i just used 4000 N for T2 on the Fx then solved for T1 then plugged that into the other equation and got the weight :)

thanks again.