Maximum Force Analysis for FB and FC Limits in a Tension System

[Ry122]

The maximum tensions that FB and FC can undergo are 5kN and 5.5kN respectively.
Determine the maximum force of FA that can be reached without exceeding these limits.
http://users.on.net/~rohanlal/Q1.jpg

This is all i have so far
0 = -FA + FBsin40
What's the next step?

 

[Snazzy]

Calculate the components of F_B. Use this to solve for the difference between F_B and F_C to calculate F_A.

 

[Ry122]

wouldnt u just calculate the x component since FC has no y component?

 

[Ry122]

Is this answer correct:
5sin40=3.2139kN
To determine tension in FC
5cos40=3.830kN
3.830kN does not exceed the maximum limit for FC therefore the maximum force of FA is 3.2139kN.

 

[Snazzy]

You already have the tension in F_C. The rope, F_A, is not straight down. The sum of the forces in the x-direction must be zero, and the sum of the forces in the y-direction must be zero.

 

[Ry122]

5.5kN is not the tension in FC it is the maximum limit.

 

[Snazzy]

It is the maximum limit of the tension in F_C and you want to find the maximum limit of the tension in F_A.

 

[Ry122]

But if FC was to have a tension of 5.5kN that would put greater tension on FB. Therefore you would have to lower FB's y component to compensate for the increase of the x component to not let the tension in FB exceed the max limit.

 

[adiputra]

I think in this case Fc does not have vertical component, so we need not to care about it. the diagram try to fool us i think. so FA(max)=FB(max)=5kN.