# Determining maximum force

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 [Broken]

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

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

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

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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.

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.

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

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.

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.

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.