For which case would the applied force be greater?

[paulimerci]

1/(x+y) < 1/(x-y)

 

[haruspex]

1/(x+y) < 1/(x-y)

Right, so is F_A or F_B the greater?

 

[paulimerci]

Right, so is F_A or F_B the greater?

F_{B is greater}

 

[haruspex]

F_{B is greater}

Right. All clear now?

 

[paulimerci]

Right. All clear now?

Yes @haruspex , Thank you!

 

[PeroK]

Right, so if $x+y>x-y$ is $\frac 1{x+y}>\frac1{x-y}$ or $\frac 1{x+y}<\frac1{x-y}$.

We also need the condition $x > y$ here.

 

[haruspex]

We also need the condition $x > y$ here.

Yes, that was specified in post #31.