# For which case would the applied force be greater?

• paulimerci
In summary: I understand N is smaller in case A and larger in case B. How do...In summary, in case A, the net force along X is negative, while in case B, the net force along Y is positive. This is because the angle of the pulling force is greater in case B than case A.
1/(x+y) < 1/(x-y)

paulimerci said:
1/(x+y) < 1/(x-y)
Right, so is FA or FB the greater?

haruspex said:
Right, so is FA or FB the greater?
FB is greater

paulimerci said:
FB is greater
Right. All clear now?

haruspex said:
Right. All clear now?
Yes @haruspex , Thank you!

haruspex said:
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.

PeroK said:
We also need the condition ##x > y## here.
Yes, that was specified in post #31.

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