For which case would the applied force be greater?

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.
  • #36
1/(x+y) < 1/(x-y)
 
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  • #37
paulimerci said:
1/(x+y) < 1/(x-y)
Right, so is FA or FB the greater?
 
  • #38
haruspex said:
Right, so is FA or FB the greater?
FB is greater
 
  • #39
paulimerci said:
FB is greater
Right. All clear now?
 
  • #40
haruspex said:
Right. All clear now?
Yes @haruspex , Thank you!
 
  • #41
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.
 
  • #42
PeroK said:
We also need the condition ##x > y## here.
Yes, that was specified in post #31.
 

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