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The normal force for the case of pushing a crate across a floor with an applied force $F_{app}$ at an angle $\theta$ above the horizontal is given by Newton's second law:

$$ \sigma F_y = 0 = n - mg + F_{app}\sin(\theta) $$

Thus:

$$ n = mg - F_{app}\sin(\theta) $$

If the angle of $F_{app}$ is below the horizontal, we get:

$$ \sigma F_y = 0 = n - mg - F_{app}\sin(\theta) $$

Thus:

$$ n = mg + F_{app}\sin(\theta) $$

If the second derivation is wrong, I want to know precisely why. I say it is correct. If correct, why can't I use angles measured from the $+x$ axis to solve problems for angles below the horizontal? Example, if the angle is $20$ degrees below the horizontal, $\theta$ is $+340$ or $-20$ when measured from the $+x$ axis. This gives:

$$ n = mg - 0.342~F_{app} $$

This is wrong because of the sign of the second term. Obviously, the normal force increases for $F_{app}$ at an angle below the horizontal.

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  • $\begingroup$ To avoid confusion, you need to clarify the angle. The normal force will be reduced if the applied force has a vertical upward component and increased if the applied force has a downward vertical component. $\endgroup$
    – Bob D
    Oct 9 '20 at 16:05
  • $\begingroup$ I attempted to clarify. I want to define all angles from the +x axis, as is often done. The problem is that when this is done after using Newton's 2nd Law (as shown in my original post) I get the wrong sign. As I see it, we should be able to use Newton's 2nd law in both cases and derive TWO different equations. Do you see an error in either of my derivations for n? $\endgroup$
    – Joe
    Oct 9 '20 at 17:33
  • $\begingroup$ It’s still not clear. A picture is worth a thousand words. Show the crate and the force and the angle the force makes with the horizontal $\endgroup$
    – Bob D
    Oct 9 '20 at 17:40
  • $\begingroup$ I will If I can find out how to upload a pic. $\endgroup$
    – Joe
    Oct 13 '20 at 1:34
  • $\begingroup$ Just click on the add a picture icon and drag and drop your JPEG file $\endgroup$
    – Bob D
    Oct 13 '20 at 5:35
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You have got your signs wrong. If $F$ is applied at an angle above the horizontal then it increases the normal force, and vice versa.

So if the angle $\theta$ is positive when above the horizontal then the normal force $N$ is

$N = mg + F_{app}\sin \theta$

If the angle $\theta$ is negative i.e. below the horizontal then you can still use this formula, remembering that $\sin \theta$ is negative if $\theta$ is in the range $(-180^0,0^0)$. Or you can replace $\theta$ with $\theta’=-\theta$ which is positive when below the horizontal, in which case

$N = mg - F_{app}\sin \theta’$

Remember that the normal force cannot become negative, so strictly speaking the correct expression is

$N = \max(mg + F_{app}\sin \theta,0)$

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  • $\begingroup$ I appreciate your response. However, I can't agree with your very first statement. A vector that is applied above the horizontal points into the first quadrant. A vector that is applied below the horizontal points into the 4th quadrant. Thus, for a force applied below the horizontal, the normal force increases. An angle of +340 wrt the + x-axis increases the normal force. If we can't agree on that, we can never agree on signs. $\endgroup$
    – Joe
    Oct 9 '20 at 15:23
  • $\begingroup$ @Joe This is the root cause of your confusion. A force applied from a direction below the horizontal has an upwards vertical component and so it will reduce the normal force. A force applied from a direction above the horizontal has a downwards vertical component and so it will increase the normal force. $\endgroup$
    – gandalf61
    Oct 9 '20 at 16:19
  • $\begingroup$ Thanks again. I agree 100% with what you just wrote. I fully understand that. You and I do not agree on what below or above the horizontal means. Serway and Vuille (11th Ed.) believe (as do I) that a force applied BELOW the horizontal points into the FOURTH quadrant. If I understand your statement, you say the opposite. $\endgroup$
    – Joe
    Oct 9 '20 at 17:28
  • $\begingroup$ @Joe You need to think in terms of the direction from which the force is applied, which is opposite to the direction in which the force vector points. The force is pushing the crate, not pulling it. $\endgroup$
    – gandalf61
    Oct 9 '20 at 17:39
  • $\begingroup$ If you can put that issue aside, do you agree that there are TWO derivations possible from Newton's second law? One for pushing one for pulling? (What one means by above or below is not my question.) Do my TWO equations for n look correct? $\endgroup$
    – Joe
    Oct 9 '20 at 17:44

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