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I was solving a banking problem when I had the following doubt. We have a mass m on a wedge having an angle A. We have to find the normal force acting on the block of mass m. I tried to decompose the force mg acting in downward direction into two components:one parallel to the direction of wedge and other in a perpendicular direction with respect to the wedge. It gave the following equation:- $$N=mgcosA$$ On the other hand if I decompose normal force into two components and I would get the following equation:- $$NcosA=mg$$ $$N=mgsecA$$

So which one is right?

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  • $\begingroup$ This is not a static equilibrium situation but a situation where the net force on the block is $mg \sin A$ down the slope with there being no net force on the block perpendicular to the slope with $N - mg \cos A = 0$. $\endgroup$
    – Farcher
    Commented Nov 29, 2016 at 9:56

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Don't forget the acceleration!

You can split $N$ into two components, yes:

$$N_y=N\cos A\qquad\text{ and }\qquad N_x=N\sin A$$

But equating the first one to gravity $w$ is wrong. If you set the upwards and downwards y-forces equal to each other, then you are applying Newton's 1st law, which is not the case here. Rather, you should use Newton's 2nd law, giving:

$$N_y-w=ma_y\quad\Leftrightarrow\quad N\cos A=ma_y+w$$

because there is an acceleration component along this y-direction. So you were missing a term in the expression.

In slope-systems it is usually easier to choose the coordinate system along the slope. This is how you got your red arrows. This is usually easier, because the y-directions then is perpendicular to the slope (and thus to the motion) so that there is no acceleration along this axis and Newton's 1st law can be used.

Otherwise in the case of your blue arrows, you have chosen a coordinate system, where there is an acceleration component along both axes. We would rather avoid that by placing the coordinate system smarter, if possible.

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  • $\begingroup$ I also had this doubt which one is correct. when mass is stationary $a_y$ should be 0 and NcosA =w. Can you say where I am wrong? $\endgroup$
    – user212727
    Commented Jan 10, 2019 at 15:03
  • $\begingroup$ @TheBroly I am not sure I understand your question. Yes, if the mass is stationary so $a_y=0$ then obviously we have: $N\cos A=w$. What is the issue with this? $\endgroup$
    – Steeven
    Commented Jan 10, 2019 at 16:57
  • $\begingroup$ ,My doubt is when mass is stationary then NcosA=w and also MgcosA=N . They both seem to be balanced. So which one is correct equation between NcosA=mg and mgcosA=N ? $\endgroup$
    – user212727
    Commented Jan 10, 2019 at 17:22
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    $\begingroup$ @TheBroly Oh, I understand now. Well, the explanation is that it is impossible. For the box to stay stationary, there must be another force - a pushing force or a friction or something like that. Without such force, the situation is simply impossible. So when you split into components and then set them equal according to Newton's 1st law, then you must include all forces - in your examples here, you are missing a force in one of the equations. The equations as they look now cannot be true both at the same time, because that means box is stationary without any other forces. $\endgroup$
    – Steeven
    Commented Jan 10, 2019 at 17:29
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It looks like the point of your question is solely “How do you find the normal?”

First, you did an excellent job decomposing the gravitational force. Students often try to make the hypotenuse of the triangle along the ramp and end up with nonsensical numbers. You instead did the right thing, making mg the hypotenuse. Good!

Second, once you do that, stop. Newton’s 1st law says that the support force will be equal to the component of the gravitational force perpendicular to the surface because there is no acceleration perpendicular to the ramp. So N = mg cos(A). (just realize they are in the opposite direction).

You are actually done now and should stop working (unless another part of the problem is asking for something). If you try to find any components of N, you will only run yourself in circles with numbers and formulas that are meaningless for what you the problem is actually trying to find and do far more work than is necessary.

Remember, often, the easy answer is the right one. (Something I say to my students who are certain that if they easily got the answer it must be wrong, then erase it and do something completely ridiculous.)

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