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One really common problem in almost all physics textbooks is that of the wedge. So let us say that a block is resting on a wedge that is inclined at angle x from the horizontal. The block has mass mg. Let us now try to find the normal force. The typical solution is find out the component of gravity in the normal force direction by using the expression cos(x)*Mg. Here the normal force is smaller than the force of gravity. But what if I equated gravity to be equal to the component of the normal force in the upright y axis (in the direction of gravity. ) Now the normal force is greater than the force of gravity. Why is the first approach correct while the second is wrong? Can someone please explain why some vectors are not broken down into components?

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  • $\begingroup$ The word "normal" means perpendicular here. Indeed, the normal force must be perpendicular to the wedge surface; otherwise the normal force on its own would accelerate the block. $\endgroup$ Jun 2, 2022 at 14:38
  • $\begingroup$ Not sure I understand. It's basic math that any vector can be decomposed into as many components as the space has dimensions. "Breaking" down vectors into components, a.k.a. projecting them, is a routine operation that mostly depends on the basis you chose to work in. $\endgroup$
    – Miyase
    Jun 2, 2022 at 14:38
  • $\begingroup$ A normal force must always be perpendicular to a surface. If you equate a "normal force" with gravity in your scenario, then you are not really dealing with the true normal force. $\endgroup$
    – Steeven
    Jun 2, 2022 at 14:45

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You are welcome to resolve the normal force into horizontal and vertical components if you'd like. If you do so, this would be the result:

enter image description here

Newton's 2nd law then takes the form $$\sum F_x = m a_x \implies -N\sin(\theta) = ma_x$$ $$\sum F_y = m a_y \implies N\cos(\theta) - mg = ma_y$$

The problem with this approach is that neither $a_x$ nor $a_y$ is going to be zero. In the standard approach to the problem, we resolve the vectors into components parallel to and normal to the face of the incline. We then require that the acceleration normal to the incline is equal to zero - otherwise the block would be accelerating into or away from the incline. Therefore, our equations become $$\sum F_\parallel = ma_\parallel \implies -mg\sin(\theta) = ma_\parallel$$ $$\sum F_\perp = ma_\perp\color{red}{\overset{!}{=}}0 \implies N-mg\cos(\theta) = 0 \implies N = mg\cos(\theta)$$

The key thing to understand is that if we're working with vectors resolved in the $\parallel$ and $\perp$ directions to the face of the incline, the constraint that the block move along the surface takes an extremely simple form - namely, that $a_\perp = 0$.


The same is not true in your proposed approach. If we work with $x$ and $y$ components instead of $\parallel$ and $\perp$ components, then the constraint that the block must move along the surface takes the form $$a_y\cos(\theta)-a_x\sin(\theta)=0$$ Going back to our equations, we find $$a_x = -\frac{N\sin(\theta)}{m} \qquad a_y = \frac{N\cos(\theta)}{m} - g$$ $$\implies \frac{N}{m}\sin^2(\theta) + \frac{N}{m}\cos^2(\theta) - g\cos(\theta) = 0$$ $$\implies \frac{N}{m} - g\cos(\theta) = 0 \implies N = mg\cos(\theta)$$

So we see that we recover the correct expression for $N$ in this way as well. However, we had to do much more work! This is a good example of the importance of choosing coordinates which simplify your problem as much as possible.

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  • $\begingroup$ I think you might have resolved the components of your vector wrongly in the picture. $\endgroup$
    – Eisenstein
    Jul 3, 2023 at 7:15
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    $\begingroup$ @Eisenstein Thanks for the catch - I corrected the relevant typos and regenerated the figure. $\endgroup$
    – J. Murray
    Jul 3, 2023 at 15:59
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The second approach is wrong. The normal force is normal to the surface on which the block rests. Since there is no acceleration in direction normal to the wedge surface: $$\sum F_{normal} = N-mg\cos{x} = 0$$ Hence $$N = mg \cos{x}$$ The normal force balances the component of gravity perpendicular to the wedge surface and hence we don't see the block sinking into the wedge. The component of gravity parallel to the wedge surface will slide, or tend to slide the block down the wedge.

All of this boils down to selecting a suitable set of axis such that the analysis is simplified.

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But what if I equated gravity to be equal to the component of the normal force in the upright y axis (in the direction of gravity. )

The origin of the gravitational force is not the component of the normal force on the y-axis. It's (classically) the attraction between the object on the wedge and the earth.

The origin of the normal force is the reaction to the component of the gravitation force acting perpendicular to the surface of the wedge, per Newton's 3r law.

You can think about this in terms of action-reaction force pairs. The component of the gravitational force acting perpendicular to the wedge being the "action" force and the equal and opposite normal force being the "reaction" force, per Newton's third law. (Although there is no need to distinguish between an "action" and "reaction" force with respect to Newton's 3rd law)

Hope this helps.

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