I happen to face this question in an exam paper. The question is phrased as:

A block of mass $m$ is released from the top of fixed inclined smooth place. If $\theta$ is the angle of inclination, then the vertical acceleration of the block is?


a) $g$

b) $g\sin^2(\theta)$

c) $g\sin(\theta)$

d) $g\sin(\theta)\cos(\theta)$

I know that the gravitational acceleration is $g$ (vertically downwards) or $g\sin(\theta)$ along the direction of the inclined plane.

But the answer key I got with the paper read that the correct answer is $g\sin^2(\theta)$. Am I missing something or is the answer key simply wrong?

I mean, I have resolved the components of gravity and no value signifies $g\sin^2(\theta)$.


closed as off-topic by user191954, Jon Custer, John Rennie, Kyle Kanos, ZeroTheHero Nov 23 '18 at 1:03

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  • $\begingroup$ You forgot about the normal force on the block due to the plane. The normal force will have a vertical component opposite to the direction of gravitational force. Use Newton's second law to form the force equation along the vertical direction. $\endgroup$ – Mitchell Aug 8 '17 at 13:14
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    $\begingroup$ So, it is accelerating at $g\sin\theta$ parallel to the plane: what is the vertical component of that acceleration? $\endgroup$ – tfb Aug 8 '17 at 13:38
  • $\begingroup$ The vertical component is g itself! I mean, gravity acts downward. I have no idea why one must consider the resolution here. I found the Normal force isn't helping me. I reached at mg - Ncos(THETA) = mgsin(THETA). Please help. $\endgroup$ – Sabesh Bharathi Aug 8 '17 at 13:53
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    $\begingroup$ What is the vertical component of a vector magnitude $a$ at an angle $\theta$ to the horizontal? $\endgroup$ – tfb Aug 8 '17 at 14:32
  • $\begingroup$ Sabesh. The vertical component isn't $g$. It would have been $g$ if gravity was the only force. $\endgroup$ – Steeven Nov 1 '18 at 12:11

I believe that the conceptual problem here is that you aren't recognizing that any vector, (in this problem, let's call it $\vec{a}_{\mathrm{net}}$) can be considered to be the sum of other vectors regardless of how you arrived at its value or what coordinates might be most convenient.

Once you have determined how $\vec{a}_{\mathrm{net}}$ relates to other quantities in the problem, you can break it down into a sum of other vectors, aligned with directions of your own choosing. Therefore, you can break $\vec{a}_{\mathrm{net}}(=\vec{a}_\parallel)$ into component vectors which are horizontal and vertical. At this point, you would be ignoring the weight vector because you've already incorporated its presence into $\vec{a}_{\mathrm{net}}$.

Bottom line concept: We don't have to be bound to any single coordinate system. We are free to resolve vectors into the coordinates we find convenient (or in the case of a test question, specified but not ordinarily used).

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    $\begingroup$ I like this: it avoids doing the 'do my homework for me' thing while providing an underlying answer. $\endgroup$ – tfb Aug 8 '17 at 16:02

The question asks about the VERTICAL acceleration.

You correctly saw that the force along the surface is $mg\sin\theta$. That gives an acceleration along the surface of $g\sin\theta$; the vertical component of that is $g\sin^2\theta$ as stated in the answer key.


In this question

$$ma=mg\sin\theta$$ where $a$ is along the inclined plane

But in question they are asking vertical acceleration so we need to take the component of $g\sinθ$

So now it will be like--

Vertical acceleration =$g\sinθ\sinθ= g\sin^2 θ$.


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