-1
$\begingroup$

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?

Options:

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

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Community, Jon Custer, John Rennie, Kyle Kanos, ZeroTheHero
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
2
$\begingroup$

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).

$\endgroup$
  • 1
    $\begingroup$ 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
1
$\begingroup$

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.

$\endgroup$
-1
$\begingroup$

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 =$g\sinθ\sinθ= g\sin^2 θ$.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.