The problem is as follows:

A block is sliding with a constant velocity over an incline. The angle of that incline is $\alpha$. What would be the acceleration of the block when the angle of the incline is $2\alpha$? (You may use $g=10\frac{m}{s^2}$).}

The alternatives given on my book are as follows:

$\begin{array}{ll} 1.&g\sin\alpha\\ 2.&g\cos\alpha\\ 3.&g\tan\alpha\\ 4.&g\cos\alpha\\ 5.&g\sin \left(2\alpha\right)\\ \end{array}$

Okay for this problem. I was not given a FBD or any sort of sketch neither a diagram. So I made one, which is shown below.

Sketch of the solution

However I'm not sure if I'm making the right interpretation, which I believe it might be trivial.

The diagram summarizes two moments stated from the problem. The first is when the angle is $\alpha$ and the block in the upper part is sliding with constant velocity. Hence no acceleration, right?.

The lower portion shows what is happening when the block is sliding when the angle is doubled. Therefore there is an acceleration. No any force is mentioned other than the gravity acting on the block so I assumed what it is shown in the second diagram.

Therefore the equation would be:


Since both masses cancel and there isn't given any sort of friction. I'm assuming the surface is frictionless.


$a= g\sin\left(2\alpha\right)$

So that would be it, alternative number $5$. But the book states that the right answer is the $3$ option. What am I doing wrong here?.

The only thing which comes to my mind is that the surface is not frictionless which for that case the coefficient of static friction (which I believe is called the angle of repose) is given as:


But in this case I don't know if it applies to this situation. But this is an answer whioh is close to what the book says. Can somebody help me?. Did I missed anything? or could be that the answer is not right?.


3 Answers 3


We know the first moment has limiting friction since the block is neither accelerating nor decelerating. So, we have that $\mu_k = \tan\alpha$. So, our net force along the incline is $$ma = mg\sin{2\alpha} - \mu N$$ $$\implies a=g\sin{2\alpha}-g\cos{2\alpha}\tan{\alpha}$$ $$ = {2g\tan\alpha\over 1+\tan^2\alpha}-g{1-\tan^2\alpha\over 1+\tan^2\alpha}\tan\alpha$$ $$=g{2\tan\alpha-\tan\alpha+\tan^3\alpha\over1+\tan^2\alpha}$$ $$= g\tan\alpha \tag 3$$


Since in the first case the velocity is constant we have

$$mg\sin \alpha-\mu_d mg\cos \alpha =0 \implies \mu_d=\tan \alpha$$


$$mg\sin (2\alpha)-\mu_d mg\cos (2\alpha) =ma \implies a=g\left(\sin (2\alpha)-\tan \alpha\cos (2\alpha)\right)=$$

$$=g\left(2\sin \alpha \cos \alpha -2\sin \alpha \cos \alpha+\tan \alpha\right)=g\tan \alpha$$

  • $\begingroup$ Thanks for that. I was a bit stuck in the final trigonometrical simplication. What I used was the power reduction identity to solve that riddle with the tangent function. But perhaps does it exist a much simpler approach?. Can you add the step in the middle of how you got to $\tan\alpha$?. I got it as well but it was a bit longer, maybe you have an easier method. $\endgroup$ Nov 25, 2019 at 11:33
  • $\begingroup$ @ChrisSteinbeckBell We have used that $$\tan \alpha\cos (2\alpha)=\tan \alpha(2\cos^2 (\alpha)-1)=2\sin \alpha \cos \alpha-\tan \alpha$$ $\endgroup$
    – user
    Nov 25, 2019 at 11:36


After determining $\mu = \tan\alpha$ when $\beta = 2\alpha$ we have the movement equation

$$ m g\sin\beta - \mu mg\cos\beta = m a $$

  • $\begingroup$ Yes I noticed this. But I had to struggle with the trigonometrical simplification. Perhaps can you offer some help with it?. I did reach the answer by using the power reduction identity but it might be an easier method. $\endgroup$ Nov 25, 2019 at 11:34
  • $\begingroup$ With a little effort on your part you can resolve the issue. $\endgroup$
    – Cesareo
    Nov 25, 2019 at 11:46

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