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I was playing a game when I started thinking about this problem:

Say, a person is sandboarding down a hill/dune, right before the slope begins, what are the conditions that must be met in order for man and his board to stay in contact with the ground, like:

  • What should be the maximum velocity of the man+board system before
    traversing down the hill for which they don't lose there contact with the ground or What should be
    the minimum velocity of the same system so that the board loses its
    contact with the ground and is in the air?
  • The maximum angle of declination of the cliff for which the system stays in touch with the ground or the relation between the angle of declination and velocity of the system?

My Approach:

enter image description here

Upon equating the Forces, we get: $$tan\theta \le\mu$$ I also tried to write the energy equation while accounting for friction but it doesn't seem to help, $$mgh+\frac{m{v_0}^2}2-mg\mu L=\frac{mv^2}2+mgx-mg\mu cot\theta (h-x) $$ which upon simplification gives (taking $tan\theta=\mu$), $$2g(h-x)=\frac{v^2-v_0^2}2+mgL\mu$$ where the initial velocity(on the horizontal surface) of the board is $v_o$, Length, which the system travels before starting its descent(on the horizontal surface) is $L$, coefficient of kinetic friction is $\mu$ and the height of the cliff is $h$.

Now, the above equation doesn't seem to put a speed limit on $v_o$ (but indeed gives a relation between $\theta$ and $v_o$, of which I am not very sure). Also, I couldn't come up with any other way to solve the problem.

I tried to search about the problem on the internet but all I got was the energy conservation principle which is used to solve problems regarding skiing but that's not exactly what I am looking for.

So, any insights about where I went wrong and other ways to solve this problem are highly appreciable.

(Since the problem is purely imaginative in nature, please feel free to suggest changes or add any other constraints which might make the problem more fun.)

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  • $\begingroup$ The torque when the system is moving on the horizontal surface due to the normal reaction is anticlockwise while that due to the weight of the system is clockwise and equal to that due to normal reaction, (that due to friction is 0). So what would be providing this clockwise torque? And why clockwise? $\endgroup$
    – Samarth
    Commented Aug 5, 2020 at 16:38
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    $\begingroup$ Can I treat the man and skateboard as block ? $\endgroup$ Commented Aug 5, 2020 at 16:40
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    $\begingroup$ Also , how do you make your diagrams , your diagrams are good and neat .+1 for that. $\endgroup$ Commented Aug 5, 2020 at 16:47
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    $\begingroup$ Thank you. I used Window's Ink workspace's whiteboard. $\endgroup$
    – Samarth
    Commented Aug 5, 2020 at 16:49

1 Answer 1

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Toplling diagram

My diagram represents the position of the block just before toppling.

To loose contact there should be a net torque . Sorry for my previous comment .

In my diagram the block is about to topple due to anti-clockwise torque .I am taking torque about COM.You can take torque about any point but since centre of mass is accelerating you need to be careful for a pseudo torque.

$\tau_{friction}>\tau_{normal}$

If this isn't true in this case the block can never topple.

$\mu Nb>Na$

$\mu>a/b$

Now $tan(\theta)= \mu$

$\therefore$

$$\theta= arctan(a/b)$$

and it's independent of the velocity of the block.

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  • $\begingroup$ Your answer gives the conditions for toppling when the block is already descending/started its descent. What I was really asking for are the conditions right before it starts its descent so it loses its contact with the ground right from the beginning and goes like a projectile. In that case, If we measure the torque about COM at the rightmost point of the horizontal surface, it'll be 0. $\endgroup$
    – Samarth
    Commented Aug 6, 2020 at 13:54
  • $\begingroup$ @Samarth The man can simply jump , that's why replaced it block. It can jump with at any velocity too lose contact. $\endgroup$ Commented Aug 6, 2020 at 14:21
  • $\begingroup$ The answer will still hold if the block is at rest. $\endgroup$ Commented Aug 6, 2020 at 14:28

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