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## Main Question or Discussion Point

Hi,

I just saw the solution to a problem I was pondering about recently and did not really understand one of the steps/assumptions they made, and therefore I hope that you guys can clarify my doubts.

The question is like this, you have a block on a frictionless inclined slope of height h and inclined x from the horizontal. The slope is on a frictionless surface as well. The block has mass m and the slope has mass M. Calculate the time taken for the block to reach the bottom of the slope.

For those who wants to know, the answer is:

sqrt[(2h(M + m sin^2 x))/((M+m) g sin^2 x)]

But I have a few doubts about the method used to solve the problem (I am talking about the method that uses Newtonian Mechanics, not Lagrangian, my physics is not that good yet XD).

In the method given, You draw the forces exerted on the block and on the slope. For the block, the forces are mg (weight) and N (reaction force from the slope). For the slope, it is N(the reaction force from the block). Then you assume that the reaction force N is constant, so after you applied Newton's Laws and resolved the accelerations into the correct axes, you can use kinematics equations.

However, I don't understand why N is constant throughout the motion. Isn't the slope moving away from the block? If it is, then N should change rit? Actually, when I was trying to solve the problem at first, I thought that there might be moments when the block might be not in contact with the slope, so I was thinking of the block making repeated elastic collisions with the slope, but it seems I am wrong.

Therefore, can someone please explain to me why N is constant, or give a formal proof? Thanks a lot.

I just saw the solution to a problem I was pondering about recently and did not really understand one of the steps/assumptions they made, and therefore I hope that you guys can clarify my doubts.

The question is like this, you have a block on a frictionless inclined slope of height h and inclined x from the horizontal. The slope is on a frictionless surface as well. The block has mass m and the slope has mass M. Calculate the time taken for the block to reach the bottom of the slope.

For those who wants to know, the answer is:

sqrt[(2h(M + m sin^2 x))/((M+m) g sin^2 x)]

But I have a few doubts about the method used to solve the problem (I am talking about the method that uses Newtonian Mechanics, not Lagrangian, my physics is not that good yet XD).

In the method given, You draw the forces exerted on the block and on the slope. For the block, the forces are mg (weight) and N (reaction force from the slope). For the slope, it is N(the reaction force from the block). Then you assume that the reaction force N is constant, so after you applied Newton's Laws and resolved the accelerations into the correct axes, you can use kinematics equations.

However, I don't understand why N is constant throughout the motion. Isn't the slope moving away from the block? If it is, then N should change rit? Actually, when I was trying to solve the problem at first, I thought that there might be moments when the block might be not in contact with the slope, so I was thinking of the block making repeated elastic collisions with the slope, but it seems I am wrong.

Therefore, can someone please explain to me why N is constant, or give a formal proof? Thanks a lot.