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If we have a ramp and a block on it, and if all the surfaces are frictionless so both the the ramp and block can move. Is the normal force different from the time that ramp is not moving? If yes can you explain why? And how can we calculate the acceleration of the ramp?

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  • $\begingroup$ Do you mean that the ramp and the block both move? This would mean the block slides down the ramp at the same time the ramp slides across the horizontal surface? $\endgroup$ – Ernie Aug 10 '15 at 15:34
  • $\begingroup$ Yes exactly.they both move $\endgroup$ – Farzin nasiri Aug 10 '15 at 16:23
  • $\begingroup$ There are two moving objects, a normal force between them, and another normal force (gravity) between the ramp and the horizontal, which reacts to the combined mass of ramp and block. As the block and the ramp move left and right, and as the gravitational force pulls neither right nor left, the total momentum of the system must net to zero: M block + M ramp = zero. Here's a link to some hints: ocw.mit.edu/courses/mechanical-engineering/…. The example combines conservation of momentum with conservation of energy. $\endgroup$ – Ernie Aug 11 '15 at 0:47
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Ah, I asked a question like this on an exam some years ago. The students didn't like this question very much...

The ramp will accelerate because the block exerts a normal force on it which is not vertical, so there is no other force which can oppose its horizontal component.

You are correct that the normal force will be different from what it would be if the ramp was held stationary. Remember that, in general, the only way we ever know the value of a normal force is by solving for it from Newton's 2nd law. It is easy to forget this since most cases we look at are trivial so we don't have to go through this process.

So, you should draw FBDs for the block and for the ramp. Don't forget what Newton's 3rd law tells you or you are likely to forget that the block exerts a normal force on the ramp. You might have to think carefully about what the most convenient coordinate axes are. Both the block and the ramp accelerate. We don't know either acceleration, but there is something we know about the acceleration of the ramp, and realizing this will point you towards the better choice of axes. Now write Newton's 2nd law (x and y components) for each of the block and the ramp. You have 4 equations and lots of unknowns. Newton's 3rd law tells you that the size of the normal force by the ramp on the block is the same as the size of the normal force by the block on the ramp, so this eliminates an unknown for you. At this point you should be down to a small enough number of unknowns that you can solve the system of equations. Good luck with that.

There are some somewhat sophisticated things you can do to simplify matters. For example, the only external forces on the system (the block+ramp) are vertical (gravity, and the normal force that the floor exerts on the ramp). So the acceleration of the center of mass of the system must point straight down. This could simplify your solution if you make clever use of it.

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This problem can be solved intuitively.

IF the ramp were fixed, the normal force perpendicular to the incline of the ramp and the underside of the block wouldn't change during the block's downward journey, and the block's acceleration would be constant. Here's an example of a block sliding on a fixed ramp: http://hyperphysics.phy-astr.gsu.edu/hbase/mincl.html.

HOWEVER, suppose you wanted the block to reach the end of the ramp in less time, without putting any additional energy into the system? You could make the ramp movable!

Why? The effective angle of incline of the block's path through space increases as the normal force between block and ramp drives the ramp out from under the block. The block's actual path through space becomes the hypotenuse of a shrinking right triangle. As the hypotenuse gets shorter, the block's acceleration increases as its actual path gets steeper, even though the incline of the moving ramp stays the same. And at the same time, the normal force between block and ramp constantly decreases, until it effectively becomes zero when the block hits the table and stops. At that point, the effective angle of the block's path would be really steep, perhaps even a right angle, and the block would tilt on to its end.

The answers to your questions are in the above paragraph.

(1) The normal force between block and ramp changes from the time that the ramp is not moving. It changes because it is the component of gravitational force that is perpendicular to the effective incline of the block's path; it's g cos theta, where theta is the angle between the block's effective path and the horizontal table on which the ramp rests. When the block hits the table and tilts off the ramp, the normal force between block and ramp becomes zero. Cos 90 degrees = 0, and F = g * 0 = 0.

(2) As the normal force constantly decreases while the block slides down an effectively increasing incline path, the acceleration of the ramp must constantly decrease. You can see this as a consequence of Newton's 2nd law of motion:

F = m * a

a = F/m

a is acceleration of the ramp

m is mass of the ramp

F is the normal force between ramp and block

The ramp's mass stays the same of course, but the normal force F decreases as the cosine of effective theta decreases.

Example:

Let's say effective theta is 30 degrees when the block is near the top, and 70 degrees near the bottom (as the ramp slides out, the block's trajectory through space steepens). Let's say the block's mass is 3 kilograms.

Normal force near top = 3 Kg * 9.8 m/s^2 * cos 30 = 25 newtons

Normal force near bottom = 3 Kg * 9.8 m/s^2 * cos 70 = 10 newtons

As effective theta increases, the slope of the block's actual path through space increases, acceleration of the block increases, and acceleration of the ramp decreases because the normal force between block and ramp decreases. When the block hits the table, the ramp's acceleration becomes zero, and the ramp continues sliding with constant velocity on the frictionless surface.

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  • $\begingroup$ I believe that the solution you linked does not take account of the motion of the ramp (no friction between ramp and horizontal surface, implied by the last line "how can we calculate the acceleration of the ramp?"). $\endgroup$ – Floris Aug 10 '15 at 14:15
  • $\begingroup$ @Floris: I assumed he meant movement of the block only, and that the ramp's acceleration may refer to the block's acceleration on the ramp, as there's no mention of a force that would cause the ramp to move, whereas gravity is understood to move the block. My assumption takes too much liberty with the language of the question. I'll ask him in a comment what his intention is. $\endgroup$ – Ernie Aug 10 '15 at 15:32
  • $\begingroup$ "so both the ramp and the block can move" seems pretty unambiguous to me. In a frictionless world, the ramp moves to the left as the block moves to the right, to conserve momentum. $\endgroup$ – Floris Aug 10 '15 at 15:33
  • $\begingroup$ @Floris: OK, I see your point now. Yes, that's reasonable. I will edit my answer. $\endgroup$ – Ernie Aug 10 '15 at 15:35
  • $\begingroup$ @Floris: I used the block's actual path through space to conclude that (1) the block's path steepens with time, (2) normal force between ramp and block decreases with time, (3) the block's acceleration increases, while the ramp's acceleration decreases, with time, (4) the ramp's acceleration becomes zero when the block hits the table. $\endgroup$ – Ernie Aug 11 '15 at 20:29

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