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Assuming all surfaces to be frictionless. If the mass of the inclined plane is large enough, could $N$ ever be equal to $mgcos\theta$. Reasons?

So far I've come up with: $mgcos\theta - N = ma$ meaning if N were to equal $mgcos\theta$, $ma = 0$. Since the surfaces are frictionless and since N is acting on the inclined plane too(it'll have a horizontal component), this isn't possible. Am I right? Where am I going wrong?

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  • $\begingroup$ I'm not sure what all is being asked: $N$ is always equal to $mg\cos\theta$ (with inverse direction) by the third law. $\endgroup$ – Guillermo Angeris May 7 '16 at 6:17
  • $\begingroup$ @GuillermoAngeris No. N and mgcos(theta) aren't action-reaction pairs. $\endgroup$ – xasthor May 7 '16 at 6:20
  • $\begingroup$ This is independent of having or not having friction. Imagine N and $mg\cos\theta$ are not equal, then we would have acceleration either outwards of the plane (which means our cube would start flying) or into the ramp (thus, our cube would pass through the plane itself), neither of which are at all physically possible. $\endgroup$ – Guillermo Angeris May 7 '16 at 6:23
  • $\begingroup$ @GuillermoAngeris The reaction to mg is given by the earth being pulled(with a force -mg). N is not necessarily equal to mgcostheta or this system above wouldn't accelerate towards the left $\endgroup$ – xasthor May 7 '16 at 6:27
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    $\begingroup$ Ah, okay, now I see; the system isn't just the block moving along the plane, but also the incline plane itself. In this case, the reaction force is never equal, as you've previously mentioned: one clear way to see this is by noting that the $x$-coordinate of the centre of mass of the system must remain unchanged (by conservation of momentum), yet the small block clearly moves forward, thus the plane must move backwards in proportion to the velocity of the small block---by a factor of $M_\text{incline}/m_\text{block}$---such that the centre of mass remains constant. $\endgroup$ – Guillermo Angeris May 7 '16 at 6:43
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For completeness, I write a short solution:

Note that we have no forces acting on the entire system along the $x$ (horizontal) direction, thus, if we consider the center of mass of the entire system, we are forced to conclude that the center of mass is constant along this direction, by conservation of momentum.

It's intuitively clear that the smaller block will slide along the given surface which implies that, since our center of mass must remain constant along this direction, the incline plane must move backwards at a velocity proportional to the small block's by the ratio of their masses; thus the normal force on the block is only equal to the downwards force on the plane (i.e. $N=mg\cos\theta$) at the limit where this ratio becomes 0 (that is, when the mass of the incline plane is infinite).

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What you need to do is first make a free body diagram(FBD) of both the wedge and the block m. Then find the forces that are acting before writing Newton's 2nd law.

Also, note that your equation: $ mgcos(\theta) - N = ma$ is wrong. Why? Because the inclined plane is moving as well. As you yourself pointed out, there is an equal and opposite normal force on the wedge too and all the surfaces are frictionless. Since block m is resting on an accelerating frame, you will first have to find an inertial frame of reference to apply Newton's laws. My suggestion is to draw the FBD, then choose a coordinate system, (horizontal and vertical axes will do just fine) and figure out how to write the net force equation keeping in mind that both the block and wedge are accelerating. (hint: use relative velocity/acceleration concept along x and y axes.)

It doesn't matter how large the mass of the wedge is. Frictionless surface must give you the hint that there is an unbalanced horizontal component of net force that will give a net horizontal acceleration.

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  • $\begingroup$ The problem is given itself in inertial frame of reference, hence Newton's laws can be used directly why use longer methods like finding relative velocity/acceleration first? $\endgroup$ – Vaibhav Singh May 7 '16 at 6:52
  • $\begingroup$ Because the wedge itself is accelerating. And the OP is not taking that into account. $\endgroup$ – Entangled_Particle May 7 '16 at 6:54

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