Suppose you have a wedge on a table with length $l$ and mass $M$. The angle between the slope and the horizontal direction is $\alpha$. You also have a point of mass on the top of the wedge with mass $m$. The distance the wedge travels is to be calculated while the point of mass gets to the bottom. There is no friction between any of the surfaces.

I write the external forces acting on the system which would be the $mg$ and $Mg$ forces (pointing downwards) acting on the point and the wedge respectively and also a froce $T$ (pointing upwards) acting on the wedge exerted by the table coming from the constraint that the wedge must not dwell into the table.

My problem is that the solutuion suggests that all vertical external forces acting on the center of mass cancel out. However, if I do a little calculation I find that the force with which the point of mass pushes against the surface of the slope is $mg\cos\alpha$. The vertical component of this pushes the table downwards, which is \begin{equation}mg\cos\alpha\cos(\pi/2-\alpha) = \frac{mg}{2}\sin(2\alpha)\end{equation}

Thus the total force with which the slope-mass system pushes the table is \begin{equation}\frac{mg}{2}\sin(2\alpha) + Mg\end{equation}

This force is equal (in magnitude) with the force $T$, but this way vertical forces certainly do not cancel out, because the sum of them is \begin{equation}T -Mg - mg = \frac{mg}{2}\sin(2\alpha) + Mg - Mg - mg = mg\left(\frac{sin(2\alpha)}{2} - 1\right)\end{equation} which is negative, thus points downwards and indicates that the center of mass moves downwards.

What is the problem in my way of thinking?


2 Answers 2


What is the problem in my way of thinking?

You forgot the normal force between the block and the wedge.

  • 4
    $\begingroup$ This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. $\endgroup$
    – Kyle Kanos
    Dec 31, 2014 at 20:56
  • 1
    $\begingroup$ Well... it's a fairly uninformative answer, but it does appear to answer the question of "What is the problem in my way of thinking?" and therefore it doesn't qualify for deletion as not being an answer. $\endgroup$
    – David Z
    Jan 1, 2015 at 0:40
  • $\begingroup$ @KyleKanos he really forgot this thing please don't demotivate me on this site, I want to be more involed with physics!! $\endgroup$
    – RE60K
    Jan 1, 2015 at 9:18
  • $\begingroup$ @ADG but the normal force you mentioned is a force acting between parts of the system and is not an external one. Thus it does not contribute to the movement of the center of mass. How is it relevant then? $\endgroup$ Jan 1, 2015 at 10:48

It turned out, that there was no problem at all, because the exercise asked the distance the wedge travels. But since it can only move horizontally the vertical acceleration of the center of mass (CM) is unimportant. The key to the solution is that all horizontal forces are inner forces, thus they can't change the acceleration of the CM, so the $x$ coordinate is unchanged. Suppose that initially the CM of the wedge is at $x_M$ and the point of mass (PM) starts at $x=0$. This way the horizontal coordinate of the CM initially is \begin{equation} x_{CM} = \frac{0m+x_M M}{m+M}\end{equation} and is unchanged.

If the wedge travelled a distance $d$ backwards, then the $x_m$ coordinate of the PM is $l-d$ where $l$ is the length of the lower side of the wedge facing the table. Due to the rigidness, the CM of the wedge will be at $x_M - d$, and thus the $x$ coordinate of the CM of the whole system will be \begin{equation}x_{CM} = \frac{(l-d)m + (x_M-d)M}{m+M}.\end{equation}

Because there are only vertical external forces, the horizontal coordinate of the CM can't accelerate and it started with zero horizontal speed so the two expressions for the $x$ coordinate of the CM must equal, and this way one may obtain an expression for the distance travelled backwards as \begin{equation}d = \frac{m}{m+M} l.\end{equation}


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