Solved Problem on Dynamics
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A cart, with mass M, moves without friction on horizontal rails with a speed equal to v0. At the front of the cart, there is a body with mass m and an initial speed equal to zero relative to the cart. What is the length of the cart so the body will not fall from it? The dimensions of the body, relative to the length of the cart, can be neglected. The coefficient of friction between the body and the cart is μ.


Problem data:
  • Speed of the cart:    v0;
  • Mass of cart:    M;
  • Initial speed of body:    v0B = 0;
  • Mass of body:    m;
  • Coefficient of friction between the body and the cart:    μ.
Problem diagram:

We choose a frame of reference on the ground, at the rear of the cart, and point to the right (Figure 1). The length of the cart is equal to L, the front of the cart is at a distance of S = L from the origin. We consider that the block of mass m was placed on the front of the cart quite smooth so that no vertical perturbations occur on the system, beyond the weight of the body and the normal reaction of the cart on the block. We assume the acceleration due to gravity is equal to g.
Figure 1

Solution

From Newton's First Law, "A body remains at rest or in a uniform motion in a straight line, unless a force changes their state" then the block tends to remain at the point L where it was placed, but as the cart moves to the right, and there is friction between the block and the cart, it acts in the block with a friction force to the right (Figure 2-A). This friction force changes the state of the rest of the body and begins to drag the block to the right with acceleration aB.
From Newton's Third Law, "Two bodies exerting forces on each other, these forces are equal in magnitude and opposite directions" thus to the action of the friction force of the cart, opposes the reaction of the friction force of the block in the cart, of the same magnitude, and directed to the left (Figure 2-B) which will produce in the cart a deceleration aC.
Figure 2

Drawing free-bodies diagrams, we have the forces that act on them, we can apply Newton's Second Law
\[ \begin{gather} \bbox[#99CCFF,10px] {\vec{F}=m\vec{a}} \tag{I} \end{gather} \]
Block:
  • \( {\vec F}_{gB} \): gravitational force on the block;
  • \( {\vec N}_{B} \): normal reaction force;
  • \( {\vec F}_{f} \): force of friction.
Figure 3

In the vertical direction, there is no motion, gravitational force \( {\vec F}_{gB} \) , and normal reaction force \( {\vec N}_{B} \) cancel.
\[ \begin{gather} N_{B}=F_{gB} \tag{II} \end{gather} \]
The gravitational force is given by
\[ \bbox[#99CCFF,10px] {F_{g}=mg} \]
applying this expression to block B
\[ \begin{gather} F_{gB}=mg \tag{III} \end{gather} \]
substituting the expression (III) into expression (II)
\[ \begin{gather} N_{B}=mg \tag{IV} \end{gather} \]
In the horizontal direction applying the expression (I), we have the force of friction as the resultant.
\[ \begin{gather} F_{at}=ma_{B} \tag{V} \end{gather} \]
the friction force is given by
\[ \bbox[#99CCFF,10px] {F_{f}=\mu N} \]
for the block, the friction force will be
\[ \begin{gather} F_{f}=\mu N_{B} \tag{VI} \end{gather} \]
equating expressions (V) and (VI)
\[ \begin{gather} \mu N_{B}=ma_{B} \tag{VII} \end{gather} \]
substituting the expression (IV) into expression (VII)
\[ \begin{gather} \mu \cancel{m}g=\cancel{m}a_{B}\\ a_{B}=\mu g \tag{VIII} \end{gather} \]
Cart:
  • \( {\vec F}_{gC} \): gravitational force on the cart;
  • \( {\vec N}_{1} \) and \( {\vec N}_{2} \): normal reaction forces;
  • \( -{\vec F}_{f} \): force of friction, \( \left|\;{\vec{F}}_{f}\;\right|=\left|\;-{\vec{F}}_{f}\;\right| \).
In the vertical direction, there is no motion, the gravitational force \( {\vec F}_{gC} \) , and normal reaction forces \( {\vec N}_{1} \) and \( {\vec N}_{2} \) cancel.
Figure 4

Note: No need to write the equation of Newton's Second Law for the vertical direction, therefore, the friction force that appears on the cart \( -{\vec F}_{f} \) It has the same magnitude of the friction force of the block, and this force depends on the normal reaction of the block \( {\vec{N}}_{B} \) and not of the normal reactions on the wheels of the cart, \( {\vec{N}}_{1} \) and \( {\vec{N}}_{2} \). If it existed friction between the wheels and the rails then this friction force would depend on the normal reactions on the wheels and mass of the cart \( F_{gC}=Mg \).

In the horizontal direction applying the expression (I) we have the friction force as the resultant
\[ \begin{gather} -F_{f}=Ma_{C} \tag{IX} \end{gather} \]
the friction force is given by the expression (VI)
\[ \begin{gather} -\mu N_{B}=Ma_{C} \tag{X} \end{gather} \]
substituting the expression (III) into expression (X)
\[ \begin{gather} -\mu mg=Ma_{C}\\ a_{C}=\frac{-{\mu mg}}{M} \tag{XI} \end{gather} \]
The block, under the action of the friction force with the cart, begins to accelerate from the rest to a final speed. The cart under the action of the friction force with the block begins to slow down to a final speed. The equation of velocity as a function of displacement is given by
\[ \bbox[#99CCFF,10px] {v^{2}=v_{0}^{2}+2a\Delta S} \]
writing this expression for the two bodies
\[ \begin{gather} v_{B}^{2}=v_{0B}^{2}+2a_{B}\Delta S_{B}\\ v_{B}^{2}=0+2\mu gL\\ v_{B}^{2}=2\mu gL \end{gather} \]
\[ \begin{gather} v_{C}^{2}=v_{0C}^{2}+2a_{C}\Delta S_{C}\\ v_{C}^{2}=v_{0}^{2}-2\frac{\mu mg}{M}L \end{gather} \]
Note: Why is the displacement of the block SB is equal to the displacement of the cart SC and equal to the length of the cart (L)?
"Forgetting" the reference frame on the ground, we will consider a point p on the rear of the cart and a point q on the front, and assuming another reference frame fixed at the point p on the cart (Figure 5). For an observer at p, looking forward it sees the block starting to move toward him with initial speed −v0, the speed magnitude will decrease until it is equal to zero, when the block reaches point p, v0>v1>v2>...>vf=0, so that the block does not fall from the cart. So the block will move all the length of the cart \( \Delta S_{B}=L \).
Attention: In Figure 5 it appears that the block moves back, in fact, the block moves forward under the action of the friction force. As seen from the reference frame in p the cart is fixed to the observer and the block and rails move back, in the same way when we have seated in a car the trees and the road signs are left behind.
Figure 5

Choosing a frame of reference on the block, q (Figure 6). For an observer on the block looking at the rear, it sees the point p starting to move towards with an initial speed v0, the magnitude of speed will decrease until the point p reaches the position of the block v0>v1>v2>...>vf=0, so that the block does not fall from the cart. So point p will move from a distance equal to the length of the cart \( \Delta S_{C}=L \)..
Figure 6

For the block does not fall from the cart, we must have the condition that when the block arrives on the rear of the cart, the final speeds of the block and cart are equal.
\[ \begin{gather} v_{B}^{2}=v_{C}^{2}\\ 2\mu gL=v_{0}^{2}-2\frac{\mu mg}{M}L\\ 2\mu gL+2\frac{\mu mg}{M}L=v_{0}^{2} \end{gather} \]
factoring the term 2μgL on the left-hand side of the equation
\[ 2\mu gL\left(1+\frac{m}{M}\right)=v_{0}^{2} \]
in the expression in parentheses, the common factor between 1 and M is M
\[ \begin{gather} 2\mu gL\left(\frac{M+m}{M}\right)=v_{0}^{2}\\ L\left(\frac{M+m}{M}\right)=\frac{v_{0}^{2}}{2\mu g}\\ L=\frac{v_{0}^{2}}{2\mu g}\left(\frac{M}{M+m}\right) \end{gather} \]
this is the minimum length so that the block does not fall from the cart, for any value greater than this block does not fall
\[ \bbox[#FFCCCC,10px] {L\geqslant \frac{v_{0}^{2}}{2\mu g}\left(\frac{M}{M+m}\right)} \]
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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .