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.
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.
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.
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.
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\vec{F}=m\vec{a}} \tag{I}
\end{gather}
\]
Block:
- \( {\vec F}_{gB} \): gravitational force on the block;
- \( {\vec N}_{B} \): normal reaction force;
- \( {\vec F}_{f} \): force of friction.
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
\[
\begin{gather}
\bbox[#99CCFF,10px]
{F_{g}=mg}
\end{gather}
\]
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
\[
\begin{gather}
\bbox[#99CCFF,10px]
{F_{f}=\mu N}
\end{gather}
\]
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)
\[
\begin{gather}
\mu \cancel{m}g=\cancel{m}a_{B}\\[5pt]
a_{B}=\mu g \tag{VIII}
\end{gather}
\]
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, \( \left|\;{\vec{F}}_{f}\;\right|=\left|\;-{\vec{F}}_{f}\;\right| \).
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 force of friction 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)
\[
\begin{gather}
-\mu mg=Ma_{C}\\[5pt]
a_{C}=\frac{-{\mu mg}}{M} \tag{XI}
\end{gather}
\]
as the block is under the action of constant acceleration, the equation of velocity as a function of time is
given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{v=v_{0}+at}
\end{gather}
\]
for the block will be
\[
\begin{gather}
v_{B}=v_{0B}+a_{B}t\\[5pt]
v_{B}=0+\mu gt\\[5pt]
v_{B}=\mu gt \tag{XII}
\end{gather}
\]
the equation for cart will be
\[
\begin{gather}
v_{C}=v_{0C}+a_{C}t\\[5pt]
v_{C}=v_{0}-\frac{\mu mg}{M}t \tag{XIII}
\end{gather}
\]
The cart will decrease speed, and the block increase the speed until the two bodies move at the same speed,
then, in a given instant we have the condition
\[
\begin{gather}
v_{B}=v_{C}\\[5pt]
\mu gt=v_{0}-\frac{\mu mg}{M}t\\[5pt]
\mu gt+\frac{\mu mg}{M}t=v_{0}\\[5pt]
\mu gt\left(1+\frac{m}{M}\right)=v_{0}
\end{gather}
\]
in the expression in parentheses, the common factor between 1 and M is M
\[
\begin{gather}
\mu gt\left(\frac{M+m}{M}\right)=v_{0}\\[5pt]
t=\frac{v_{0}}{\mu g}\left(\frac{M}{M+m}\right) \tag{XIV}
\end{gather}
\]
The equation of displacement as a function of time with constant acceleration is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{S=S_{0}+v_{0}t+\frac{a}{2}t^{2}}
\end{gather}
\]
for the block
\[
\begin{gather}
S_{B}=S_{0B}+v_{0B}t+\frac{a_{B}}{2}t^{2}
\end{gather}
\]
In Figure 1, we see the initial position of the block is on the front of the cart
\( S_{0B}=L \)
and using the initial speed
of the block given in the problem and the acceleration calculated in (VI)
\[
\begin{gather}
S_{B}=L+0.t+\frac{\mu g}{2}t^{2}\\[5pt]
S_{B}=L+\frac{\mu g}{2}t^{2} \tag{XV}
\end{gather}
\]
substituting the value of the time obtained in (XIV) into the expression (XV)
\[
\begin{gather}
S_{B}=L+\frac{\mu g}{2}\left[\frac{v_{0}}{\mu g}\left(\frac{M}{M+m}\right)\right]^{2}\\[5pt]
S_{B}=L+\frac{\mu g}{2}\frac{v_{0}^{2}}{\mu^{2}g^{2}}\left(\frac{M}{M+m}\right)^{2}\\[5pt]
S_{B}=L+\frac{v_{0}^{2}}{2\mu g}\left(\frac{M}{M+m}\right)^{2} \tag{XVI}
\end{gather}
\]
The equation for the cart is given by
\[
\begin{gather}
S_{C}=S_{0C}+v_{0C}t+\frac{a_{C}}{2}t^{2}
\end{gather}
\]
In Figure 1, we see the initial position of the cart is at the origin of the reference frame,
S0C=0, and using the initial speed of the cart given in the problem and the
acceleration calculated in (IX)
\[
\begin{gather}
S_{C}=0+v_{0}t-\frac{1}{2}\frac{\mu mg}{M}t^{2}\\[5pt]
S_{C}=v_{0}t-\frac{1}{2}\frac{\mu mg}{M}t^{2} \tag{XVII}
\end{gather}
\]
substituting the value of time obtained in (XIV) into the expression (XVII)
\[
\begin{gather}
S_{C}=v_{0}\frac{v_{0}}{\mu g}\left(\frac{M}{M+m}\right)-\frac{1}{2}\frac{\mu mg}{M}\left[\frac{v_{0}}{\mu g}\left(\frac{M}{M+m}\right)\right]^{2}\\[5pt]
S_{C}=\frac{v_{0}^{2}}{\mu g}\left(\frac{M}{M+m}\right)-\frac{1}{2}\frac{\mu mg}{M}\frac{v_{0}^{2}}{\mu^{2}g^{2}}\left(\frac{M}{M+m}\right)^{2}\\[5pt]
S_{C}=\frac{v_{0}^{2}}{\mu g}\left(\frac{M}{M+m}\right)-\frac{1}{2}\frac{m}{M}\frac{v_{0}^{2}}{\mu g}\left(\frac{M}{M+m}\right)^{2} \tag{XVIII}
\end{gather}
\]
So that the block does not fall from the cart we must have the following condition
\[
\begin{gather}
S_{B}-S_{C}\geqslant 0
\end{gather}
\]
\[
\begin{gather}
L+\frac{v_{0}^{2}}{2\mu g}\left(\frac{M}{M+m}\right)^{2}-\left[\frac{v_{0}^{2}}{\mu g}\left(\frac{M}{M+m}\right)-\frac{1}{2}\frac{m}{M}\frac{v_{0}^{2}}{\mu g}\left(\frac{M}{M+m}\right)^{2}\right]\geqslant 0\\[5pt]
L+\frac{v_{0}^{2}}{2\mu g}\left(\frac{M}{M+m}\right)^{2}-\frac{v_{0}^{2}}{\mu g}\left(\frac{M}{M+m}\right)+\frac{m}{M}\frac{v_{0}^{2}}{2\mu g}\left(\frac{M}{M+m}\right)^{2}\geqslant 0
\end{gather}
\]
factoring the term
\( \frac{v_{0}^{2}}{\mu g}\left(\frac{M}{M+m}\right) \)
on the left-hand side of inequation
\[
\begin{gather}
L+\frac{v_{0}^{2}}{\mu g}\left(\frac{M}{M+m}\right)\left[\frac{1}{2}\left(\frac{M}{M+m}\right)-1+\frac{m}{M}\frac{1}{2}\left(\frac{M}{M+m}\right)\right]\geqslant 0\\[5pt]
L+\frac{v_{0}^{2}}{\mu g}\left(\frac{M}{M+m}\right)\left[\frac{1}{2}\left(\frac{M}{M+m}\right)-1+\frac{1}{2}\left(\frac{m}{M+m}\right)\right]\geqslant 0\\[5pt]
L+\frac{v_{0}^{2}}{\mu g}\left(\frac{M}{M+m}\right)\left[\frac{M}{2(M+m)}+\frac{m}{2(M+m)}-1\right]\geqslant 0\\[5pt]
L+\frac{v_{0}^{2}}{\mu g}\left(\frac{M}{M+m}\right)\left[\frac{\cancel{M+m}}{2\cancel{(M+m)}}-1\right]\geqslant 0\\[5pt]
L+\frac{v_{0}^{2}}{\mu g}\left(\frac{M}{M+m}\right)\left[\frac{1}{2}-1\right]\geqslant 0
\end{gather}
\]
in the expression between brackets, the Least Common Multiple (LCM) between 2 and 1 is 2
\[
\begin{gather}
L+\frac{v_{0}^{2}}{\mu g}\left(\frac{M}{M+m}\right)\left[\frac{1-2}{2}\right]\geqslant 0\\[5pt]
L+\frac{v_{0}^{2}}{\mu g}\left(\frac{M}{M+m}\right)\left[-{\frac{1}{2}}\right]\geqslant 0\\[5pt]
L-\frac{v_{0}^{2}}{2\mu g}\left(\frac{M}{M+m}\right)\geqslant 0
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{L\geqslant \frac{v_{0}^{2}}{2\mu g}\left(\frac{M}{M+m}\right)}
\end{gather}
\]
Note: As seen in the problem, the cart has an initial speed of v0, and the
block is placed on the cart with a speed equal to zero (Figure 5-A). Under the action of the friction
force, the cart decelerates and the block accelerates until both reach the same speed. At this moment the
block has the same speed as the cart relative to the ground, and speed equal to zero relatives to the cart
itself (the block stop slipping). The block seems to slip back from the cart, but in fact, the friction
force makes the block advance relative to the ground, the cart, however, advances faster.
In Figure 5-B, the instant the speeds of the bodies are equal the cart is at a point of the trajectory SC and the block in the point SB. If the difference of the positions is positive SB−SC>0, this means that the block stopped slipping at any point on the cart.
Figure 5
In Figure 5-C, if the difference of the positions is equal to zero SB−SC=0, this means that the block stopped slipping at the rear end of the cart, where the reference was taken to the cart.
In Figure 5-D, if the difference of the positions is negative SB−SC<0, this means that the block continued to slip and fell from the cart (the cart was not long enough for the speeds to equate).
Attention: Figure 5-D is just a scheme of what happens to the block, the equations used to describe only the straight motion, not the movement of the block falling from the cart.
In Figure 5-B, the instant the speeds of the bodies are equal the cart is at a point of the trajectory SC and the block in the point SB. If the difference of the positions is positive SB−SC>0, this means that the block stopped slipping at any point on the cart.
In Figure 5-C, if the difference of the positions is equal to zero SB−SC=0, this means that the block stopped slipping at the rear end of the cart, where the reference was taken to the cart.
In Figure 5-D, if the difference of the positions is negative SB−SC<0, this means that the block continued to slip and fell from the cart (the cart was not long enough for the speeds to equate).
Attention: Figure 5-D is just a scheme of what happens to the block, the equations used to describe only the straight motion, not the movement of the block falling from the cart.
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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 .