Solved Problem on Dynamics

A cart with mass M is attached by a rope to a load with mass m. In the initial instant, the cart has speed v₀ and moves to the left in a horizontal plane. Determine:
a) The interval of time until the cart stop;
b) The displacement until the cart stop.
Considering the cord inextensible and negligible mass, there is no friction in the horizontal plane and on the pulley, and assume the acceleration due to gravity is equal to g.

Problem data:

Mass of cart: M;
Initial speed of cart: v₀;
Mass of load: m;
Acceleration due to gravity: g.

Problem diagram:

We choose a frame of reference pointing to the right, and with origin from the point where the cart is initially. We choose the acceleration in the direction in which the load is descending.

Figure 1

Solution

Drawing free-bodies diagrams, we have the forces that act in each of them, we apply Newton's Second Law

\[ \begin{gather} \bbox[#99CCFF,10px] {\vec{F}=m\vec{a}} \tag{I} \end{gather} \]

Cart:

Vertical direction:

\( {\vec F}_{gM} \): gravitational force on the cart;
\( {\vec N}_{M} \): normal reaction force due to the contact between the wheels and the surface.

Horizontal direction:

\( \vec{T} \): tension force on the rope.

In the vertical direction, there is no motion, the gravitational force \( {\vec F}_{gM} \), and normal reaction force \( {\vec N}_{M} \) cancel.

Figure 2

In the horizontal direction applying the expression (I)

\[ \begin{gather} T=Ma \tag{II} \end{gather} \]

Load:

\( {\vec F}_{gm} \): gravitational force on the load;
\( \vec{T} \): tension force on the rope.

In the horizontal direction, no forces are acting. In the vertical direction, applying the expression (I)

\[ \begin{gather} F_{gm}-T=ma \tag{III} \end{gather} \]

Figure 3

a) Equations (II) and (III) can be written as a system of two equations to two unknowns, T and a

\[ \begin{gather} \left\{ \begin{array}{l} T=Ma\\ F_{gm}-T=ma \end{array} \right. \end{gather} \]

ddding the two equations

\[ \begin{gather} \frac{ \begin{aligned} \cancel{T}=Ma\\ F_{gm}-\cancel{T}=ma \end{aligned} } {F_{gm}=\left(M+m\right)a}\\ a=\frac{F_{gm}}{M+m} \end{gather} \]

The gravitational force is given by

\[ \bbox[#99CCFF,10px] {F_{g}=mg} \]

for gravitational force on the load

\[ F_{gm}=mg \]

substituting this value in the above expression for acceleration

\[ \begin{gather} a=\frac{mg}{M+m} \tag{IV} \end{gather} \]

From the initial situation, it is understood that the cart was launched by some force to the left. At a given moment this force stopped acting, and when it started counting the time the speed had magnitude v₀. At the instant in which the launching force stops only the tension force acts on the rope, due to the load, it acts on the cart with an acceleration that opposes the motion (deceleration), thus the cart moves to the right.
The cart moves initially against the orientation of the trajectory, so its speed is negative −v₀, the instant it stops its final speed will be zero (v = 0).
The equation of velocity as a function of time is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {v=v_{0}+at} \tag{V} \end{gather} \]

substituting the expression (IV) and values of the initial and final speeds into the expression (V)

\[ \begin{gather} 0=-v_{0}+\frac{mg}{M+m}t\\ \frac{mg}{M+m}t=v_{0} \end{gather} \]

\[ \bbox[#FFCCCC,10px] {t=v_{0}\frac{(M+m)}{mg}} \]

b) Using the equation of displacement as a function of time with constant acceleration

\[ \begin{gather} \bbox[#99CCFF,10px] {S=S_{0}+v_{0}t+\frac{a}{2}t^{2}} \tag{VI} \end{gather} \]

substituting the initial position, the acceleration found in (IV), and the result of the previous item for the time interval into (VI)

\[ \begin{gather} S=0-v_{0}\left[v_{0}\frac{(M+m)}{mg}\right]+\frac{1}{2}\frac{mg}{(M+m)}\left[v_{0}\frac{(M+m)}{mg}\right]^{2}\\[10pt] S=-v_{0}^{2}\frac{(M+m)}{mg}+\frac{1}{2}\frac{mg}{(M+m)}v_{0}^{2}\frac{(M+m)^{2}}{m^{2}g^{2}}\\[10pt] S=-v_{0}^{2}\frac{(M+m)}{mg}+\frac{1}{2}v_{0}^{2}\frac{(M+m)}{mg} \end{gather} \]

factoring the term \( v_{0}^{2}\frac{(M+m)}{mg} \) on the right-hand side of the equation

\[ S=v_{0}^{2}\frac{(M+m)}{mg}\left(-1+\frac{1}{2}\right) \]

multiplying and dividing by 2 the factor −1 in parentheses

\[ \begin{gather} S=v_{0}^{2}\frac{(M+m)}{mg}\left(\frac{2}{2}\times(-1)+\frac{1}{2}\right)\\ S=v_{0}^{2}\frac{(M+m)}{mg}\left(\frac{-2+1}{2}\right) \end{gather} \]

\[ \bbox[#FFCCCC,10px] {S=-{\frac{1}{2}}v_{0}^{2}\frac{(M+m)}{mg}} \]

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .