Solved Problem on Dynamics
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In the system shown in the figure, pulleys 1 and 2 are massless and without friction, pulley 1 is fixed
and pulley 2 is mobile. Block A has a mass of 11 kg and block B rises with an acceleration
of 1 m/s2. Determine the acceleration of block A, the mass of block B, and the
tension force on the rope. Assume the acceleration due to gravity is equal to 10 m/s2.
Problem data:
- Mass of body A: mA = 11 kg;
- Acceleration of body B: aB = 1 m/s2;
- Acceleration due to gravity: g = 10 m/s2.
The gravitational force of the body A
\( {\vec F}_{gA} \)
produces a tension force
\( \vec{T} \)
on the rope, it is transferred by the rope to the other side of the pulley attached to the ceiling. This
tension is transmitted by the rope passing through the second pulley (mobile) to the other side of the
pulley. To balance these two traces we have on the rope that leaves the pulley a tension equal to the
\( 2\vec{T} \),
this tension acts on the rope that holds the body B where acts the gravitational force
\( {\vec F}_{gB} \)
(Figure 1).
Solution
When the block A drops a distance h a point C in the rope descends h
(Figure 2), as this pulley is fixed on the ceiling, the rope on the other side of the pulley rises
h. As the second pulley is free, a point D on the rope should rise h, from this
value
\( \frac{h}{2} \)
is the point displacement, relative to the pulley, and the other
\( \frac{h}{2} \)
are resulting from the pulley rising itself. So we have the condition
\[
\begin{gather}
\Delta S_{A}=2\Delta S_{B} \tag{I}
\end{gather}
\]
From Kinematics, we have the equation of displacement as a function of time with a constant acceleration
\[
\begin{gather}
\bbox[#99CCFF,10px]
{S=S_{0}+v_{0}t+\frac{a}{2}t^{2}}
\end{gather}
\]
\[
\begin{gather}
S-S_{0}=v_{0}t+\frac{a}{2}t^{2}
\end{gather}
\]
as
\( \Delta S=S-S_{0} \)
\[
\begin{gather}
\Delta S=v_{0}t+\frac{a}{2}t^{2} \tag{II}
\end{gather}
\]
Writing the expression (II) for each of the blocks
\[
\begin{gather}
\Delta S_{A}=v_{0A}t+\frac{a_{A}}{2}t^{2} \tag{III-a}\\[10pt]
\Delta S_{B}=v_{0B}t+\frac{a_{B}}{2}t^{2} \tag{III-b}
\end{gather}
\]
assuming that blocks begin the motion from rest, their speeds will be zero,
\( v_{0A}=v_{0B}=0 \),
substituting these values the expressions (III-a) and (III-b) reduces to
\[
\begin{gather}
\Delta S_{A}=0\times t+\frac{a_{A}}{2}t^{2}\\[5pt]
\Delta S_{A}=\frac{a_{A}}{2}t^{2} \tag{IV-a}\\[10pt]
\Delta S_{B}=0\times t+\frac{a_{B}}{2}t^{2}\\[5pt]
\Delta S_{B}=\frac{a_{B}}{2}t^{2} \tag{IV-b}
\end{gather}
\]
substituting expressions (IV-a) and (IV-b) in the condition (I)
\[
\begin{gather}
\frac{a_{A}}{2}t^{2}=2\frac{a_{B}}{2}t^{2}\\[5pt]
\frac{a_{A}}{2}t^{2}=a_{B}t^{2}\\[5pt]
a_{A}\cancel{t^{2}}=2a_{B}\cancel{t^{2}}\\[5pt]
a_{A}=2a_{B}
\end{gather}
\]
substituting the value of the acceleration for body B
\[
\begin{gather}
a_{A}=2\times 1
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{a_{A}=2\;\text{m/s}^{2}}
\end{gather}
\]
Drawing a free-body diagram, we have the forces that act in each of them, and we apply
Newton's Second Law
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\vec{F}=m\vec{a}} \tag{V}
\end{gather}
\]
Body A:
- \( {\vec F}_{gA} \): gravitational force on body A;
- \( \vec{T} \): tension force on the rope.
\[
\begin{gather}
F_{gA}-T=m_{A}a_{A} \tag{VI}
\end{gather}
\]
Body B:
- \( {\vec F}_{gB} \): gravitational force on body B;
- \( \vec{T} \): tension force on the rope.
\[
\begin{gather}
2T-F_{gB}=m_{B}a_{B} \tag{VII}
\end{gather}
\]
The equations (VI) and (VII) can be written as a system of two equations with two variables, T and mB
\[
\left\{
\begin{matrix}
F_{gA}-T=m_{A}a_{A}\\[5pt]
2T-F_{gB}=m_{B}a_{B}
\end{matrix}
\right.
\]
multiplying the first equation by 2, and adding the two equations
\[
\begin{gather}
\qquad\;\;\left\{
\begin{array}{l}
\;F_{gA}-T=m_{A}a_{A}\qquad (\times 2)\\
\;2T-F_{gB}=m_{B}a_{B}
\end{array}
\right.\\[10pt]
\frac{
\begin{matrix}
\;2F_{gA}-\cancel{2T}=2m_{A}a_{A}\\
\;\cancel{2T}-F_{gB}=m_{B}a_{B}
\end{matrix}
}
{2F_{gA}-F_{gB}=2m_{A}a_{A}+m_{B}a_{B}} \tag{VIII}
\end{gather}
\]
The gravitational force is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{F_{g}=mg} \tag{IX}
\end{gather}
\]
Applying the expression (IX) to bodies A and B
\[
\begin{gather}
F_{gA}=m_{A}g \tag{X-a}
\end{gather}
\]
\[
\begin{gather}
F_{gB}=m_{B}g \tag{X-b}
\end{gather}
\]
substituting expressions (X-a) and (X-b) into expression (VIII)
\[
\begin{gather}
2m_{A}g-m_{B}g=2m_{A}a_{A}+m_{B}a_{B}\\[5pt]
m_{B}a_{B}+m_{B}g=2m_{A}g-2m_{A}a_{A}
\end{gather}
\]
factoring the term −mB on the left-hand side of the equation and
2mA on the right-hand side
\[
\begin{gather}
m_{B}\left(a_{B}+g\right)=2m_{A}\left(g-a_{A}\right)\\[5pt]
m_{B}=\frac{2m_{A}\left(g-a_{A}\right)}{\left(a_{B}+g\right)}
\end{gather}
\]
substituting the values given in the problem and the acceleration of the body A found above
\[
\begin{gather}
m_{B}=\frac{2\times 11\times \left(10-2\right)}{\left(1+10\right)}\\[5pt]
m_{B}=\frac{22\times \left(8\right)}{11}\\[5pt]
m_{B}=2\times 8
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{m_{B}=16\;\text{kg}}
\end{gather}
\]
Solving the expression (VI) for the tension force and substituting the gravitational force of the body
A by the value given in (X-a)
\[
\begin{gather}
T=m_{A}g-m_{A}a_{A}\\[5pt]
T=11\times 10-11a\times\\[5pt]
T=110-22\\[5pt]
T=88\;\text{N}
\end{gather}
\]
The tension force on the rope attached to body A will be
T = 88 N
and the tension on the rope attached to body B will be
2T = 2\times 88 = 176 N.
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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 .