Solved Problem on Dynamics
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A rope passes through a fixed pulley in the ceiling, at one end there is a block of
mA = 36 kg, and at the other end a pulley 2. By this second pulley passes a rope in
whose ends are mB = 16 kg and mC = 8 kg (this system is a double
Atwood machine). Calculate the accelerations of the masses and the tension force on the rope. Neglect
the masses and frictions on the pulleys, assuming the acceleration due to gravity is equal to
10 m/s2.
Problem data:
- Mass of body A: mA = 36 kg;
- Mass of body B: mB = 16 kg;
- Mass of body C: mC = 8 kg;
- Acceleration due to gravity: g = 10 m/s2.
The gravitational force of body C
\( {\vec F}_{gC} \)
produces a tension force
\( \vec{T} \)
on the rope, this is transferred by the rope to the other side of the mobile pulley, this tension acts
on the rope that holds the body B where acts the gravitational force
\( {\vec F}_{gB} \).
To balance these two tensions on the rope of the mobile pulley acts the tension force
\( 2\vec{T} \).
This tension is transmitted by the rope, that passes through the second pulley (fixed), to the other
side of the pulley where acts the gravitational force
\( {\vec F}_{gA} \)
due to body A. To balance these two tensions, we have on the rope on the fixed pulley a tension
equal to
\( 4\vec{T} \)
(Figure 1).
Note: If we began through the body A, on the rope we would have a tension force
\( \vec{T} \),
transmitted to the other side of the fixed pulley, and balanced by tension
\( 2\vec{T} \)
on the ceiling. In the mobile pulley, we would have the tension
\( \vec{T} \)
transmitted by the rope and divided into
\( \frac{\vec{T}}{2} \)
for bodies B and C.
Solution
As block A has a mass greater than the sum of the other blocks, it descends with acceleration
\( {\vec{a}}_{A} \),
the rope passing through pulley 1 pulls the pulley with the same acceleration (Figure 2). In the second
pulley, block B has a mass greater than block C, then block B descents with
\( {\vec{a}}_{B} \)
acceleration and block C climbs with acceleration
\( {\vec{a}}_{C} \)
(these accelerations are relative to pulley 1).
Drawing a free-body diagram, we have the forces that act in each of them, and we apply Newton's Second Law
Drawing a free-body diagram, we have the forces that act in each of them, and we apply Newton's Second Law
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\vec{F}=m\vec{a}} \tag{I}
\end{gather}
\]
Body A:
We choose the positive direction in the same direction of acceleration (Figure 3). In the horizontal direction, no forces are acting. In the vertical direction applying the expression (I)
- \( {\vec F}_{gA} \): gravitational force on body A;
- \( \vec{T} \): tension force on the rope.
We choose the positive direction in the same direction of acceleration (Figure 3). In the horizontal direction, no forces are acting. In the vertical direction applying the expression (I)
\[
\begin{gather}
F_{gA}-2T=m_{A}a_{A} \tag{II}
\end{gather}
\]
Body B:
- \( {\vec F}_{gB} \) : gravitational force on body B;
- \( \vec{T} \) : tension force on the rope.
\[
\begin{gather}
F_{gB}-T=m_{B}a_{B} \tag{III}
\end{gather}
\]
Body C:
- \( {\vec F}_{gC} \): gravitational force on body C;
- \( \vec{T} \): tension force on the rope.
\[
\begin{gather}
T-F_{gC}=m_{C}a_{C} \tag{IV}
\end{gather}
\]
Note: Equations (II), (III), and (IV) can be written as a system of three equations to
four unknowns (aA, aB, aC and T), as there
are more unknowns than equations that system will be undetermined.
We consider the movements of blocks B and C relative to the mobile pulley and the fixed pulley. We choose a frame of reference RI on the fixed pulley, an inertial frame of reference, and a frame of reference RN in the mobile pulley, a non-inertial reference frame (Figure 6).
Pulley 2 has the same acceleration as block A, so the acceleration of the reference frame on this pulley is itself from the acceleration of block A
\[
\begin{gather}
a_{{R}_{\text{N}}}=a_{A} \tag{V}
\end{gather}
\]
The blocks B and C have, relative to the non-inertial reference frame, the same
acceleration
\( \vec{a} \),
in which one rises from one side the other descends from the other side.Acceleration relative to the inertial frame of reference will be the sum of the acceleration of the block relative to the non-inertial reference frame with the acceleration of the non-inertial reference frame itself.
Body B:
\[
\begin{gather}
a_{B}=a_{B/R_{\text{N}}}+a_{R_{\text{N}}} \tag{VI}
\end{gather}
\]
the acceleration relative to the non-inertial reference frame will be
\[
\begin{gather}
a_{B/R_{\text{N}}}=-a \tag{VII}
\end{gather}
\]
substituting expressions (V) and (VII) into expression (VI)
\[
\begin{gather}
a_{B}=-a+a_{A} \tag{VIII}
\end{gather}
\]
Body C:
\[
\begin{gather}
a_{C}=a_{C/R_{\text{N}}}+a_{R_{\text{N}}} \tag{IX}
\end{gather}
\]
the acceleration relative to the non-inertial reference frame will be
\[
\begin{gather}
a_{C/R_{\text{N}}}=a \tag{X}
\end{gather}
\]
substituting expressions (V) and (X) into expression (IX)
\[
\begin{gather}
a_{C}=a+a_{A} \tag{XI}
\end{gather}
\]
Adding the expressions (VIII) and (XI)
\[
\begin{gather}
\frac{
\begin{matrix}
a_{B}=-a+a_{A}\\
a_{C}=a+a_{A}
\end{matrix}}
{a_{B}+a_{C}=0+2a_{A}}\\[5pt]
2a_{A}=a_{B}+a_{C} \tag{XII}
\end{gather}
\]
Equations (II), (III), (IV), and (XII) can be written as a system of four equations to four unknowns
(aA, aB, aC and T)
\[
\begin{gather}
\left\{
\begin{matrix}
\;F_{gA}-2T=m_{A}a_{A}\\
\;F_{gB}-T=m_{B}a_{B}\\
\;T-F_{gC}=m_{C}a_{C}\\
\;2a_{A}=a_{B}+a_{C} \tag{XIII}
\end{matrix}
\right.
\end{gather}
\]
solving the first, second, and third equations of system (XIII) for accelerations
aA, aB, and aC, and substituting them in the fourth
equation
\[
\begin{gather}
a_{A}=\frac{F_{gA}-2T}{m_{A}} \tag{XIV}
\end{gather}
\]
\[
\begin{gather}
a_{B}=\frac{F_{gB}-T}{m_{B}} \tag{XV}
\end{gather}
\]
\[
\begin{gather}
a_{C}=\frac{T-F_{gC}}{m_{C}} \tag{XVI}
\end{gather}
\]
\[
\begin{gather}
2\left(\frac{F_{gA}-2T}{m_{A}}\right)=\frac{F_{gB}-T}{m_{B}}+\frac{T-F_{gC}}{m_{C}}\\[5pt]
\frac{2 F_{gA}-4T}{m_{A}}=\frac{F_{gB}-T}{m_{B}}+\frac{T-F_{gC}}{m_{C}}
\end{gather}
\]
The gravitational force is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{F_{g}=mg}
\end{gather}
\]
Writing the gravitational forces of bodies A, B and C
\[
\begin{gather}
F_{gA}=m_{A}g \tag{XVII-a}
\end{gather}
\]
\[
\begin{gather}
F_{gB}=m_{B}g \tag{XVII-b}
\end{gather}
\]
\[
\begin{gather}
F_{gC}=m_{C}g \tag{XVII-c}
\end{gather}
\]
substituting these expressions
\[
\begin{gather}
\frac{2m_{A}g-4T}{m_{A}}=\frac{m_{B}g-T}{m_{B}}+\frac{T-m_{C}g}{m_{C}}
\end{gather}
\]
the common factor between mA, mB and mC is
mA.mB.mC
\[
\begin{gather}
\frac{2m_{A}m_{B}m_{C}g-4Tm_{B}m_{C}}{\cancel{m_{A}m_{B}m_{C}}}=\frac{m_{A}m_{B}m_{C}g-Tm_{A}m_{C}+Tm_{A}m_{B}-m_{A}m_{B}m_{C}g}{\cancel{m_{A}m_{B}m_{C}}}\\[5pt]
2m_{A}m_{B}m_{C}g-4Tm_{B}m_{C}=-Tm_{A}m_{C}+Tm_{A}m_{B}\\[5pt]
2m_{A}m_{B}m_{C}g=-Tm_{A}m_{C}+Tm_{A}m_{B}+4Tm_{B}m_{C}\\[5pt]
2m_{A}m_{B}m_{C}g=T(-m_{A}m_{C}+m_{A}m_{B}+4m_{B}m_{C})\\[5pt]
T=\frac{2m_{A}m_{B}m_{C}g}{-m_{A}m_{C}+m_{A}m_{B}+4m_{B}m_{C}}
\end{gather}
\]
substituting numeric values data in the problem
\[
\begin{gather}
T=\frac{2\times 36\times 16\times 8\times 10}{-36\times 8+36\times 16+4\times 16\times 8}\\[5pt]
T=\frac{92160}{800}\\[5pt]
T=115,2\;\text{N}
\end{gather}
\]
The tension force on the rope attached to bodies B and C will be
T = 115.2 N,
the tension force on the rope attached to body A and the pulley 2 will be
2T = 2×115.2 = 230.4 N
and the tension force on the rope attached to the pulley 1, fixed on the ceiling, will be
4T = 4×115.2 = 460.8 N .Substituting gravitational forces given by expressions in (XIV) in expressions (XIV), (XV), and (XVI), the tension obtained above, and the problem data, we have the accelerations of the blocks
\[
\begin{gather}
a_{A}=\frac{m_{A}g-2T}{m_{A}}\\[5pt]
a_{A}=\frac{36\times 10-2\times 115.2}{36}\\[5pt]
a_{A}=\frac{360-230.4}{36}\\[5pt]
a_{A}=\frac{129.6}{36}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{a_{A}=3.6\;\text{m/s}^{2}}
\end{gather}
\]
\[
\begin{gather}
a_{B}=\frac{m_{B}g-T}{m_{B}}\\[5pt]
a_{B}=\frac{16\times 10-115.2}{16}\\[5pt]
a_{B}=\frac{160-115.2}{16}\\[5pt]
a_{B}=\frac{44.8}{16}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{a_{B}=2.8\;\text{m/s}^{2}}
\end{gather}
\]
\[
\begin{gather}
a_{C}=\frac{T-m_{C}g}{m_{C}}\\[5pt]
a_{C}=\frac{115.2-8\times 10}{8}\\[5pt]
a_{C}=\frac{115.2-80}{8}\\[5pt]
a_{C}=\frac{35.2}{8}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{a_{C}=4.4\;\text{m/s}^{2}}
\end{gather}
\]
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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 .