Solved Problem on Dynamics
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FR
PT-BR
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In an Atwood machine, the two bodies, at rest on a horizontal surface, are connected by a rope of
negligible mass that passes over a frictionless pulley of negligible mass. The masses weight
mA = 24 kg and mB = 40 kg. Determine the body accelerations when:
a) F = 400 N;
b) F = 720 N;
c) F = 1200 N.
a) F = 400 N;
b) F = 720 N;
c) F = 1200 N.
Problem data:
- Mass of body A: mA = 24 kg;
- Mass of body B: mB = 40 kg;
- Acceleration due to gravity: g = 9.8 m/s2.
We choose a frame of reference oriented positively upward, in the same direction as the force \( \vec F \).
The force applied to a pulley is equally distributed between the two sides (Figure 1-A). So, the magnitude of the force on each side of the pulley will be \( \frac{\vec F}{2} \).
Assuming the rope has a negligible mass, it only transmits the force of the pulley to the bodies. Thus, the component of force \( \vec F \) over each body will be \( \frac{\vec F}{2} \) (Figure 1-B).
Drawing a free-bodies diagram for each block.
-
Body A (Figure 2):
- \( \dfrac{\vec{F}}{2} \): transmitted force from the pulley;
- \( {\vec W}_{\small A} \): weight of the body A.
-
Body B (Figura 2):
- \( \dfrac{\vec F}{2} \): transmitted force from the pulley;
- \( {\vec W}_{\small B} \): weight of body B.
Solution
Applying Newton's Second Law
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\vec F=m\vec a}
\end{gather}
\]
- Body A:
\[
\begin{gather}
\frac{F}{2}-W_{\small A}=m_{\small A}a_{\small A} \tag{III}
\end{gather}
\]
- Body B:
\[
\begin{gather}
\frac{F}{2}-W_{\small B}=m_{\small B}a_{\small B} \tag{II}
\end{gather}
\]
The weight is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{W=mg}
\end{gather}
\]
we have for bodies A and B
\[
\begin{gather}
W_{\small A}=m_{\small A}g \tag{III}
\end{gather}
\]
\[
\begin{gather}
W_{\small B}=m_{\small B}g \tag{IV}
\end{gather}
\]
substituting equation (III) into equation (I)
\[
\begin{gather}
\frac{F}{2}-m_{\small A}g=m_{\small A}a_{\small A} \tag{V}
\end{gather}
\]
substituting equation (IV) into equation (II)
\[
\begin{gather}
\frac{F}{2}-m_{\small B}g=m_{\small B}a_{\small B} \tag{VI}
\end{gather}
\]
a) If \( F=400\;\text{N} \), the acceleration of body A is given by equation (V)
\[
\begin{gather}
a_{\small A}=\frac{\dfrac{400\;\mathrm N}{2}-(24\;\mathrm kg)\times\left(9.8\;\mathrm{\frac{m}{s^2}}\right)}{24\;\mathrm{kg}}\\[5pt]
a_{\small A}=-{\frac{35.2\;\mathrm{\frac{\cancel{kg}.m}{s^2}}}{24\;\mathrm{\cancel{kg}}}}\\[5pt]
a_{\small A}=-1.5\;\mathrm{m/s^2}
\end{gather}
\]
For body B, using the equation (VI)
\[
\begin{gather}
a_{\small B}=\frac{\dfrac{400\;\mathrm N}{2}-(40\;\mathrm{kg})\times\left(9.8\;\mathrm{\frac{m}{s^2}}\right)}{40\;\mathrm{kg}}\\[5pt]
a_{\small B}=-{\frac{192\;\mathrm{\frac{\cancel{kg}.m}{s^2}}}{40\;\mathrm{\cancel{kg}}}}\\[5pt]
a_{\small B}=-4.8\;\mathrm{m/s^2}
\end{gather}
\]
As the accelerations are negative, the bodies must move against the direction of the frame (downward),
but as they are on a surface, they remain at rest, and their accelerations are zero
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{a_{\small A}=a_{\small B}=0}
\end{gather}
\]
b) If \( F=720\;\text{N} \), the acceleration of body A is given by equation (V)
\[
\begin{gather}
a_{\small A}=\frac{\dfrac{720\;\mathrm N}{2}-(24\;\mathrm{kg})\times\left(9.8\;\mathrm{\frac{m}{s^2}}\right)}{24\;\mathrm{kg}}\\[5pt]
a_{\small A}=\frac{124.8\;\mathrm{\frac{\cancel{kg}.m}{s^2}}}{24\;\mathrm{\cancel{kg}}}\\[5pt]
a_{\small A}=5.2\;\mathrm{m/s^2}
\end{gather}
\]
For body B, using the equation (VI)
\[
\begin{gather}
a_{\small B}=\frac{\dfrac{720\;\mathrm N}{2}-(40\;\mathrm{kg})\times\left(9.8\;\mathrm{\frac{m}{s^2}}\right)}{40\;\mathrm{kg}}\\[5pt]
a_{\small B}=-{\frac{32\;\mathrm{\frac{\cancel{kg}.m}{s^2}}}{40\;\mathrm{\cancel{kg}}}}\\[5pt]
a_{\small B}=-0.8\;\mathrm{m/s^2}
\end{gather}
\]
Body A has the acceleration
As the acceleration of body B is negative, it must move against the orientation of the frame (downward), but as it is on a surface, it remains at rest, and its acceleration will be zero
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{a_{\small A}=5.2\;\mathrm{m/s^2}}
\end{gather}
\]
As the acceleration of body B is negative, it must move against the orientation of the frame (downward), but as it is on a surface, it remains at rest, and its acceleration will be zero
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{a_{\small B}=0}
\end{gather}
\]
c) If \( F=1200\;\text{N} \), the acceleration of body A is given by the equation (IV)
\[
\begin{gather}
a_{\small A}=\frac{\dfrac{1200\;\mathrm N}{2}-(24\;\mathrm{kg})\times\left(9.8\;\mathrm{\frac{m}{s^2}}\right)}{24\;\mathrm{kg}}\\[5pt]
a_{\small A}=\frac{364.8\;\mathrm{\frac{\cancel{kg}.m}{s^2}}}{24\;\mathrm{\cancel{kg}}}\\[5pt]
a_{\small A}=15.2\;\mathrm{m/s^2}
\end{gather}
\]
For body B, using the equation (VI)
and body B has acceleration
\[
\begin{gather}
a_{\small B}=\frac{\dfrac{1200\;\mathrm N}{2}-(40\;\mathrm{kg})\times\left(9.8\;\mathrm{\frac{m}{s^2}}\right)}{40\;\mathrm{kg}}\\[5pt]
a_{\small B}=\frac{208\;\mathrm{\frac{\cancel{kg}.m}{s^2}}}{40\;\mathrm{\cancel{kg}}}\\[5pt]
a_{\small B}=4.3\;\mathrm{m/s^2}
\end{gather}
\]
Body A has acceleration
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{a_{\small A}=15.2\;\mathrm{m/s^2}}
\end{gather}
\]
and body B has acceleration
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{a_{\small B}=4.3\;\mathrm{m/s^2}}
\end{gather}
\]
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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 .