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 weights of the masses are
ma = 24 kg and mb = 40 kg. Determine the body accelerations when:
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.
Problem diagram:
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;
- \( {\vec W}_a \): weight of the body A.
-
Body B (Figure 3):
- \( \dfrac{\vec F}{2} \): transmitted force from the pulley;
- \( {\vec W}_b \): weight of body B.
Solution:
Applying Newton's Second Law
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\vec F=m\vec a}
\end{gather}
\]
\[
\begin{gather}
\frac{F}{2}-W_a=m_aa_a \tag{I}
\end{gather}
\]
\[
\begin{gather}
\frac{F}{2}-W_b=m_ba_b \tag{II}
\end{gather}
\]
The weight is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{W=mg}
\end{gather}
\]
we have for bodies, A and B
\[
\begin{gather}
W_a=m_ag \tag{III} \\[10pt]
W_b=m_bg \tag{IV}
\end{gather}
\]
substituting equation (III) into equation (I)
\[
\begin{gather}
\frac{F}{2}-m_ag=m_aa_a \tag{V}
\end{gather}
\]
substituting equation (IV) into equation (II)
\[
\begin{gather}
\frac{F}{2}-m_bg=m_ba_b \tag{VI}
\end{gather}
\]
a) If
\( F=400\;\text N \),
the acceleration of body A is given by equation (V)
\[
\begin{gather}
a_a=\frac{\dfrac{400\;\mathrm N}{2}-(24\;\mathrm kg)\times\left(9.8\;\mathrm{\frac{m}{s^2}}\right)}{24\;\mathrm{kg}} \\[5pt]
a_a=-{\frac{35.2\;\mathrm{\frac{\cancel{kg}.m}{s^2}}}{24\;\mathrm{\cancel{kg}}}} \\[5pt]
a_a=-1.5\;\mathrm{m/s^2}
\end{gather}
\]
For body B, using the equation (VI)
\[
\begin{gather}
a_b=\frac{\dfrac{400\;\mathrm N}{2}-(40\;\mathrm{kg})\times\left(9.8\;\mathrm{\frac{m}{s^2}}\right)}{40\;\mathrm{kg}} \\[5pt]
a_b=-{\frac{192\;\mathrm{\frac{\cancel{kg}.m}{s^2}}}{40\;\mathrm{\cancel{kg}}}} \\[5pt]
a_b=-4.8\;\mathrm{m/s^2}
\end{gather}
\]
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
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{a_a=a_b=0}
\end{gather}
\]
b) If
\( F=720\;\text N \),
the acceleration of body A is given by equation (V)
\[
\begin{gather}
a_a=\frac{\dfrac{720\;\mathrm N}{2}-(24\;\mathrm{kg})\times\left(9.8\;\mathrm{\frac{m}{s^2}}\right)}{24\;\mathrm{kg}} \\[5pt]
a_a=\frac{124.8\;\mathrm{\frac{\cancel{kg}.m}{s^2}}}{24\;\mathrm{\cancel{kg}}} \\[5pt]
a_a=5.2\;\mathrm{m/s^2}
\end{gather}
\]
For body B, using the equation (VI)
\[
\begin{gather}
a_b=\frac{\dfrac{720\;\mathrm N}{2}-(40\;\mathrm{kg})\times\left(9.8\;\mathrm{\frac{m}{s^2}}\right)}{40\;\mathrm{kg}} \\[5pt]
a_b=-{\frac{32\;\mathrm{\frac{\cancel{kg}.m}{s^2}}}{40\;\mathrm{\cancel{kg}}}} \\[5pt]
a_b=-0.8\;\mathrm{m/s^2}
\end{gather}
\]
Body
A has the acceleration
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{a_a=5.2\;\mathrm{m/s^2}}
\end{gather}
\]
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
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{a_b=0}
\end{gather}
\]
c) If
\( F=1200\;\text N \),
the acceleration of body A is given by the equation (IV)
\[
\begin{gather}
a_a=\frac{\dfrac{1200\;\mathrm N}{2}-(24\;\mathrm{kg})\times\left(9.8\;\mathrm{\frac{m}{s^2}}\right)}{24\;\mathrm{kg}} \\[5pt]
a_a=\frac{364.8\;\mathrm{\frac{\cancel{kg}.m}{s^2}}}{24\;\mathrm{\cancel{kg}}} \\[5pt]
a_a=15.2\;\mathrm{m/s^2}
\end{gather}
\]
For body
B, using the equation (VI)
\[
\begin{gather}
a_b=\frac{\dfrac{1200\;\mathrm N}{2}-(40\;\mathrm{kg})\times\left(9.8\;\mathrm{\frac{m}{s^2}}\right)}{40\;\mathrm{kg}} \\[5pt]
a_b=\frac{208\;\mathrm{\frac{\cancel{kg}.m}{s^2}}}{40\;\mathrm{\cancel{kg}}} \\[5pt]
a_b=4.3\;\mathrm{m/s^2}
\end{gather}
\]
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
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{a_{\small B}=4.3\;\mathrm{m/s^2}}
\end{gather}
\]
ES
FR
PT-BR
