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Solved Problem on Dynamics
 Portugu√™s     English


In an Atwood machine, the two bodies at rest on a horizontal surface, are connected by a cord, of negligible mass, that passes over a frictionless pulley of negligible mass. The masses weight mA = 24 kg and mB = 40 kg and the free-fall acceleration g = 10 m/s2. Find body accelerations when:
a) F = 400 N;
b) F = 720N;
c) F = 1200 N.
Atwood machine subjected to force F with masses of mA=24 kg and mB=40 kg.


Problem data
  • mass of body A:    mA = 24 kg;
  • mass of body B:    mB = 40 kg;
  • free fall acceleration:    g = 10 m/s2.
Problem diagram

The frame of reference is oriented positively upward, in the same direction of the force \( \vec F \).
The force applied to a pulley is evenly distributed between the two sides (figure 1-A), so the magnitude of the force on each side of the pulley will be \( \frac{\vec F}{2} \).

(A) Force (F) applied to lift Atwood machine is evenly distributed on both sides of the pulley; (B) The force distributed on both sides of the pulley is transmitted by the rope to the block where it is attached.
figure 1

Solution

Assuming the rope has a negligible mass it only transmits the force of the pulley to the bodies, thus the component of the force \( \vec F \) over each body will be \( \frac{\vec F}{2} \) (figur3 1-B).
Drawing a free-bodies diagram for each block, we have
Body A
  • \( \frac{\vec{F}}{2} \): transmitted force from the pulley;
  • \( \vec{P}_{\text{A}} \): weight of the body A.
The weight of the body A is given by
Block A subjected to force F / 2 and weight PA.
figure 2
\[ P_{\text{A}}=m_{\text{A}}g \tag{I} \]
Body B
  • \( \frac{\vec{F}}{2} \): transmitted force from the pulley;
  • \( \vec{P}_{\text{B}} \): weight of body B.
The weight of the body B is given by
Block B subjected to force F / 2 and weight PB.
figure 3
\[ P_{\text{B}}=m_{\text{B}}g \tag{II} \]
Using Newton's Second Law
\[ \bbox[#99CCFF,10px] {\vec{F}=m\vec{a}} \]
we have for body A
\[ \frac{F}{2}-P_{\text{A}}=m_{\text{A}}a_{\text{A}} \tag{III} \]
substituting expression (I) into (III)
\[ \frac{F}{2}-m_{\text{A}}g=m_{\text{A}}a_{\text{A}} \tag{IV} \]
in the same way for body B
\[ \frac{F}{2}-P_{\text{B}}=m_{\text{B}}a_{\text{B}} \tag{V} \]
substituting expression (II) into (V)
\[ \frac{F}{2}-m_{\text{B}}g=m_{\text{B}}a_{\text{B}} \tag{VI} \]

a) If \( F=400\;\text{N} \), the acceleration of body A will be given by expression (IV)
\[ \frac{400}{2}-24 \times 10=24 a_{\text{A}}\\ 200-240=24a_{\text{A}}\\ 24a_{\text{A}}=-40\\ a_{\text{A}}=-{\frac{40}{24}}\\ a_{\text{A}}=-1.7\ \text{m}/s^{2} \]
For body B we have the expression (VI)
\[ \frac{400}{2}-40 \times 10=40a_{\text{B}}\\ 200-400=40a_{\text{B}}\\ 40a_{\text{B}}=-200\\ a_{\text{B}}=-{\frac{200}{40}}\\ a_{\text{B}}=-5\ \text{m}/s^{2} \]
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

\[ \bbox[#FFCCCC,10px] {a_{\text{A}}=a_{\text{B}}=0} \]
Blocks at rest on the surface, with zero accelerations (aA = aB = 0), the force F is not sufficient to raise the system.
figure 4

b) If \( F=720\;\text{N} \), the acceleration of body A will be given by expression (IV)
\[ \frac{720}{2}-24 \times 10=24a_{\text{A}}\\ 360-240=24a_{\text{A}}\\ 24a_{\text{A}}=120\\ a_{\text{A}}=\frac{120}{24}\\ a_{\text{A}}=\;5\ \text{m}/s^{2} \]
For body B we have the expression (VI)
\[ \frac{720}{2}-40 \times 10=40a_{\text{B}}\\ 360-400=40a_{\text{B}}\\ 40a_{\text{B}}=-40\\ a_{\text{B}}=-{\frac{40}{40}}\\ a_{\text{B}}=-1\ \text{m}/s^{2} \]
The body A has the acceleration

\[ \bbox[#FFCCCC,10px] {a_{\text{A}}=5\ \text{m}/s^{2}} \]

As the acceleration of the 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

\[ \bbox[#FFCCCC,10px] {a_{\text{B}}=0} \]
The pulley rising under the action of force F pulling block A with acceleration aA = 5 m/s^2 while block B remains at rest on the surface.
figure 5
 
c) If \( F=1200\;\text{N} \), the acceleration of body A will be given by the expression (IV)
\[ \frac{1200}{2}-24 \times 10=24a_{\text{A}}\\ 600-240=24a_{\text{A}}\\ 24a_{\text{A}}=360\\ a_{\text{A}}=\frac{360}{24}\\ a_{\text{A}}=\;15\ \text{m}/s^{2} \]
For body B we have the expression (VI)
\[ \frac{1200}{2}-40 \times 10=40a_{\text{B}}\\ 600-400=40a_{\text{B}}\\ 40a_{\text{B}}=200\\ a_{\text{B}}=\frac{200}{40}\\ a_{\text{B}}=5\ \text{m}/s^{2} \]
Body A has acceleration

\[ \bbox[#FFCCCC,10px] {a_{\text{A}}=15\ \text{m}/s^{2}} \]

And body B has acceleration

\[ \bbox[#FFCCCC,10px] {a_{\text{B}}=5\ \text{m}/s^{2}} \]
Atwood machine rising under the action of force F with block A rising with acceleration aA = 15 m/s^2 d block B rising with acceleration aB = 5 m/s^2.
figure 6

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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 .