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Solved Problem on Dynamics
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Two bodies of masses mA = 6 kg and mB = 4 kg are on a frictionless horizontal surface. A horizontal force of constant magnitude equal to 25 N is applied to push the two bodies. Find the acceleration acquired by the set and the magnitude of the contact force between the bodies.
Blocks of mass mA=6 kg and mB=4 kg subjected to a force of 25 N.


Problem diagram

We use a frame of reference oriented to the right in the same direction of the applied force, this produces an acceleration a on the system.

Problem data
  • mass of body A:    mA = 6 kg;
  • mass of body B:    mB = 4 kg;
  • force applied to the system:    F = 25 N.
Acceleration produced by force (F) on the blocks.
figure 1
Solution

Drawing a free-body diagram for each body, we have for the body A
Horizontal direction
  • \( \vec{F} \) : force applied to the body;
  • \( -\vec{f} \) : normal force of body B on A.
Vertical direction
  • \( \vec{N}_{\text{A}} \) : normal force of the surface on the body;
  • \( \vec{P}_{\text{A}} \) : weight of body A.
Block A under the action of the weight (WA), the normal force (NA), contact force (-f), and an external force (F).
figure 2
In the vertical, direction there is no motion, normal force and weight cancel out.
In the horizontal, direction using Newton's Second Law
\[ \bbox[#99CCFF,10px] {\vec{F}=m\vec{a}} \]
\[ F-f=m_{\text{A}}a \tag{I} \]
For body B, we have

Horizontal direction
  • \( \vec{{f}} \): force of body A on B.
Vertical direction
  • \( {\vec{{N}}}_{\text{B}} \): normal force of the surface on the body;
  • \( {\vec{{P}}}_{\text{B}} \): weight of body B.
Block B under the action of the weight (WB), the normal force (NB), and contact force (f).
figure 3
In the vertical direction, there is no motion, normal force and weight cancel out.
In the horizontal direction applying Newton's Second Law
\[ {f=m_{\text{B}}a} \tag{II} \]
Equations (I) and (II) can be written as a system of linear equations with two variables (f and a)
\[ \left\{ \begin{array}{rr} F-f&=m_{\text{A}}a\\ f&=m_{\text{B}}a \end{array} \right. \]
substituting (II) into (I) we have the acceleration
\[ F-m_{\text{B}}a=m_{\text{A}}a\\ F=m_{\text{A}}a+m_{\text{B}}a \]
factoring the acceleration
\[ F=a\;(\;m_{\text{A}}+m_{\text{B}}\;)\\ a=\frac{F}{m_{\text{A}}+m_{\text{B}}}\\ a=\frac{25}{6+4}\\ a=\frac{25}{10} \]
\[ \bbox[#FFCCCC,10px] {a=2.5\;\text{m/s}^{2}} \]

Note: Since the two bodies form a set subjected to the same force, both have the same acceleration, the system behaves as if it were a single body with total mass given by the sum of the masses of the two bodies A and B.
 
Substituting the acceleration found in expression (II) we have the force of contact between the bodies
\[ f=4\times2.5 \]
\[ \bbox[#FFCCCC,10px] {f=10\;\text{N}} \]

Note: In the same way we could substitute the acceleration in the expression (I) to obtain the contact force, in this case:
\[ 25-f=6\times2.5\Rightarrow 25-f=15\Rightarrow f=25-15\Rightarrow f=10\;\text{N} \]
 

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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 .