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.
Problem data:
- Mass of body A: mA = 6 kg;
- Mass of body B: mB = 4 kg;
- Force applied to the system: F = 25 N.
Problem diagram:
We choose a frame of reference oriented to the right in the same direction as the applied force
\( \vec F \)
and produce an acceleration a on the system.
Drawing a
Free-Body Diagram for each body
-
Body A:
-
Horizontal direction:
- \( \vec F \) : force applied to the body;
- \( -\vec f \) : reaction force of body B on A.
-
Vertical direction:
- \( {\vec N}_{\small A} \) : normal reaction force of the surface on the body;
- \( {\vec W}_{\small A} \) : weight of body A.
-
Body B:
-
Horizontal direction:
- \( \vec f \): action force of body A on B.
-
Vertical direction:
- \( {\vec N}_{\small B} \): normal reaction force of the surface on the body;
- \( {\vec W}_{\small B} \): weight of body B.
Solution
Applying
Newton's Second Law
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\vec{F}=m\vec{a}}
\end{gather}
\]
In the vertical direction, there is no motion, the normal force and weight cancel out.
In the horizontal direction
\[
\begin{gather}
F-f=m_{\small A}a \tag{I}
\end{gather}
\]
In the vertical direction, there is no motion, the normal force and weight cancel out.
In the horizontal direction
\[
\begin{gather}
{f=m_{\small B}a} \tag{II}
\end{gather}
\]
Equations (I) and (II) can be written as a system of linear equations with two variables
(
f and
a)
\[
\left\{
\begin{array}{rr}
F-f&=m_{\small A}a\\
f&=m_{\small B}a
\end{array}
\right.
\]
substituting equation (II) into equation (I), we have the acceleration
\[
\begin{gather}
F-m_{\small B}a=m_{\small A}a\\[5pt]
F=a(m_{\small A}+m_{B})\\[5pt]
a=\frac{F}{m_{\small A}+m_{B}}\\[5pt]
a=\frac{25\;\mathrm N}{(6\;\mathrm kg)+(4\;\mathrm kg)}\\[5pt]
a=\frac{25\;\mathrm{\frac{\cancel{kg}.m}{s^2}}}{10\;\mathrm{\cancel{kg}}}
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{a=2.5\;\mathrm{m/s^2}}
\end{gather}
\]
Note: since the two bodies form a set subjected to the same force, and 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 equation (II), we have the force of contact between the bodies
\[
\begin{gather}
f=(4\;\mathrm{kg})\left(2.5\;\mathrm{\frac{m}{s^2}}\right)
\end{gather}
\]
\[
\begin{gather}
\bbox[#FFCCCC,10px]
{f=10\;\mathrm N}
\end{gather}
\]
Note: in the same way, we could substitute the acceleration in the equation (I) to obtain the
force of contact, in this case:
\[ (25\;\mathrm N)-f=(6\;\mathrm{kg})\left(2.5\;\mathrm{\frac{m}{s^2}}\right)\Rightarrow (25\;\mathrm N)-f=15\;\mathrm N\Rightarrow f=10\;\mathrm N \]