Newton's equation of motion

Question:
The objects A and B have masses, 20.0 kg and 10.0 kg, respectively. They are placed on a frictionless surface. The force (29.4 N) is acted as shown and these objects move together.
(1) Find the acceleration of the objects.
(2) What is the force exerting B from A?

Answer:
From the second figure, the forces exerting B from A and vice versa are equal due to the action and re-action pair.

(1) You can set up the Newton's equation for object A:
\[ \sum F_A = F - N = m_A a\]
The equation for object B is:
\[ \sum F_B = N = m_B a\]
Notice that the acceleration of each object is the same since they move together. Solve for the acceleration from them. Plug $N=m_B
a$ into $F - N = m_A a$.
\[ F - m_B a = m_A a \\
    (m_A + m_B)a = F  \\
   a = \frac{F}{m_A + m_B} \]
Therefore,
\[ a = \frac{29.4 \mathrm{N}}{20.0 \mathrm{kg} + 10.0 \mathrm{kg}} = 0.98 \mathrm{m/s^2}\]

(2) From the solution in (1), we can find the force exerting B from A, which is $N$ in the figure. Thus,
\[ N = m_B a = 10.0\mathrm{kg}\times 0.98\mathrm{m/s^2} = 9.8\mathrm{N}\]

Powered by Hirophysics.com