Two blocks of mass m₁ and m₃ , connected by a rod of mass m₂ , are sitting on a low friction surface, and you push to the left on the right block (mass m₁ ) with a constant force of magnitude F(Figure 4.57).

(a) What is the acceleration$\frac{\mathbf{d}{\mathbf{v}}_{\mathbf{x}}}{\mathbf{d}\mathbf{t}}$ of the blocks?

(b) What is the vector force ${\overrightarrow{\mathbf{F}}}_{\mathbf{n}\mathbf{e}\mathbf{t}\mathbf{3}}$exerted by the rod on the block of mass m₃ ? ($\left|{\overrightarrow{F}}_{\mathrm{net}3}\right|$ is approximately equal to the compression force in the rod near its left end.)What is the vector force${\overrightarrow{\mathbf{F}}}_{\mathbf{net}\mathbf{1}}$exerted by the rod on the block of massm₁? $\left|{\overrightarrow{\mathbf{F}}}_{\mathbf{n}\mathbf{e}\mathbf{t}\mathbf{1}}\right|$is approximately equal to the compression force in the rod near its right end.)

(c) Suppose that instead of pushing on the right block (mass m₁ ), you pull to the left on the left block (mass m₃) with a constant force of magnitude F . Draw a diagram illustrating this situation. Now what is the vector force exerted by the rod on the block of mass m₃?

• The mass of Block 1 is m₁
• The mass of Block 2 is m₂
• The mass of a rod is m₃
• The magnitude of the constant force acting on Block 1 is F .

Acceleration is defined as the rate of change velocity with respect to time.

Vector force is defined as the force which has both magnitude and direction.

Acceleration,

$\begin{array}{rcl}a& =& \frac{d{v}_x}{dt}\\ & =& \frac{\Sigma F}{m}\\ \overrightarrow{a}& =& -\frac{\overrightarrow{F}}{m_1+m_2+m_3}\end{array}$

Hence, the acceleration of the blocks is$\frac{\overrightarrow{\mathbf{F}}}{{\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}\mathbf{+}{\mathbf{m}}_{\mathbf{3}}}$

The equation of motion can be written as:

$\Sigma m\frac{d{v}_x}{dt}=\Sigma F$

Mass 3 is applying force then:

${\overrightarrow{F}}_{net3}=-\frac{m_3\overrightarrow{F}}{m_1+m_2+m_3}$

Hence, the vector force${\overrightarrow{\mathbf{F}}}_{\mathbf{n}\mathbf{e}\mathbf{t}\mathbf{3}}$exerted by the rod on the block of massis

$\mathbf{-}\frac{{\mathbf{m}}_{\mathbf{3}}\mathbf{F}}{{\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}\mathbf{+}{\mathbf{m}}_{\mathbf{3}}}\mathbf{}\mathbf{.}$

The vector force ${\overrightarrow{F}}_{net3}$exerted by the rod on the block of massm₁,

${\overrightarrow{F}}_{net3}=-\frac{(m_3+m_2)\overrightarrow{F}}{m_1+m_2+m_3}$

Hence, the vector force${\overrightarrow{F}}_{net1}$exerted by the rod on the block of mass m₁ is$\frac{\mathbf{-}\mathbf{(}{\mathbf{m}}_{\mathbf{3}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}\mathbf{)}\overrightarrow{\mathbf{F}}}{{\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}\mathbf{+}{\mathbf{m}}_{\mathbf{3}}}$ .

The vector force ${\overrightarrow{\mathrm{F}}}_{\mathrm{net}3}$, when you pull to the left on the left block (massm₃)

${\overrightarrow{\mathrm{F}}}_{\mathrm{net}3}=-\frac{m_3\overrightarrow{F}}{m_1+m_2+m_3}$

Hence, the vector force${\overrightarrow{\mathbf{F}}}_{\mathbf{net}\mathbf{3}}$ when you pull to the left on the left block (mass m₃) is$\mathbf{-}\frac{{\mathbf{m}}_{\mathbf{3}}\overrightarrow{\mathbf{F}}}{{\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}\mathbf{+}{\mathbf{m}}_{\mathbf{3}}}$