Question

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?

Short answer

$$\begin{gathered} \mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{}\mathbf{}\frac{\mathbf{-}\mathbf{F}}{{\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}\mathbf{+}{\mathbf{m}}_{\mathbf{3}}} \\ \mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{-}\frac{{\mathbf{m}}_{\mathbf{3}}\mathbf{F}}{{\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}\mathbf{+}{\mathbf{m}}_{\mathbf{3}}}\mathbf{,}\frac{\mathbf{-}\mathbf{(}{\mathbf{m}}_{\mathbf{3}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}\mathbf{)}\mathbf{F}}{{\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}\mathbf{+}{\mathbf{m}}_{\mathbf{3}}} \\ \mathbf{\left(}\mathit{c}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{-}\frac{{\mathbf{m}}_{\mathbf{3}}\overrightarrow{\mathbf{F}}}{{\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}\mathbf{+}{\mathbf{m}}_{\mathbf{3}}} \end{gathered}$$

Step-by-step solution

Identification of the given data

• 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 .

Definition of Acceleration and Vector Force

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.

(a) Determination of Acceleration of the blocks

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

(b) Determination of Vector Force F→net3 exerted by the rod on the block of mass m3 and m1

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

(c) Determination of Vector Force F→net3, when you pull to the left on the left block (mass m3)

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