Q91P - University Physics with Modern Physics

Question

In terms of m₁, m₂, and g, find the acceleration of each block in Fig. P5.91. There is no friction anywhere in the system.

Step-by-Step Solution

Answer

The acceleration of each block are $\frac{{\mathrm{m}}_2\mathrm{g}}{4{\mathrm{m}}_1+{\mathrm{m}}_2}$ and $\frac{2{\mathrm{m}}_2\mathrm{g}}{4{\mathrm{m}}_1+{\mathrm{m}}_2}$ respectively.

Step 1: Identification of given data

The given data can be listed below as:

• The value of the first mass is m₁.
• The value of the first mass is m₂.

Step 2: Significance of the acceleration

The acceleration is described as the division of the velocity of an object with respect to time. The acceleration is also directly proportional to the force exerted by an object.

Step 3: Determination of the acceleration

The free body diagram of the system has been drawn below in four parts:

The equation of the force in the direction is expressed as:

 

 $\mathrm{T}={\mathrm{m}}_1{\mathrm{a}}_1$                                                                                                             …(i)

 

Here, T is the tension, m₁ is the mass of the first object and a₁ is the acceleration of the first object. 

 

The equation of the force in the y direction is expressed as:

 

 $$\begin{gathered} 2\mathrm{T}-{\mathrm{m}}_2\mathrm{g}=-{\mathrm{m}}_2{\mathrm{a}}_2 \\ \mathrm{T}=\frac{{\mathrm{m}}_2\mathrm{g}-{\mathrm{m}}_2{\mathrm{a}}_2}{2} \end{gathered}$$                                                                                      …(ii)

 

Here, m₂ is the mass of the second object, g is the acceleration due to gravity and a₂ is the acceleration of the second object. 

 

Equating the equation (i) and (ii).

 

$\begin{array}{rcl}{\mathrm{m}}_1{\mathrm{a}}_1& =& \frac{{\mathrm{m}}_2\mathrm{g}-{\mathrm{m}}_2{\mathrm{a}}_2}{2}\\ 2{\mathrm{m}}_1{\mathrm{a}}_1& =& {\mathrm{m}}_2\mathrm{g}-{\mathrm{m}}_2{\mathrm{a}}_2\end{array}$                                                                                            …(iii)

 

According to the diagram, the equation of the length is expressed as:

 

l=x+2y

 

Here, l is the total cord’s length, x and y are the distances between the first and the second mass from the pulley. Here, the length is fixed but the distance between the masses are not fixed.

 

Substitute x₁ for x and x₂ for y in the above equation.

 

data-custom-editor="chemistry" $\mathrm{l}={\mathrm{x}}_1+2{\mathrm{x}}_2$

 

Differentiating the above equation with respect to time.

 

data-custom-editor="chemistry" $0=-{\mathrm{v}}_1+2{\mathrm{v}}_2$

 

Here, v₁ and v₂ are the velocities of the first and the second mass respectively. 

 

Differentiating the above equation with respect to time.

 

data-custom-editor="chemistry" $\begin{array}{rcl}0& =& -{\mathrm{a}}_1+2{\mathrm{a}}_2\\ {\mathrm{a}}_2& =& \frac{{\mathrm{a}}_1}{2}\end{array}$

 

Here, a₁ and a₂ are the acceleration of the first and the second mass respectively. 

 

Substitute the values from the above equation in the equation (iii).

 

data-custom-editor="chemistry" $\begin{array}{rcl}4{\mathrm{m}}_1{\mathrm{a}}_2& =& {\mathrm{m}}_2\mathrm{g}-{\mathrm{m}}_2{\mathrm{a}}_2\\ {\mathrm{a}}_2\left(4{\mathrm{m}}_1+{\mathrm{m}}_2\right)& =& {\mathrm{m}}_2\mathrm{g}\\ {\mathrm{a}}_2& =& \frac{{\mathrm{m}}_2\mathrm{g}}{4{\mathrm{m}}_1+{\mathrm{m}}_2}\end{array}$

 

Substitute $\frac{{\mathrm{a}}_1}{2}$ for a₂ in the above equation.

 

data-custom-editor="chemistry" $$\begin{gathered} \frac{{\mathrm{a}}_1}{2}=\frac{{\mathrm{m}}_2\mathrm{g}}{4{\mathrm{m}}_1+{\mathrm{m}}_2} \\ {\mathrm{a}}_1=\frac{2{\mathrm{m}}_2\mathrm{g}}{4{\mathrm{m}}_1+{\mathrm{m}}_2} \end{gathered}$$

 

Thus, the acceleration of each block are data-custom-editor="chemistry" $\frac{{\mathrm{m}}_2\mathrm{g}}{4{\mathrm{m}}_1+{\mathrm{m}}_2}$ and data-custom-editor="chemistry" $\frac{2{\mathrm{m}}_2\mathrm{g}}{4{\mathrm{m}}_1+{\mathrm{m}}_2}$ respectively.