Q114CP - University Physics with Modern Physics

Question

Double Atwood’s Machine. In Fig. P5.114 masses  ${\mathbf{m}}_{\mathbf{1}}$ and ${\mathbf{m}}_{\mathbf{2}}$ are connected by a light string A over a light, frictionless pulley B. The axle of pulley B is connected by a light string C over a light, frictionless pulley D to a mass ${\mathbf{m}}_{\mathbf{3}}$ . Pulley D is suspended from the ceiling by an attachment to its axle. The system is released from rest. In terms of  ${\mathbf{m}}_{\mathbf{1}}\mathbf{,}{\mathbf{m}}_{\mathbf{2}}\mathbf{,}{\mathbf{m}}_{\mathbf{3}\mathbf{,}}$ and g, what are (a) the acceleration of block ${\mathbf{m}}_{\mathbf{3}}$ ; (b) the acceleration of pulley B; (c) the acceleration of block ${\mathbf{m}}_{\mathbf{1}}$ ; (d) the acceleration of block  ${\mathbf{m}}_{\mathbf{2}}$; (e) the tension in string A; (f) the tension in string C? (g) What do your expressions give for the special case of  ${\mathbf{m}}_{\mathbf{1}}\mathbf{=}{\mathbf{m}}_{\mathbf{2}}$ and  ${\mathbf{m}}_{\mathbf{3}}\mathbf{=}{\mathbf{m}}_{\mathbf{1}}\mathbf{+}{\mathbf{m}}_{\mathbf{2}}$? Is this reasonable?

Step-by-Step Solution

Answer

(A) The acceleration of block ${\mathrm{m}}_3$ is  $\mathrm{g}=\left(\frac{-4{\mathrm{m}}_1{\mathrm{m}}_2+{\mathrm{m}}_2{\mathrm{m}}_3+{\mathrm{m}}_1{\mathrm{m}}_3}{4{\mathrm{m}}_1{\mathrm{m}}_2+{\mathrm{m}}_2{\mathrm{m}}_3+{\mathrm{m}}_1{\mathrm{m}}_3}\right)$.

(B) The acceleration of pulley B is $\mathrm{g}=\left(\frac{-4{\mathrm{m}}_1{\mathrm{m}}_2-{\mathrm{m}}_2{\mathrm{m}}_3-{\mathrm{m}}_1{\mathrm{m}}_3}{4{\mathrm{m}}_1{\mathrm{m}}_2+{\mathrm{m}}_2{\mathrm{m}}_3+{\mathrm{m}}_1{\mathrm{m}}_3}\right)$.

(C) The acceleration of block  ${\mathrm{m}}_1$ is $\mathrm{g}=\left(\frac{4{\mathrm{m}}_1{\mathrm{m}}_2-3{\mathrm{m}}_2{\mathrm{m}}_3+{\mathrm{m}}_1{\mathrm{m}}_3}{4{\mathrm{m}}_1{\mathrm{m}}_2+{\mathrm{m}}_2{\mathrm{m}}_3+{\mathrm{m}}_1{\mathrm{m}}_3}\right)$.

(D) The acceleration of block ${\mathrm{m}}_2$ is $\mathrm{g}=\left(\frac{4{\mathrm{m}}_1{\mathrm{m}}_2-3{\mathrm{m}}_2{\mathrm{m}}_3+{\mathrm{m}}_1{\mathrm{m}}_3}{4{\mathrm{m}}_1{\mathrm{m}}_2+{\mathrm{m}}_2{\mathrm{m}}_3+{\mathrm{m}}_1{\mathrm{m}}_3}\right)$ .

(E) The tension in string A is $\left(\frac{4{\mathrm{m}}_1{\mathrm{m}}_2{\mathrm{m}}_3}{4{\mathrm{m}}_1{\mathrm{m}}_2+{\mathrm{m}}_2{\mathrm{m}}_3+{\mathrm{m}}_1{\mathrm{m}}_3}\right)$.

(F) The tension in string C is $\left(\frac{8{\mathrm{m}}_1{\mathrm{m}}_2{\mathrm{m}}_3}{4{\mathrm{m}}_1{\mathrm{m}}_2+{\mathrm{m}}_2{\mathrm{m}}_3+{\mathrm{m}}_1{\mathrm{m}}_3}\right)$.

(G)  For the give condition accelerations are equal to zero, and ${\mathrm{T}}_{\mathrm{A}}=\mathrm{mg}$, and ${\mathrm{T}}_{\mathrm{C}}=2 \mathrm{mg}$ .

Step 1: Concept/Significance of pulley

Pulley is made of simple metallic or wooden material. It is a simple machine which consists of wheel and a rope, and mainly used for lifting heavy loads.

Step 2: Identification of given data:

• The masses are ${\mathrm{m}}_1,{\mathrm{m}}_2 \mathrm{and} {\mathrm{m}}_3$.

Step 3: (a) Find the acceleration of block m 3 :

Draw the free-body diagram for the masses ${\text{m}}_1,m_2 and m_3$ .

Let the acceleration of  ${\mathrm{m}}_1,{\mathrm{m}}_2$ and  ${\mathrm{m}}_3$. are  $a_1,a_2$ and $a_3$ respectively.

 

Use the Newton’s second law to find the force acting on block $m_1$.

 $\sum {\mathrm{F}}_{\mathrm{y}}={\mathrm{m}}_1{\mathrm{a}}_1$

${\mathrm{m}}_1\mathrm{g}-{\mathrm{T}}_{\mathrm{A}}={\mathrm{m}}_1{\mathrm{a}}_1$                                                                                                     ….. (1)

Here, ${\mathrm{T}}_{\mathrm{A}}$  is tension in the block  A, and g is acceleration due to gravity.

 

Use the Newton’s second law to find the force acting on block ${\mathrm{m}}_2$.

$\sum {\mathrm{F}}_{\mathrm{y}}={\mathrm{m}}_2{\mathrm{a}}_2$ 

${\mathrm{m}}_2\mathrm{g}-{\mathrm{T}}_{\mathrm{A}}={\mathrm{m}}_2{\mathrm{a}}_2$                                                                                                    ….. (2)

Here, ${\mathrm{T}}_{\mathrm{A}}$ is tension in the block B.

Use the Newton’s second law to find the force acting on block $m_3$.

$\sum {\mathrm{F}}_{\mathrm{y}}={\mathrm{m}}_3{\mathrm{a}}_3$ 

${\mathrm{m}}_3\mathrm{g}-{\mathrm{T}}_{\mathrm{C}}={\mathrm{m}}_3{\mathrm{a}}_3$                                                                                                   ….. (3)

Here, ${\mathrm{T}}_{\mathrm{C}}$ is tension in the block C .

 

Draw the free-body diagram of block B.

From the above figure,

$$\begin{gathered} 2{\mathrm{T}}_{\mathrm{A}}-{\mathrm{T}}_{\mathrm{C}}=0 \\ {\mathrm{T}}_{\mathrm{C}}=2{\mathrm{T}}_{\mathrm{A}} \end{gathered}$$  

 

The acceleration of the pulley B is given by,

$$\begin{gathered} a_B=-a_3 \\ \frac{a_1+a_2}{2}=-a_3 \end{gathered}$$ 

$a_1+a_2=-2a_3$                                                                                                        ….. (4) 

 

From the equation (1), the acceleration of  $m_1$ is,

$m_{1}g-T_A=m_1{a}_1$ 

$a_1=g-\left(\frac{T_A}{m_1}\right)$                                                                                                        ….. (5) 

 

Similarly, the accelerations of $m_2$  and $m_3$ are derived from equations (2) and (3).

${\mathrm{a}}_2=\mathrm{g}-\left(\frac{{\mathrm{T}}_{\mathrm{A}}}{{\mathrm{m}}_2}\right)$ 

And,

$$\begin{gathered} a_3=g-\left(\frac{T_C}{m_3}\right) \\ {T}_C=m_3(g-a_3) \end{gathered}$$ 

                                                                                                    ….. (7)

 

Substitute the value of $a_1$ and $a_2$ in equation (4).

$$\begin{gathered} 2{\mathrm{a}}_3=\left[\mathrm{g}-\left(\frac{{\mathrm{T}}_{\mathrm{A}}}{{\mathrm{m}}_1}\right)+\mathrm{g}-\left(\frac{{\mathrm{T}}_{\mathrm{A}}}{{\mathrm{m}}_2}\right)\right] \\ =-\left[2\mathrm{g}-{\mathrm{T}}_{\mathrm{A}}\left(\frac{1}{{\mathrm{m}}_1}+\frac{1}{{\mathrm{m}}_2}\right)\right] \end{gathered}$$ 

 

Since $T_A=\frac{T_C}{2}$ then,

$2{\mathrm{a}}_3=-\left[2\mathrm{g}-\left(\frac{{\mathrm{T}}_{\mathrm{C}}}{2}\right)\left(\frac{1}{{\mathrm{m}}_1}+\frac{1}{{\mathrm{m}}_2}\right)\right]$ 

 

Substitute the value of  $T_C$ in the above equation.

$$\begin{gathered} 2a_3=-\left[2g-\left(\frac{m_3\left(g-a_3\right)}{2}\right)\left(\frac{1}{m_1}+\frac{1}{m_2}\right)\right] \\ =g\left(\frac{-4m_1{m}_2+m_2{m}_3+m_1{m}_3}{4m_1{m}_2+m_2{m}_3+m_1{m}_3}\right) \end{gathered}$$ 

 

Therefore, the acceleration of block $m_3$ is $=g\left(\frac{-4m_1{m}_2+m_2{m}_3+m_1{m}_3}{4m_1{m}_2+m_2{m}_3+m_1{m}_3}\right)$.

Step 4: (b) Find the acceleration of pulley B:

Find the acceleration of pulley B as follows.

$$\begin{gathered} a_B=-a_3 \\ =g\left(\frac{4m_1{m}_2-m_2{m}_3-m_1{m}_3}{4m_1{m}_2+m_2{m}_3+m_1{m}_3}\right) \end{gathered}$$ 

 

Therefore, the acceleration of pulley B is  $g\left(\frac{4m_1{m}_2-m_2{m}_3-m_1{m}_3}{4m_1{m}_2+m_2{m}_3+m_1{m}_3}\right)$.

Step 5: (c) Find the acceleration of block m 1 :

From equation (5),

$$\begin{gathered} a_1=g-\left(\frac{T_A}{m_1}\right) \\ =g-\left(\frac{T_C}{2m_1}\right) \\ =g-\left(\frac{m_3\left(g-a_3\right)}{2m_1}\right) \\ =g-g\left[\frac{m_3}{2m_1}-\frac{m_3}{2m_1}\left(g\frac{-4m_1{m}_2+m_2{m}_3+m_1{m}_3}{4m_1{m}_2+m_2{m}_3+m_1{m}_3}\right)\right] \\ {a}_1=g\left(\frac{4m_1{m}_2-3m_2{m}_3+m_1{m}_3}{4m_1{m}_2+m_2{m}_3+m_1{m}_3}\right) \end{gathered}$$ 

 

 

Therefore, the acceleration of block $m_1$  is  $g\left(\frac{4m_1{m}_2-3m_2{m}_3+m_1{m}_3}{4m_1{m}_2+m_2{m}_3+m_1{m}_3}\right)$.

Step 6: (d) Find the acceleration of block m 2 :

From equation (6),

$$\begin{gathered} a_2=g-\left(\frac{T_A}{m_2}\right) \\ =g-\left(\frac{T_C}{2m_2}\right) \\ =g-g\left[\frac{m_3}{2m_2}-\frac{m_3}{2m_3}\left(g\left(\frac{-4m_1{m}_2+m_2{m}_3+m_1{m}_3}{4m_1{m}_2+m_2{m}_3+m_1{m}_3}\right)\right)\right] \\ {a}_2=g\left(\frac{4m_1{m}_2-3m_2{m}_3+m_1{m}_3}{4m_1{m}_2+m_2{m}_3+m_1{m}_3}\right) \end{gathered}$$ 

 

 

Therefore, the acceleration of block $m_2$  is $g\left(\frac{4m_1{m}_2-3m_2{m}_3+m_1{m}_3}{4m_1{m}_2+m_2{m}_3+m_1{m}_3}\right)$.

Step 7: (e) Find the tension in string A:

The tension in string A is calculated by using equation (5),

$$\begin{gathered} T_A=m_1\left(g-a_1\right) \\ =m_1\left(g-g\left(\frac{4m_1{m}_2-3m_2{m}_3+m_1{m}_3}{4m_1{m}_2+m_2{m}_3+m_1{m}_3}\right)\right) \\ =g\left(\frac{4m_1{m}_2{m}_3}{4m_1{m}_2+m_2{m}_3+m_1{m}_3}\right) \end{gathered}$$ 

 

Therefore, the tension in string A is $g\left(\frac{4m_1{m}_2{m}_3}{4m_1{m}_2+m_2{m}_3+m_1{m}_3}\right)$ .

Step 8: (f) Find the tension in string C:

Calculate the tension in string C as follows.

$$\begin{gathered} T_C=2T_A \\ =2\left(g\left(\frac{4m_1{m}_2{m}_3}{4m_1{m}_2+m_2{m}_3+m_1{m}_3}\right)\right) \\ =g\left(\frac{8m_1{m}_2{m}_3}{4m_1{m}_2+m_2{m}_3+m_1{m}_3}\right) \end{gathered}$$ 

 

Therefore, the tension in string C  is $g\left(\frac{8m_1{m}_2{m}_3}{4m_1{m}_2+m_2{m}_3+m_1{m}_3}\right)$.

Step 9: (g) Find the expressions for the special case of m 1 = m 2 and m 3 = m 1 + m 2 , and find whether it is reasonable:

If $m_1=m_2=m$  and  $m_1=m_2=2m$, then the numerator of each acceleration will be zero, and tension is given by,

$$\begin{gathered} {\mathrm{T}}_{\mathrm{A}}=\frac{4{\mathrm{m}}^2\left(2\mathrm{m}\right)}{8{\mathrm{m}}^2}\mathrm{g} \\ =\mathrm{mg} \end{gathered}$$ 

 

And,

$$\begin{gathered} {\mathrm{T}}_{\mathrm{C}}=\frac{8{\mathrm{m}}^2\left(2\mathrm{m}\right)}{8{\mathrm{m}}^2}\mathrm{g} \\ =2\mathrm{mg} \end{gathered}$$ 

 

Therefore, for the give condition accelerations are equal to zero, and ${\mathrm{T}}_{\mathrm{A}}=\mathrm{mg}$ , and  ${\mathrm{T}}_C=2\mathrm{mg}$.