Q 170 Question: Double Atwood鈥檚 Mach... [FREE SOLUTION]

Chapter 1: Q 170 (page 170)

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

Short Answer

Expert verified

(A) The acceleration of block $m_{3}$ is $g (\frac{- 4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}}{4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}})$ .

(B) The acceleration of pulley B is $g (\frac{4 m_{1} m_{2} - m_{2} m_{3} - m_{1} m_{3}}{4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}})$ .

(C) The acceleration of block $m_{1}$ is $g (\frac{4 m_{1} m_{2} - 3 m_{2} m_{3} + m_{1} m_{3}}{4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}})$ .

(D) The acceleration of block $m_{2}$ is $g (\frac{4 m_{1} m_{2} - 3 m_{2} m_{3} + m_{1} m_{3}}{4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}})$ .

(E) The tension in string A is $g (\frac{4 m_{1} m_{2} m_{3}}{4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}})$ .

(F) The tension in string C is $g (\frac{8 m_{1} m_{2} m_{3}}{4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}})$ .

(G) For the give condition accelerations are equal to zero, and $T_{A} = m g$ , and $T_{C} = 2 m g$ .

Step by step solution

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.

Identification of given data:

The masses are $m_{1}$ , $m_{2}$ , and $m_{3}$ .

(a) Find the acceleration of block m3:

Draw the free-body diagram for the masses $m_{1}$ , $m_{2}$ , and $m_{3}$ .

Let the acceleration of $m_{1}$ , $m_{2}$ , and $m_{3}$ . are $a_{1}$ , $a_{2}$ , and $a_{3}$ respectively.

Use the Newton鈥檚 second law to find the force acting on block $m_{1}$ .

$鈭� F_{y} = m_{1} a_{1}$

$m_{1} g - T_{A} = m_{1} a_{1}$ 鈥�.. (1)

Here, $T_{A}$ is tension in the block A, and g is acceleration due to gravity.

Use the Newton鈥檚 second law to find the force acting on block $m_{2}$ .

$鈭� F_{y} = m_{2} a_{2}$

$m_{2} g - T_{A} = m_{2} a_{2}$ 鈥�.. (2)

Here, $T_{A}$ is tension in the block B.

Use the Newton鈥檚 second law to find the force acting on block $m_{3}$ .

$鈭� F_{y} = m_{3} a_{3}$

$m_{3} g - T_{C} = m_{3} a_{3}$ 鈥�.. (3)

Here, $T_{C}$ is tension in the block C.

Draw the free-body diagram of block B.

From the above figure,

$2 T_{A} - T_{C} = 0 T_{C} = 2 T_{A}$

The acceleration of the pulley B is given by,

$a_{B} = - a_{3} \frac{a_{1} + a_{2}}{2} = - a_{3}$

$a_{1} + a_{2} = - 2 a_{3}$ 鈥�.. (4)

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

$m_{1} g - T_{A} = m_{1} a_{1}$

$a_{1} = g - (\frac{T_{A}}{m_{1}})$ 鈥�.. (5)

Similarly, the accelerations of $m_{2}$ and $m_{3}$ are derived from equations (2) and (3).

$a_{2} = g - (\frac{T_{A}}{m_{2}})$

And,

$a_{3} = g - (\frac{T_{C}}{m_{3}})$

$T_{C} = m_{3} (g - a_{3})$ 鈥�.. (7)

Substitute the value of $a_{1}$ and $a_{2}$ in equation (4).

$\begin{array}{rcl} 2 a_{3} & = & [g - (\frac{T_{A}}{m_{1}}) + g - (\frac{T_{A}}{m_{2}})] \\ = & - [2 g - T_{A} (\frac{1}{m_{1}} + \frac{1}{m_{2}})] \end{array}$

Since $T_{A} = \frac{T_{C}}{2}$ then,

$2 a_{3} = - [2 g - (\frac{T_{C}}{2}) (\frac{1}{m_{1}} + \frac{1}{m_{2}})]$

Substitute the value of $T_{C}$ in the above equation.

$\begin{array}{rcl} 2 a_{3} & = & - [2 g - (\frac{m_{3} (g - a_{3})}{2}) (\frac{1}{m_{1}} + \frac{1}{m_{2}})] \\ = & g (\frac{- 4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}}{4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}}) \end{array}$

Therefore, the acceleration of block $m_{3}$ is $g (\frac{- 4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}}{4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}})$ .

(b) Find the acceleration of pulley B:

Find the acceleration of pulley B as follows.

$\begin{array}{rcl} a_{B} & = & - a_{3} \\ = & g (\frac{4 m_{1} m_{2} - m_{2} m_{3} - m_{1} m_{3}}{4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}}) \end{array}$

Therefore, the acceleration of pulley B is $g (\frac{4 m_{1} m_{2} - m_{2} m_{3} - m_{1} m_{3}}{4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}})$ .

(c) Find the acceleration of block m1:

From equation (5),

$\begin{array}{rcl} a_{1} & = & g - (\frac{T_{A}}{m_{1}}) \\ = & g - (\frac{T_{C}}{2 m_{1}}) \\ = & g - (\frac{m_{3} (g - a_{3})}{2 m_{1}}) \\ = & g - g [\frac{m_{3}}{2 m_{1}} - \frac{m_{3}}{2 m_{1}} (g \frac{- 4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}}{4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}})] \end{array}$

$a_{1} = g (\frac{4 m_{1} m_{2} - 3 m_{2} m_{3} + m_{1} m_{3}}{4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}})$

Therefore, the acceleration of block $m_{1}$ is $g (\frac{4 m_{1} m_{2} - 3 m_{2} m_{3} + m_{1} m_{3}}{4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}})$ .

(d) Find the acceleration of block m2:

From equation (6),

$\begin{array}{rcl} a_{2} & = & g - (\frac{T_{A}}{m_{2}}) \\ = & g - (\frac{T_{C}}{2 m_{2}}) \\ = & g - (\frac{m_{3} (g - a_{3})}{2 m_{2}}) \\ = & g - g [\frac{m_{3}}{2 m_{2}} - \frac{m_{3}}{2 m_{3}} (g (\frac{- 4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}}{4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}}))] \end{array}$

$a_{2} = g (\frac{4 m_{1} m_{2} - 3 m_{2} m_{3} + m_{1} m_{3}}{4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}})$

Therefore, the acceleration of block $m_{2}$ is $g (\frac{4 m_{1} m_{2} - 3 m_{2} m_{3} + m_{1} m_{3}}{4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}})$ .

(e) Find the tension in string A:

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

$\begin{array}{rcl} T_{A} & = & m_{1} (g - a_{1}) \\ = & m_{1} (g - g (\frac{4 m_{1} m_{2} - 3 m_{2} m_{3} + m_{1} m_{3}}{4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}})) \\ = & g (\frac{4 m_{1} m_{2} m_{3}}{4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}}) \end{array}$

Therefore, the tension in string A is $g (\frac{4 m_{1} m_{2} m_{3}}{4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}})$ .

(f) Find the tension in string C:

Calculate the tension in string C as follows.

$\begin{array}{rcl} T_{C} & = & 2 T_{A} \\ = & 2 (g (\frac{4 m_{1} m_{2} m_{3}}{4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}})) \\ = & g (\frac{8 m_{1} m_{2} m_{3}}{4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}}) \end{array}$

Therefore, the tension in string C is $g (\frac{8 m_{1} m_{2} m_{3}}{4 m_{1} m_{2} + m_{2} m_{3} + m_{1} m_{3}})$ .

(g) Find the expressions for the special case of m1=m2 and m3=m1+m2, and find whether it is reasonable:

If $m_{1} = m_{2} = m$ and $m_{3} = m_{1} + m_{2} = 2 m$ , then the numerator of each acceleration will be zero, and tension is given by,

$\begin{array}{rcl} T_{A} & = & \frac{4 m^{2} (2 m)}{8 m^{2}} g \\ = & m g \end{array}$

And,

$\begin{array}{rcl} T_{C} & = & \frac{8 m^{2} (2 m)}{8 m^{2}} g \\ = & 2 m g \end{array}$

Therefore, for the give condition accelerations are equal to zero, and $T_{A} = m g$ , and $T_{C} = 2 m g$ .

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Short Answer

Step by step solution

Concept/Significance of pulley:

Identification of given data:

(a) Find the acceleration of block m3:

(b) Find the acceleration of pulley B:

(c) Find the acceleration of block m1:

(d) Find the acceleration of block m2:

(e) Find the tension in string A:

(f) Find the tension in string C:

(g) Find the expressions for the special case of m1=m2 and m3=m1+m2, and find whether it is reasonable:

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