Question

Let m, = 1 kg, m2 = 2 kg and m2 = 3 kg in figure (5-E12).
Find the accelerations of m,, m, and ms. The string
from the upper pulley to m, is 20 cm when the system
is released from rest. How long will it take before m,
strikes the pulley?
Note:- solve the question by using constraints concept
​

Answer 1

Free Body Diagrams of each mass $\sf{m_1,\ m_2,\ m_3}$ are shown below.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(0,0){\circle*{1}}\multiput(0,0)(5,-5){2}{\vector(0,1){10}}\put(0,0){\vector(0,-1){10}}\put(-6,-2){ \sf{m_1} }\put(-9,-11){ \sf{10\ N} }\put(-6,9){ \sf{T_1} }\put(7,-1){ \sf{a_1} }\put(30,0){\circle*{1}}\multiput(30,0)(5,-5){2}{\vector(0,1){10}}\put(30,0){\vector(0,-1){10}}\put(24,-2){ \sf{m_2} }\put(21,-11){ \sf{20\ N} }\put(24,9){ \sf{T_2} }\put(37,-1){ \sf{a_2} }\end{picture}$

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(30,0){\circle*{1}}\multiput(30,0)(5,5){2}{\vector(0,-1){10}}\put(30,0){\vector(0,1){10}}\put(24,-2){ \sf{m_3} }\put(21,-11){ \sf{30\ N} }\put(24,9){ \sf{T_2} }\put(37,-1){ \sf{a_3} }\end{picture}$

And free body diagram of the pulley connecting masses $\sf{m_2}$ and $\sf{m_3}$ is shown below.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\put(30,0){\circle*{1}}\multiput(27.5,0)(5,0){2}{\vector(0,-1){10}}\put(30,0){\vector(0,1){10}}\put(40,5){\vector(0,-1){10}}\multiput(21,-11)(14,0){2}{ \sf{T_2} }\put(24,9){ \sf{T_1} }\put(30,0){\circle{5}}\put(42,0){ \sf{a'} }\end{picture}$

Here,

• $\sf{m_1=1\ kg}$

• $\sf{m_2=2\ kg}$

• $\sf{m_3=3\ kg}$

From FBD of $\sf{m_1,}$

$\longrightarrow\sf{T_1-10=m_1a_1}$

$\longrightarrow\sf{T_1-10=a_1\quad\quad\dots(1)}$

From FBD of $\sf{m_2,}$

$\longrightarrow\sf{T_2-20=m_2a_2}$

$\longrightarrow\sf{T_2-20=2a_2\quad\quad\dots(2)}$

From FBD of $\sf{m_3,}$

$\longrightarrow\sf{30-T_2=m_3a_3}$

$\longrightarrow\sf{30-T_2=3a_3\quad\quad\dots(3)}$

From FBD of the pulley (assuming pulley to be massless),

$\longrightarrow\sf{2T_2-T_1=0}$

$\longrightarrow\sf{T_1=2T_2\quad\quad\dots(4)}$

On applying constraint equation to the fixed pulley,

$\longrightarrow\sf{a'-a_1=0}$

$\longrightarrow\sf{a'=a_1\quad\quad\dots(5)}$

And on applying constraint equation to the pulley connecting $\sf{m_2}$ and $\sf{m_3,}$

$\longrightarrow\sf{a_3-2a'-a_2=0}$

$\longrightarrow\sf{2a'=a_3-a_2\quad\quad\dots(6)}$

Now we have 6 equations with 6 unknowns. On solving them we get,

$\longrightarrow\sf{\underline{\underline{a_1=\dfrac{190}{29}\ m\ s^{-2}\ (i.e.,\ upwards)}}}$

$\longrightarrow\sf{\underline{\underline{a_2=-\dfrac{170}{29}\ m\ s^{-2}\ (i.e.,\ downwards)}}}$

$\longrightarrow\sf{\underline{\underline{a_3=\dfrac{210}{29}\ m\ s^{-2}\ (i.e.,\ downwards)}}}$

Now we have to calculate the time taken for the block $\sf{m_1}$ to strike the upper pulley from rest.

$\sf{s=20\ cm=0.2\ m}$

$\sf{u=0\ m\ s^{-1}}$

$\sf{a=a_1=\dfrac{190}{29}\ m\ s^{-2}}$

Then the time is given by,

$\longrightarrow\sf{s=ut+\dfrac{1}{2}at^2}$

$\longrightarrow\sf{0.2=(0)t+\dfrac{1}{2}\cdot\dfrac{190}{29}\cdot t^2}$

$\longrightarrow\sf{0.2=\dfrac{95}{29}\,t^2}$

$\longrightarrow\sf{t=\sqrt{\dfrac{0.2\times29}{95}}}$

$\longrightarrow\sf{\underline{\underline{t=0.25\ s}}}$

Answer 2

Average acceleration is greatest in interval 2

Average speed is greatest in interval 3

v is positive in intervals 1, 2, and 3

a is positive in intervals 1 and 3 and negative in interval 2

a = 0 at A, B, C, D

Acceleration is given by the slope of the speed-time graph. In the given case, it is given by the slope of the speed-time graph within the given interval of time.

Since the slope of the given speed-time graph is maximum in interval 2, average acceleration will be the greatest in this interval.

Height of the curve from the time-axis gives the average speed of the particle. It is clear that the height is the greatest in interval 3. Hence, average speed of the particle is the greatest in interval 3.

In interval 1:

The slope of the speed-time graph is positive. Hence, acceleration is positive. Similarly, the speed of the particle is positive in this interval.

In interval 2:

The slope of the speed-time graph is negative. Hence, acceleration is negative in this interval. However, speed is positive because it is a scalar quantity.

In interval 3:

The slope of the speed-time graph is zero. Hence, acceleration is zero in this interval. However, here the particle acquires some uniform speed. It is positive in this interval.

Points A, B, C, and D are all parallel to the time-axis. Hence, the slope is zero at these points. Therefore, at points A, B, C, and D, acceleration of the particle is zero.