Question

Two masses $m_1$ and $m_2$ are connected by means of a light string, that passes over a light pulley as shown in the figure. If $m_1=2kg$ and $m_2=5kg$ and a vertical force $F$ is applied on the pulley, then find the acceleration of the masses and that of the pulley when.
$\left(a\right)F=35N,\left(b\right)F=70N,\left(c\right)F=140N.$

Answer 1

Since string is massless and friction is absent, hence tension in
the string is same every where.
$\left(a\right)$Let acceleration of the pulley be $a_p$. For ${
a }_{ p }$ to be non-zero
$T>m_{1}g$ ...$\left(1\right)$
$\Rightarrow T>2g$ ...$\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$,we get
$F>2\times \left(2g\right)\Rightarrow F>40N$
Therefore, when $F=35N$
$a_P=0$ and hence $a_1=a_2=0$
$\left(b\right)$ As mass of the pulley is negligible
$F=2T=0$
$\Rightarrow T=F/2$ $\Rightarrow T=35N$
To lift $m_2,$
$T\ge m_{2}g$ $T\ge 50N$
Therefore block $m_2$ will not move
$\Rightarrow T{m}_{1}g=m_1{a}_1$ $\Rightarrow 15=2a_1$
$\Rightarrow a_1=\frac{15}{2}m/s^2$
Constraint equation
$y_P+y_P-y_1=constant$
$\Rightarrow {2y}_P-y_1=c$ $\Rightarrow
2\dfrac { { d }^{ 2 }{ y }_{ p } }{ { dt }^{ 2 } } -\dfrac { { d }^{ 2 }{ y }_{ 1 } }{ { dt }^{ 2 } } =0$
$a_P=\frac{a_1}{2}=\frac{15}{4}m/s^2$
$\left(c\right)$ When $F=140N$
$T=70N$
$\Rightarrow T-m_{1}g=m_1{a}_1$ ...$\left(1\right)$
$\Rightarrow 70N-20N=2\times a_1$ $\Rightarrow a_1=25m/s^2$
$T-m_{2}g=m_2{a}_2$
$\Rightarrow 70N-50N=5a_2$ $\Rightarrow a_2=4m/s^2$
Constraint equation
$y_p-y_2+y_p-y_1=c$
$\Rightarrow 2y_p-y_1-y_2=c$
$\Rightarrow \frac{d^2{y}_p}{{dt}^2}-\frac{d^2{y}_1}{{dt}^2}-\frac{d^2{y}_2}{{dt}^2}=0$
$\Rightarrow a_p=\frac{a_1+a_2}{2}=\frac{29}{2}m/s^2$

Answers:

$a)=0$

$b)=\frac{15}{4}m/s^2$

$c)=\frac{29}{2}m/s^2$