In the figure, two identical particles each of mass m are tied together with an inextensible massless string. This is pulled at its centre with a constant force F . If the whole system lies on a smooth horizontal plane, then the acceleration of each particle towards each other at the instant shown in the figure is

Question

In the figure, two identical particles each of mass m are tied together with an inextensible massless string. This is pulled at its centre with a constant force F . If the whole system lies on a smooth horizontal plane, then the acceleration of each particle towards each other at the instant shown in the figure is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-l6mmty8lppe5.png" /> (A) (√3 /2 )(F/m) (B) (1/2 √3)(F/m) (C) (2/√3)(F/m) (D) √3(F/m)

Tardigrade Admin · Accepted Answer

2Tcos 30o=F ⇒ T=(F/2 cos 30o)=(F/√3) Solution

Sigma Fy=Tcos 60o=(F/2 √3) Sigma Fy=may ⇒ ay=(F/2 √3 m)

$\frac{3}{2} \frac{F}{m}$

$\frac{1}{2 3} \frac{F}{m}$

$\frac{2}{3} \frac{F}{m}$

$3 \frac{F}{m}$

$2 T cos 3 0^{o} = F$
$\Rightarrow T = \frac{F}{2 cos 3 0 ^{o}} = \frac{F}{3}$

$Σ F_{y} = T cos 6 0^{o} = \frac{F}{2 3}$
$Σ F_{y} = m a_{y}$
$\Rightarrow a_{y} = \frac{F}{2 3 m}$

23​​mF​

23​1​mF​

3​2​mF​

3​mF​

2Tcos30o=F ⇒T=2cos30oF​=3​F​ ΣFy​=Tcos60o=23​F​ ΣFy​=may​ ⇒ay​=23​mF​

$\frac{3}{2} \frac{F}{m}$

$\frac{1}{2 3} \frac{F}{m}$

$\frac{2}{3} \frac{F}{m}$

$3 \frac{F}{m}$

$2 T cos 3 0^{o} = F$
$\Rightarrow T = \frac{F}{2 cos 3 0 ^{o}} = \frac{F}{3}$

$Σ F_{y} = T cos 6 0^{o} = \frac{F}{2 3}$
$Σ F_{y} = m a_{y}$
$\Rightarrow a_{y} = \frac{F}{2 3 m}$