Physics - NLM Question with Solution | TestHub

Cable constraint: $v_B=v_A\cos \theta$, and differentiating, $a_B=a_A\cos \theta -v_A\sin \theta \frac{d\theta }{dt}$

With $h=0.5$m, $\theta =30^{\circ }$, $v_A=1$m/s: $\frac{d\theta }{dt}=\frac{v_A}{h}{\sin }^2\theta =\frac{1}{0.5}\times \frac{1}{4}=0.5$rad/s

So $a_B=\frac{\sqrt{3}}{2}{a}_A-\frac{1}{4}$

For A (18 kg): $T\cos 30^{\circ }=18a_A$; for B (9 kg): $90-T=9a_B$

Substituting: $10-\frac{T}{9}=\frac{T}{24}-\frac{1}{4}$

$T=\frac{738}{11}\approx 67$N