Question

Two masses of 3 kg and 5 kg are connected by a light string that passes over a smooth pulley. Determine the tension of the string

Answer 1

Explanations & Calculations

• Since the pulley is friction less, the system tends to break equilibrium along the side where there is a net force & accelerates.
• Here 5 kg side is greater, hence there is a net force of $\small 5g-3g=2g$ along that side.(downwards the 5kg side pulling the thread down).
• Applying Newton's second law towards that side, the needed parameters could be found.
• Try to draw a figure and apply the law for both masses up & down.
• Therefore,

$\qquad\qquad \begin{aligned} \small \darr F &=\small ma\\ \small 5g-T&=\small 5a\cdots(1)\\ \small \uparrow T-3g&=\small 3a\cdots(2)\\\\ \small (1)\div(2), \small \frac{5g-T}{T-3g}&=\small \frac{5}{3}\\ \small 30g&=\small 8T\\ \small T&=\small \frac{30\times9.8ms^{-2}}{8}\\ &\small \bold{36.75\,N} \end{aligned}$