Question

Objects whose masses are 300 gm and 200 gm, respectively, are attached to opposite

ends of a light string passing over a light frictionless pulley, Find (a) the acceleration

or the system and (b) the tension in the string.

Answer 1

Explanations & Calculations

• Consider the sketch attached.

• Additional to know: since the pully is frictionless, it does not rotate as the string pass over it, the string just slides on it.

• What happens here is, the higher mass goes down & the lighter mass up under the same acceleration as connected by the string.
• To calculate the needed quantities, apply Newton's second law on both masses along their moving directions.

• Knowns & unknowns are

$\qquad\qquad \begin{aligned} \small m_1&=\small 300g=0.3kg\\ \small m_2 &=\small 200g=0.2kg\\ \small a&=\small ?\\ \small T&=\small ? \end{aligned}$

• Then,

$\qquad\qquad \begin{aligned} \small \downarrow F&=\small m_1a\\ \small 0.3g-T&=\small 0.3a\cdots(1)\\\\ \small \uparrow F&=\small m_2a\\ \small T-0.2g&=\small 0.2a\cdots(2)\\\\ \end{aligned}$

• By (1) + (2) we can get the acceleration

$\qquad\qquad \begin{aligned} \small 0.3g-\cancel{T}+\cancel{T}-0.2g&=\small 0.3a+0.2a\\ \small 0.1g&=\small 0.5a\\ \small a&=\small \frac{0.1g}{0.5}=\frac{g}{5}=\frac{9.8ms^{-2}}{5}\\ &=\small \bold{1.96\,ms^{-2}} \end{aligned}$

• Since the acceleration is known, by substituting that value in either of the two equations, we can find the tension in the string during the motion.

$\qquad\qquad \begin{aligned} \small T-0.2g&=\small 0.2\times 1.96\\ \small T&=\small 0.2g+0.392\\ &=\small 0.2\times 9.8ms^{-2}+0.392\,ms^{-2}\\ &=\small \bold{2.35\,N} \end{aligned}$