Question

A bob is hanging over a pulley inside a car through a string. The second end of the string is in the hand of a person standing in the car. The car is moving with constant acceleration 'a' directed horizontally as shown in figure. Other end of the string is pulled with constant acceleration 'a' (relative to car) vertically. The tension in the string is equal to :

• A. $m\sqrt{g^2 + a^2}$
• B. $m\sqrt{g^2 + a^2} - ma$
• C. $m\sqrt{g^2 + a^2} + ma$
• D. $m(g + a)$

Answer 1

Answer: (B)

$m\sqrt{g^2 + a^2} - ma$

Answer 2

In the car's frame, the bob experiences gravity ($\vec{F}_g = -mg\hat{j}$) and a fictitious force ($\vec{F}_{fict} = -ma\hat{i}$). The resultant of these forces is $\vec{F}_{res} = -ma\hat{i} - mg\hat{j}$, with magnitude $|\vec{F}_{res}| = m\sqrt{a^2 + g^2}$. The bob accelerates upwards relative to the car with acceleration $a$ along the string ($\vec{a}'_{bob} = a\hat{u}$). Applying Newton's second law in the car's frame: $\vec{T} + \vec{F}_g + \vec{F}_{fict} = m\vec{a}'_{bob}$. Substituting $\vec{T} = -T\hat{u}$, we get $-T\hat{u} - ma\hat{i} - mg\hat{j} = ma\hat{u}$. Rearranging gives $-ma\hat{i} - mg\hat{j} = (T+ma)\hat{u}$. Thus, $T+ma = |\vec{F}_{res}| = m\sqrt{a^2 + g^2}$, leading to $T = m\sqrt{a^2 + g^2} - ma$.