Question
In adjacent figure a boy, on a horizontal platform A, kept on a smooth horizontal surface, holds a rope attached to a box B. Boy pulls the rope with a constant force of 50N. The coefficient of friction between boy and platform is 0.5. (Mass of boy = 80 kg, mass of platform = 120kg and mass of box = 100 kg)
Options
Solution
Sol. Both block & boy move together.
⇒ (80 + 120)a = 50
⇒ a = $\frac{1}{4}$ 
f = 120 × a = t
f = $\frac{120}{4}$ = 30 N
For no relative slipping
common a = $\frac{50}{(80 + 120)} = \frac{1}{4}$m/s2
∴ fr = 120 × a = 30 N
fmax μN = 0.5 × 800 = 40N
⇒ No relative slipping ⇒ aA = $\frac{1}{4}$
aB = $\frac{50}{100} = \frac{1}{2}$ m/s2
(A) w.r.t. B ⇒ 
aAB = $\frac{1}{2}\frac{1}{4} = \frac{3}{4} \Rightarrow v_{AB} = 0 + \frac{3}{4} \times 4$ = 3 m/s
(B) ${\bar{v}}_{boy/P} = 0$ ( No relative slipping)
(C) fr = 30 N
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