The free body diagram of the given system is shown above. Let us consider that the 15kg block is moving downward with the acceleration a.
From figure (a)
T + ma − mg = 0
T + 15a − 15g = 0
⇒ T = 15g − 15a……. (i)
From the figure (c)
T1 − mg − ma = 0
T1 − 5g − 5a = 0
⇒ T1 = 5g + 5a ………. (ii)
From figure (b)
T = (T1 + 5a + μR) = 0
⇒ T − (5g + 5a + 5a +μR) = 0 ……. (iii)
(where R = μg)
From Equations (i) and (ii),
15g − 15a = 5g + 10a + 0.2 (5g)
⇒ 25a = 90
⇒ a = 3.6 m/s2
From Equation (iii),
T = 5 × 10 + 10 × 3.6 + 0.2 × 5 × 10
T= 96 N in the left string.
From Equation (ii),
T1 = 5g + 5a
= 5 × 10 + 5 × 36
= 50 + 18
T1= 68 N in the right string.
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