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Question
The friction co-efficient between the table and the block shown in the following figure is 0.2. Find the tensions in the two strings.
Solution
Consider that a 15 kg object is moving downward with an acceleration a.
From the above diagram,
T + m1a − m1g = 0
T + 15a − 15g = 0
⇒ T = 15g − 15a (1)
Now,
T1 − m2g − m2a = 0
T1 − 5g − 5a = 0
⇒ T1 = 5g + 5a (2)
Again,
T − (T1 + 5a + m2R) = 0
⇒ T − (5g + 5a + 5a +m2R) = 0 (3)
(where R = μg)
From Equations (1) and (2),
15g − 15a = 5g + 10a + 0.2 (5g)
⇒ 25a = 90 [g = 10 m/s2]
⇒ a = 3.6 m/s2
From Equation (3),
T = 5 × 10 + 10 × 3.6 + 0.2 × 5 × 10 = 96 N in the left string.
From Equation (2),
T1 = 5g + 5a
= 5 × 10 + 5 × 36
= 50 + 18
= 68 N in the right string.
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