Question

The friction co-efficient between the table and the block shown in figure (6−E4) is 0.2. Find the tensions in the two strings.

Figure

Answer 1

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.