Question

The friction coefficient between the table and the block shown in figure is 0.2, find the tensions in the two strings.

Answer 1

Let the Tension in 15 kg block string be T and the tension in the string of 5 kg block be T'.

For 15 kg block,

15g - T = 15a

⇒ T = 15g - 15a

∴ T = 15(g - a)

For 5 kg block,

T' - 5g = 5a

⇒ T' = 5g + 5a

Now, for 5 kg block in between,

T - T' - Fric. = 5a

⇒ 15(g - a) - 5g - 5a - 0.2 × 5 × 10 = 5a

∴  15g - 15a - 5g - 5a - 10 = 5a

⇒ 10g - 20a -5a = 10

⇒ 10g - 25a = 10

⇒ 25a = 90

⇒ a = 90/25

⇒ a = 18/5 m/s².

∴ T = 15(g - a)

⇒ T = 15(10 - 18/5)

⇒ T = 150 - 54

∴ T = 96 N.

T' = 5g + 5a

⇒ T' = 50 + 18 × 1

∴ T' = 68 N.

Hope it helps.

Answer 2

Let the Tension in 15 kg block string be T and the tension in the string of 5 kg block be T'.

For 15 kg block,

15g - T = 15a

⇒ T = 15g - 15a

∴ T = 15(g - a)

For 5 kg block,

T' - 5g = 5a

⇒ T' = 5g + 5a

Now, for 5 kg block in between,

T - T' - Fric. = 5a

⇒ 15(g - a) - 5g - 5a - 0.2 × 5 × 10 = 5a

∴  15g - 15a - 5g - 5a - 10 = 5a

⇒ 10g - 20a -5a = 10

⇒ 10g - 25a = 10

⇒ 25a = 90

⇒ a = 90/25

⇒ a = 18/5 m/s².

∴ T = 15(g - a)

⇒ T = 15(10 - 18/5)

⇒ T = 150 - 54

∴ T = 96 N.

T' = 5g + 5a

⇒ T' = 50 + 18 × 1

∴ T' = 68 N.

Hope it helps.