Question

79.
Two blocks of masses 5 kg and 3 kg connected
by a rope of mass 2 kg are resting on a frictionless
floor as shown in the following figure. If a constant
force of 100 N is applied to 5 kg block, then the
tension in the rope at points A, B and C are
respectively given by:
100 N
60°
IC BA
3 kg
5 kg
2 kg
(1) 15 N, 20 N, 25 N (2) 25 N, 20 N, 15 N
(3) 20 N, 20 N, 20 N (4) 50 N, 50 N, 50 N​

Answer 1

Given:

Two blocks of masses 5 kg and 3 kg connected

by a rope of mass 2 kg are resting on a frictionless floor as shown in the following figure.

To find:

Tension in the rope at A , B and C.

Calculation:

Net acceleration be a:

$\boxed{ \rm{a = \dfrac{total \: force}{total \: mass} }}$

$= > \rm{a = \dfrac{100}{(5 + 3 + 2)} }$

$= > \rm{a = \dfrac{100}{10} }$

$= > \rm{a = 10 \: m {s}^{ - 2} }$

So, Let Tension at A be T_{A}

$\therefore \: T_{A} = (3 + 2) \times a$

$= > \: T_{A} = 5 \times 10$

$\boxed{= > \: T_{A} = 50 \: newton}$

Let tension at B be T_{B}

$\therefore \: T_{B} = \{3 + (2 \times \dfrac{1}{2} ) \} \times a$

$= > \: T_{B} = \{3 + 1\} \times 10$

$= > \: T_{B} = 4\times 10$

$\boxed{ = > \: T_{B} = 40 \: newton}$

Ley tension at C be T_{C}:

$\therefore \: T_{C} = 3 \times a$

$= > \: T_{C} = 3 \times 10$

$\boxed{ = > \: T_{C} = 30 \: newton}$

Refer to attached diagram.

Answer 2

Question:-

Two blocks of masses 5 kg and 3 kg connected

by a rope of mass 2 kg are resting on a frictionless floor as shown in the following figure. If a constant force of 100 N is applied to 5 kg block, then the tension in the rope at points A, B and C are respectively given by:

Answer:-

Let the tension at point A, B and C be T₁, T₂ and T₃ respectively and the acceleration be a.

Refer to the attachment

For finding the tension in the massed string, since the mass of the whole string is 2kg and the mass is uniformly distributed, we have assumed that we are cutting the string into two equal halves (left part and right part), masses of each of these halves is 1kg, and then we are considering each of them individually and drawing their F.B.D.

From the attachment, we get the following four equations:-

• T₃ = 3a ---------- (i)
• 100 - T₁ = 5a ---------- (ii)
• T₂ - T₃ = a ------------ (iii)
• T₁ - T₂ = a ----------- (iv)

(i) + (ii) + (iii) + (iv) :-

T₃ + 100 - T₁ + T₂ - T₃ + T₁ - T₂ = 3a + 5a + a + a

=> 100 = 10a

=> a = 10m/s²

In (i) :-

T₃ = 3a

=> T₃ = 3(10)

=> T₃ = 30 N (tension at C)

In (ii) :-

100 - T₁ = 5a

=> T₁ = 100 - 5(10)

=> T₁ = 50 N (tension at A)

In (iii) :-

T₂ - T₃ = a

=> T₂ = a + T₃

=> T₂ = 10 + 30

=> T₂ = 40 N (tension at B)

Ans. Tension at A = 50 N, Tension at B = 40 N, Tension at C = 30 N