Question

78. The surface shown in diagram is frictionless and T,
is 30 N, then the value of F is :-
F
T2
T
N
1
60°
3 kg
12 kg
15 kg
TTT
(1) 30 N
(3) 120 N
(2) 60 N
(4) 69.36 N

Answer 1

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Answer 2

Given:

A system of 3 masses ( 3kg, 12 kg, and 15 kg) on a frictionless surface as shown in the diagram.

Force F acting on 15 kg block at 60° from the horizontal.

T₁ = 30N and T₂ are the tension forces

To Find:

The value of force F

Solution:

Since there is no frictional force, we can use the equation of forces on individual blocks.

The effective component of F = F cos 60° = F/2

Let a be the acceleration of the system

The force equation for 15kg block:

F/2 - T₁ = 15 X a           - (1)

The force equation for 12 kg block:

T₁ - T₂ = 12 X a            -(2)

The force equation for 12 kg block:

T₂ = 3 X a                     -(3)

Substituting (3) in (2)

30 - 3a = 12a

or 15a = 30

or a = 2m/s²

Substituting the value of acceleration in (1),

F/2 - 30 = 15 X 2

or F/2 = 60N

or F = 120N

Hence, option (3) 120N is the value of F.