Question

Q6 A mass 10 kg is suspended by a rope of length 2m from the ceiling. A force 100 N in horizontal direction is applied at the mid-point of rope as shown in figure. The angle subtended by rope with the vertical in equilibrium will be (g = 10 m/s2 and neglect the mass of rope) → 100 N A 53° 10kg B 37° D 30° e 45°​

Answer 1

Answer:

For horizontal equilibrium-

T

3

=T

1

sinθ

For vertical equilibrium-

T

2

=T

1

cosθ

Taking ratios-

T

2

T

3

=tanθ

And for equilibrium of body T

2

=6kgwt=6×10=60N

⟹tanθ=

6

5

or θ=tan

−1

6

5

Explanation:

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