Question

A mass of 10kg is suspended by a rope of length 2.8 m from a ceiling. A force of 98N is applied at the midpoint of the rope is shown in figure, the angle which the rope makes with the vertical in equilibrium is (A) 30⁰ (B) 60⁰
(C) 45⁰ (D) 90⁰

Answer 1

Answer:

45

Explanation:

Given A mass of 10 kg is suspended by a rope of length 2.8 m from a ceiling. A force of 98 N is applied at the midpoint of the rope is shown in figure, the angle which the rope makes with the vertical in equilibrium is

We know that

T cos θ = T1

             = 10 x g

           = 10 x 9.8

          = 98

According to question we have

T sin θ = 98

So T sin θ /T cos θ = 98 / 98

Tan θ = 1

 Θ = 45 degree

Answer 2

Answer:

The angle which the rope makes with the vertical in equilibrium is 45°

(C) is correct option.

Explanation:

Given that,

Mass = 10 kg

Length = 2.8 m

Force = 98 N

In equilibrium position ,

The tension in the string has two component.

Horizontal component = T sin θ

Vertical component = T cos θ

Horizontal component is balanced by the force 98 N.

Vertical component is balance by the weight of the object

$T\sin\theta=F$....(I)

$T\cos\theta=mg$....(II)

We need to calculate the angle which the rope makes with the vertical in equilibrium

Divided equation (I) by (II)

$\dfrac{T\sin\theta}{T\cos\theta}=\dfrac{F}{mg}$

Put the value into the formula

$\tan\theta=\dfrac{98}{10\times9.8}$

$\theta=\dfrac{98}{10\times9.8}$

$\tan\theta=1$

$\tan\theta=\tan45^{\circ}$

$\theta=45^{\circ}$

Hence, The angle which the rope makes with the vertical in equilibrium is 45°