Question

A car is speeding up a horizontal road with acceleration a. Consider the following situations in the car: (i) A ball suspended from the ceiling by a string is maintaining a constant angle with the vertical. Find this angle. (ii) A block is kept on a smooth incline and does not slip on the incline. Find the angle of the incline with the horizontal.

Answer 1

given :

Let the pendulum (formed by the ball and the string) make angle θ with the vertical.

From the free-body diagram,

Tcosθ − mg = 0

Tcosθ = mg

⇒T=mg/cos θ             …i

ma − T sin θ = 0

⇒ ma = T sin θ

⇒T=ma/sin θ                 ….ii

⇒tan θ=a/g

⇒θ=tan⁻¹ a/g

So, the angle formed by the ball with the vertical is

tan⁻¹[ a/g.]

(ii) Let the angle of the incline be θ.

From the diagram,

⇒ ma cos θ = mg sin θ

sin θ/cos θ=a/g

tan θ=(a/g)

θ=tan⁻¹(a/g)

So, the angle of incline is  tan⁻¹(a/g)

Answer 2

ANSWER :

i) Suppose the pendulum make θ angle with the vertical.

Let,

m = mass of the pendulum

T cos θ - mg = 0

⇒ T cos θ = mg

⇒ T = mg / cos θ ....... ( i )

T sin θ = ma

⇒ T = ma / sin θ ........ ( ii )

From ( i ) and ( ii ),

mg / cos θ = ma / sin θ

⇒ tan θ = a / g

⇒ θ = tan⁻¹ ( a / g )

The angle is tan⁻¹ ( a / g ) with the vertical.

ii) m = mass of the block

Suppose the angle of incline is θ

From the 3rd diagram,

ma cos θ - mg sin θ = 0

⇒ ma cos θ = mg sin θ

⇒ sin θ / cos θ = a / g

⇒ tan θ = a / g

⇒ θ = tan⁻¹ ( a / g )

The angle of incline with the horizontal is tan⁻¹ ( a / g ).