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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.
Solution
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
\[\Rightarrow T = \frac{mg}{\cos \theta} . . . \left( i \right)\]
And, ma − T sin θ = 0
⇒ ma = T sin θ
\[\Rightarrow T = \frac{ma}{\sin \theta} . . . . \left( ii \right)\]
\[ \Rightarrow \tan \theta = \frac{a}{g}\]
\[ \Rightarrow \theta = \tan^{- 1} \frac{a}{g}\]
So, the angle formed by the ball with the vertical is \[\tan^{- 1} \left( \frac{a}{g} \right)\]
(ii) Let the angle of the incline be θ.
From the diagram,
⇒ ma cos θ = mg sin θ
\[\frac{\sin \theta}{\cos \theta} = \frac{a}{g}\]
\[\tan \theta = \frac{a}{g}\theta = \tan^{- 1} \left( \frac{a}{g} \right)\]
So, the angle of incline is \[\tan^{- 1} \left( \frac{a}{g} \right)\]