A car accelerates on a horizontal road with an acceleration a . Two situations are considered inside the car. First, a ball is suspended from the ceiling by a string and maintains a constant angle with the vertical.

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Correct answer: C. ^ -1 ((a)/(g)) and ^ -1 ((a)/(g))

In the frame of the accelerating car, a pseudo force

ma

acts on the objects in the direction opposite to the acceleration.

For the suspended ball, the forces acting are tension

T

, weight

mg

, and pseudo force

ma

.

Resolving forces along the horizontal and vertical directions for equilibrium:

T \sin \theta = ma

T \cos \theta = mg

Dividing the two equations, we get

\tan \theta = \dfrac{a}{g} \Rightarrow \theta = \tan^{-1} \left(\dfrac{a}{g}\right).<br><br>For the block on the smooth incline, the forces acting are normal reaction

N

, weight

mg

, and pseudo force

ma

.<br><br>For the block to not slip, the net force along the incline must be zero.<br><br>The component of gravity down the incline is

mg \sin

\alpha and the component of the pseudo force up the incline is

ma \cos

\alpha.<br><br>Equating these components:<br><br>

mg \sin

\alpha =

ma \cos

\alpha<br><br>

\tan

\alpha = \dfrac{a}{g} \Rightarrow \alpha = \tan^{-1} \left(\dfrac{a}{g}\right).<br><br>Both angles are

\tan^{-1} \left(\dfrac{a}{g}\right).

A car accelerates on a horizontal road with an acceleration a . Two...

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