Question 45: Laws of Motion

Question

A car is moving on a plane inclined at 30$$^\circ$$ to the horizontal with an acceleration of 10 ms
$$-$$2
parallel to the plane upward. A bob is suspended by a string from the roof of the car. The angle in degrees which the string makes with the vertical is ______________. (Take g = 10 ms
$$-$$2
)

Accepted Answer

Given,
Angle of inclination, $$	heta$$ = 30$$^\circ$$
Acceleration, a = 10 ms
$$-$$2
Acceleration due to gravity, g = 10 ms
$$-$$2
According to the question the car and bob is as shown below,
Here, F' is the pseudo force acting on the bob when we considered it from car's frame and T is the tension on the string.
In equilibrium, $$\Sigma$$F
x
= 0 and $$\Sigma$$F
y
= 0
$$\Rightarrow$$ F' cos30$$^\circ$$ = T sin$$\alpha$$
$${{ma\cos 30^\circ } \over {\sin \alpha }} = T$$ ...... (i)
where, m is the mass of the bob.
$$F'\sin 30^\circ  + mg = T\cos \alpha $$
$$ \Rightarrow ma\sin 30^\circ  + mg = {{ma\cos 30^\circ } \over {\sin \alpha }}(\cos \alpha )$$ [$$\because$$ using Eq. (i)]
$$ \Rightarrow a\sin 30^\circ  + g = {{a\cos 30^\circ } \over {\sin \alpha }}(\cos \alpha )$$
$$10 	imes {1 \over 2} + 10 = {{10 	imes \sqrt 3 } \over 2}\cot \alpha $$
$$ \Rightarrow {1 \over 2} + 1 = {{\sqrt 3 } \over 2}\cot \alpha  \Rightarrow {3 \over 2} = {{\sqrt 3 } \over 2}\cot \alpha $$
or, $$\cot \alpha  = \sqrt 3 $$
$$ \Rightarrow \alpha  = 30^\circ $$

Laws of Motion

A car is moving on a plane inclined at 30$^\circ to the horizontal with an acceleration of 10 ms -2 parallel to the plane upward. A bob is suspended by a string from the roof of the car. The angle in degrees which the string makes with the vertical is ______________. (Take g = 10 ms -$2 )

Integer / Numeric Answer Type