A car is moving on a plane inclined at 30° to the horizontal with an acceleration of 10 m s$$^{-2}$$ 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 $$g = 10$$ m s$$^{-2}$$)

We consider the motion from the point of view of an observer sitting inside the car. Since the car is accelerating, its interior is a non-inertial frame, so in addition to the real forces we must introduce a pseudo (fictitious) force on every mass in the direction opposite to the car’s acceleration.

The given data are

$$a = 10 \text{ m s}^{-2}$$ (acceleration of the car upward along the plane) $$g = 10 \text{ m s}^{-2}$$ (acceleration due to gravity, vertically downward) $$\alpha = 30^\circ \quad$$ (inclination of the plane to the horizontal)

The real force on the bob is its weight $$\vec{W}=m\vec{g},$$ vertically downward.

The pseudo force is $$\vec{F}_p = -m\vec{a},$$ opposite to the car’s acceleration, i.e. down the plane. Because the car’s acceleration is parallel to the plane and up the slope, the pseudo force is parallel to the plane and down the slope, of magnitude $$|\vec{F}_p| = ma = 10m.$$

We now add these two vectors to obtain the effective force (or effective acceleration) that the bob “feels” in the non-inertial frame:

$$\vec{g}_{\text{eff}} = \vec{g} + \left(-\vec{a}\right).$$

To combine them, we express the pseudo force in components relative to the vertical:

• The direction “down the slope” makes an angle of $$\alpha = 30^\circ$$ below the horizontal. The angle between this direction and the vertical is therefore $$90^\circ - 30^\circ = 60^\circ.$$
• Hence the pseudo force of magnitude $$10m$$ has
   vertical component $$10m\cos 60^\circ,$$
   horizontal (transverse) component $$10m\sin 60^\circ.$$

Writing the magnitudes only (the factor m cancels everywhere),

Vertical component of $$\vec{g}_{\text{eff}}$$: $$g_{\text{vert}} = 10 + 10\cos 60^\circ = 10 + 10\left(\dfrac12\right) = 10 + 5 = 15.$$

Horizontal (lateral) component of $$\vec{g}_{\text{eff}}$$: $$g_{\text{horiz}} = 10\sin 60^\circ = 10\left(\dfrac{\sqrt3}{2}\right) = 5\sqrt3.$$

The string aligns itself along the direction of $$\vec{g}_{\text{eff}},$$ so the angle $$\theta$$ that the string makes with the true vertical is determined by

$$\tan\theta = \dfrac{g_{$$ horiz $$}}{g_{$$ vert $$}} = \dfrac{5\sqrt3}{15} = \dfrac{\sqrt3}{3} = \dfrac1{\sqrt3}.$$

We know the standard trigonometric value

$$\tan 30^\circ = \dfrac1{\sqrt3}.$$

Therefore,

$$\theta = 30^\circ.$$

So, the answer is $$30^\circ$$.