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 \(\dots\dots\dots\). (Take \( g = 10 \, ms^{-2} \))

Hint:

Using vector subtraction \( \vec{g}_{eff} = \vec{g} - \vec{a} \) is often safer than resolving components along the incline, as it directly gives the angle with respect to standard horizontal/vertical axes.

Answer 1

Correct Answer: 30

Step 1: Understanding the Concept:
In an accelerated frame of reference (the car), a pseudo force acts on the bob in a direction opposite to the car's acceleration. The string will align itself along the direction of the net effective gravity (\( \vec{g}_{eff} = \vec{g} - \vec{a} \)).
Step 2: Key Formula or Approach:
We resolve the gravitational acceleration \( \vec{g} \) and car acceleration \( \vec{a} \) into components to find the angle \( \theta \) with the vertical.
Step 3: Detailed Explanation:
Let's define a coordinate system where \( y \) is vertical (upward positive) and \( x \) is horizontal.
1. Acceleration of car (\( \vec{a} \)): Moves up the incline at \( 30^\circ \).
\[ a_x = 10 \cos 30^\circ = 5\sqrt{3} \, ms^{-2} \]
\[ a_y = 10 \sin 30^\circ = 5 \, ms^{-2} \]
2. Gravity (\( \vec{g} \)):
\[ g_x = 0, \quad g_y = -10 \, ms^{-2} \]
3. Effective Acceleration (\( \vec{g}' = \vec{g} - \vec{a} \)):
\[ g'_x = 0 - 5\sqrt{3} = -5\sqrt{3} \, ms^{-2} \]
\[ g'_y = -10 - 5 = -15 \, ms^{-2} \]
4. Angle with vertical (\( \theta \)):
The string hangs in the direction of \( \vec{g}' \). The angle \( \theta \) it makes with the vertical is:
\[ \tan \theta = \frac{|g'_x|}{|g'_y|} = \frac{5\sqrt{3}}{15} = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} \]
\[ \theta = 30^\circ \]
Step 4: Final Answer:
The angle the string makes with the vertical is \( 30^\circ \).