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Step-by-Step Solution
Step 1: Identify the forces acting on the bob
In the non-inertial frame of the car (accelerating upward along the inclined plane), the bob experiences:
Its weight: mg , acting vertically downward.
A pseudo force: ma , acting opposite to the car's acceleration. Since the car accelerates up the inclined plane at an angle 30° to the horizontal, we resolve this pseudo force into components in the vertical and horizontal directions as needed.
Tension in the string: T , acting along the string at an angle \alpha with the vertical (this angle is what we want to find).
Step 2: Resolve the pseudo force
The car accelerates parallel to the inclined plane at a = 10\,\text{m/s}^2 . The plane itself is inclined at 30^\circ to the horizontal. Thus, the pseudo force on the bob (in the car's frame) has magnitude ma and is directed down the plane (i.e., opposite the car’s motion).
If we resolve this force into components along the horizontal and vertical:
Horizontal component: ma \cos(30^\circ)
Vertical component: ma \sin(30^\circ)
Step 3: Set up equilibrium conditions
In the accelerating frame of the car, the bob is in equilibrium. Hence, the net force along the horizontal and vertical directions must be zero.
Let \alpha be the angle made by the string with the vertical.
(a) Horizontal direction
The horizontal component of tension must balance the horizontal component of the pseudo force:
T \sin \alpha = ma \cos(30^\circ)
T = \frac{ma \cos(30^\circ)}{\sin \alpha}
(b) Vertical direction
The vertical component of tension must balance both the weight ( mg ) and the vertical component of the pseudo force ( ma \sin(30^\circ) ):
T \cos \alpha = mg + ma \sin(30^\circ).
Step 4: Substitute and simplify
From the horizontal equilibrium, we have
T = \frac{ma \cos(30^\circ)}{\sin \alpha}.
Substituting into the vertical equilibrium equation:
\frac{ma \cos(30^\circ)}{\sin \alpha} \cos \alpha = mg + ma \sin(30^\circ).
Factor out m (common) and plug in a = 10\,\text{m/s}^2 , g = 10\,\text{m/s}^2 , and \sin(30^\circ) = \tfrac12 , \cos(30^\circ) = \tfrac{\sqrt{3}}{2} :
\frac{10 \cdot \tfrac{\sqrt{3}}{2}}{\sin \alpha} \cos \alpha = 10 + 10 \times \tfrac12.
Right side: 10 + (10 \times \tfrac12) = 10 + 5 = 15.
Therefore,
\frac{10 \times \tfrac{\sqrt{3}}{2}}{\sin \alpha} \cos \alpha = 15.
Simplify to get
\frac{10\,\sqrt{3}}{2} \cdot \frac{\cos \alpha}{\sin \alpha} = 15
\quad \Rightarrow \quad
5\,\sqrt{3} \cot \alpha = 15.
Hence,
\cot \alpha = \frac{15}{5\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}.
Therefore,
\alpha = 30^\circ.
Step 5: Final Answer
The angle made by the string with the vertical is 30^\circ .
