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(B) Consider the frame of reference of the trolley. The trolley is accelerating down the incline with acceleration $a = g \sin 	heta$.To analyze the

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$\alpha = 	heta$

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(B) Consider the frame of reference of the trolley. The trolley is accelerating down the incline with acceleration $a = g \sin 	heta$.To analyze the forces on the bob of mass $m$ inside the trolley,we apply a pseudo force $F_p = ma = mg \sin 	heta$ acting up the incline.The forces acting on the bob are:$1$. Gravitational force $mg$ acting vertically downwards.$2$. Pseudo force $mg \sin 	heta$ acting parallel to the incline upwards.$3$. Tension $T$ in the string.In the equilibrium position relative to the trolley,the net force on the bob is zero.Resolving forces perpendicular to the string,we have the component of gravitational force $mg \sin \alpha$ balancing the component of the pseudo force $mg \sin 	heta \cos \alpha$. However,a simpler way is to resolve forces along the direction perpendicular to the string: $mg \sin \alpha = (mg \sin 	heta) \cos \alpha$ is incorrect. Let's resolve forces along the direction perpendicular to the string: The component of $mg$ perpendicular to the string is $mg \sin \alpha$ and the component of the pseudo force $mg \sin 	heta$ perpendicular to the string is $(mg \sin 	heta) \cos \alpha$.Equating these,$mg \sin \alpha = mg \sin 	heta \cos \alpha \implies 	an \alpha = \sin 	heta$. This is not matching the options. Re-evaluating: The effective acceleration is the vector sum of $g$ (downwards) and $a$ (up the incline). The string aligns with the effective gravity. The angle $\alpha$ with the vertical is given by $	an \alpha = \frac{a \cos 	heta}{g - a \sin 	heta}$. Substituting $a = g \sin 	heta$,we get $	an \alpha = \frac{g \sin 	heta \cos 	heta}{g - g \sin^2 	heta} = \frac{g \sin 	heta \cos 	heta}{g \cos^2 	heta} = 	an 	heta$. Thus,$\alpha = 	heta$.

$A$ trolley is accelerating down an incline of angle $\theta$ with acceleration $g \sin \theta$. Which of the following is correct? ($\alpha$ is the constant angle made by the string with the vertical.)

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