A trolley is accelerating down an incline of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$\alpha = 	heta$

Vedclass · Answer

(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.)

Similar Questions

$A$ body of mass $1.0 \, kg$ is falling with an acceleration of $10 \, m/s^2$. Its apparent weight will be ......... $kg \, wt$ $(g = 10 \, m/s^2)$.

$A$ lift is accelerating downward with an acceleration $a$. $A$ man inside the lift throws a ball upward with an acceleration $a_0$ relative to the lift $(a_0 < a)$. What is the acceleration of the ball as observed by an observer on the Earth?

An elevator car,whose floor-to-ceiling distance is $2.7 \, m$,starts ascending with a constant acceleration of $1.2 \, m/s^2$. $2 \, s$ after the start,a bolt begins falling from the ceiling of the car. The free-fall time of the bolt is ......... $s$.

$A$ person in an elevator accelerating upwards with an acceleration of $2\,ms^{-2}$,tosses a coin vertically upwards with a speed of $20\,ms^{-1}$. After how much time will the coin fall back into his hand? (Take $g = 10\,ms^{-2}$)

$A$ person of $60 \,kg$ mass is in a lift which is coming down such that the man exerts a force of $150 \,N$ on the floor of the lift. Then the acceleration of the lift is $\left(g=10 \,ms^{-2}\right)$ (in $\,ms^{-2}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module