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Question

(B) In the frame of the rocket,the object experiences a pseudo force $ma$ downwards,where $a = 2g.$ The total effective acceleration acting on the mas

Vedclass · Accepted Answer

$	an 	heta$

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(B) In the frame of the rocket,the object experiences a pseudo force $ma$ downwards,where $a = 2g.$ The total effective acceleration acting on the mass $m$ is $g_{eff} = g + a = g + 2g = 3g$ downwards.The forces acting on the mass $m$ along the inclined plane are:$1$. The component of the effective weight $mg_{eff} \sin 	heta = 3mg \sin 	heta$ acting down the incline.$2$. The frictional force $f$ acting up the incline.The forces perpendicular to the inclined plane are:$1$. The normal force $N = mg_{eff} \cos 	heta = 3mg \cos 	heta$.$2$. The component of the effective weight $mg_{eff} \cos 	heta = 3mg \cos 	heta$.For the mass to remain at rest,the frictional force must balance the component of the effective weight acting down the incline:$f = 3mg \sin 	heta$Since $f \le \mu N,$ the minimum coefficient of friction $\mu_{min}$ is given by:$f = \mu_{min} N$$3mg \sin 	heta = \mu_{min} (3mg \cos 	heta)$$\mu_{min} = \frac{3mg \sin 	heta}{3mg \cos 	heta} = 	an 	heta$

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