A monkey of mass m is climbing a rope hanging | vedclass

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(C) Let the friction force exerted by the rope on the monkey be $f_r$ acting in the upward direction.The forces acting on the monkey are:$1$. Weight $

Vedclass · Accepted Answer

Upward,$F=m(g+a)$

Vedclass · Answer

(C) Let the friction force exerted by the rope on the monkey be $f_r$ acting in the upward direction.The forces acting on the monkey are:$1$. Weight $mg$ acting downward.$2$. Friction force $f_r$ acting upward.According to Newton's second law of motion,the net force on the monkey is equal to the product of its mass and acceleration:$f_r - mg = ma$Rearranging the equation to solve for $f_r$:$f_r = ma + mg$$f_r = m(g + a)$Since the monkey is accelerating upward,the friction force must act in the upward direction to support the monkey and provide the necessary acceleration.Therefore,the friction force is $m(g+a)$ in the upward direction.

$A$ monkey of mass $m$ is climbing a rope hanging from the roof with acceleration $a$. The coefficient of static friction between the body of the monkey and the rope is $\mu$. Find the direction and value of the friction force on the monkey.

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