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A solid sphere of mass m is lying at rest on a rough horizontal surface. The coefficient of friction between the ground and sphere is $\mu$. The maximum value of F, so that the sphere will not slip is equal to

\[\begin {array} {1 1} (a)\;\frac{7}{5} \mu mg & \quad (b)\;\frac{4}{7} \mu mg \\ (c)\;\frac{5}{7} \mu mg & \quad  (d)\;\frac{7}{2} \mu mg \end {array}\]

2 Answers

$Ff= Macm$--------(1)
$f_e= I \alpha$
$f= \large\frac{2}{5}$$ MR^2.\large \frac{\alpha}{R}$
$f= \large\frac{2}{5} Ma _{cm}$------(2)[$ a_{cm}=Rd$ for rolling]
$\mu mg \geq \large\frac{2}{5} $$m a_{cm}$
$a_{cm} \leq \large\frac{5}{2}$$ \mu g$
From (1) and (2)
$F= M \bigg[ 1+ \large\frac{2}{5} \bigg]a_{cm}$
$\quad= \large\frac{7}{5}$$ M a_{cm}$
$F_{max }= \large\frac{7}{5}$$ M \large\frac{5}{2}$$\mu g$
$\qquad= \large\frac{7}{2}$$ M \mu g$
answered Dec 5, 2013 by meena.p

balance torque 


μμmg = 2mR^2a/5R         ------1



F - μmg = ma       ----2

from 1 and 2 force is 7μmg/2

answered Apr 19 by jainnanu010

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