A cone of radius r and height h rests on a ro | vedclass

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(C) For the cone to slide,the applied force $F$ must exceed the limiting friction: $F > \mu mg$. For the cone to topple,the torque about the edge of t

Vedclass · Accepted Answer

$\mu = \frac{r}{h}$

Vedclass · Answer

(C) For the cone to slide,the applied force $F$ must exceed the limiting friction: $F > \mu mg$. For the cone to topple,the torque about the edge of the base must be zero. The center of mass of a solid cone is at a height $h/4$ from the base. The torque due to the force $F$ applied at the vertex (height $h$) is $	au_F = F \cdot h$. The torque due to gravity $mg$ acting at the center of mass (distance $r$ from the edge) is $	au_g = mg \cdot r$. Toppling occurs when $	au_F > 	au_g$,i.e.,$F \cdot h > mg \cdot r$,or $F > \frac{mgr}{h}$. The cone slides before it topples if the force required to slide is less than the force required to topple: $\mu mg Thus,$\mu The largest value of $\mu$ for which the cone will slide before it topples is $\mu = \frac{r}{h}$.

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