Question-186645

Question Number 186645 by ajfour last updated on 07/Feb/23

Commented by ajfour last updated on 07/Feb/23

W a l l s a r e f r i c t i o n l e s s . B a l l s (s o l i d)

h a v e t h e s a m e d e n s i t y ρ . F i n d

m i n i m u m v a l u e o f m a x i m u m

s t a t i c f r i c t i o n c o e f f i c i e n t a t t h e

g r o u n d f o r t h e l a r g e r b a l l s u c h

t h a t s t a t i c s i t u a t i o n c o u l d p r e v a i l .

Answered by ajfour last updated on 07/Feb/23

\sin θ = \frac{\sqrt{2} (b - a)}{b + a}

I f J b e n o r m a l r e a c t i o n m u t u a l t o

t h e b a l l s .

N = (M + m) g = \frac{4 π ρ g (b^{3} + a^{3})}{3}

f = J \sin θ & J \cos θ = m g

\Rightarrow μ (M + m) g = m g \tan θ

μ_{s, m a x} ⩾ (\frac{a^{3}}{b^{3} + a^{3}}) \frac{\sqrt{2} (b - a)}{\sqrt{6 a b - (a^{2} + b^{2})}}

I f \frac{b}{a} = p

μ_{s, m a x} ⩾ \frac{\sqrt{2} (p - 1)}{(p^{3} + 1) \sqrt{6 p - p^{2} - 1}}

Answered by mr W last updated on 07/Feb/23

Commented by mr W last updated on 07/Feb/23

\sin θ = \frac{\sqrt{2} (b - a)}{a + b}

N_{1} = m_{1} g \tan θ

N_{3} = (m_{1} + m_{2}) g

f = μ N_{3} = N_{1}

μ = \frac{m_{1} g \tan θ}{(m_{1} + m_{2}) g}

= \frac{a^{3}}{a^{3} + b^{3}} \times \frac{\sqrt{2} (b - a)}{\sqrt{{(a + b)}^{2} - 2 {(b - a)}^{2}}}

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