Question-108133

Question Number 108133 by ajfour last updated on 14/Aug/20

Commented by ajfour last updated on 14/Aug/20

R o d A B i n s i d e a s m o o t h s p h e r e

o f r a d i u s R, F i n d N o r m a l

r e a c t i o n s a t A a n d B .

Answered by mr W last updated on 15/Aug/20

Commented by mr W last updated on 15/Aug/20

l e t λ = \frac{m}{M}

A C = \frac{M \times \frac{a + b}{2} + m \times a}{M + m} = \frac{(1 + 2 λ) a + b}{2 (1 + λ)}

C B = \frac{a + (1 + 2 λ) b}{2 (1 + λ)}

\cos γ = \frac{a + b}{2 R}

\Rightarrow γ = \cos^{- 1} {\frac{a + b}{2 R}}

\frac{\sin (α + γ)}{\sin α} = \frac{A O}{A C} = \frac{2 (1 + λ) R}{(1 + 2 λ) a + b}

\cos γ + \frac{\sin γ}{\tan α} = \frac{2 (1 + λ) R}{(1 + 2 λ) a + b}

\Rightarrow α = \tan^{- 1} {\frac{\sqrt{1 - {(\frac{a + b}{2 R})}^{2}}}{\frac{2 (1 + λ) R}{(1 + 2 λ) a + b} - \frac{a + b}{2 R}}}

\Rightarrow β = \tan^{- 1} {\frac{\sqrt{1 - {(\frac{a + b}{2 R})}^{2}}}{\frac{2 (1 + λ) R}{a + (1 + 2 λ) b} - \frac{a + b}{2 R}}}

\frac{N_{1}}{\sin β} = \frac{(M + m) g}{\sin 2 γ}

\Rightarrow N_{1} = \frac{\sin β (M + m) g}{\sin 2 γ}

\Rightarrow N_{2} = \frac{\sin α (M + m) g}{\sin 2 γ}

Commented by ajfour last updated on 15/Aug/20

S o q u i c k, a n d s u p e r b s o l u t i o n,

t h a n k s a l o t S i r .

Commented by otchereabdullai@gmail.com last updated on 15/Aug/20

The world mathematician / physicist

I always become very happy when i see

your solution to a question

Answered by ajfour last updated on 15/Aug/20

Commented by ajfour last updated on 15/Aug/20

N \cos (α + θ) = F \cos (α - θ) \dots (i)

p = \sqrt{R^{2} - {(\frac{a + b}{2})}^{2}}

\cos α = \frac{a + b}{2 R}

N \sin (α + θ) + F \sin (α - θ) = (M + m) g

\dots \dots (i i)

T o r q u e b a l a n c e a b o u t c e n t r e o f s p h e r e

m g (\frac{b - a}{2} - p \tan θ) \cos θ = M g (p \sin θ)

\Rightarrow \tan θ = \frac{m (b - a)}{2 p (M + m)}

N o w u s i n g (i) i n (i i)

N {\sin (α + θ) + \sin (α - θ) \frac{\cos (α + θ)}{\cos (α - θ)}}

= (M + m) g

\Rightarrow N = \frac{(M + m) g \cos (α - θ)}{\sin 2 α} &

F = \frac{(M + m) g \cos (α + θ)}{\sin 2 α}

Commented by mr W last updated on 15/Aug/20

p e r f e c t s o l u t i o n!

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