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Question Number 174351 by ajfour last updated on 30/Jul/22

	Answered by mr W last updated on 01/Aug/22

	Commented by mr W last updated on 01/Aug/22

	$\tan ϕ = μ$ $T = m_{1} g$ $2 T \cos ϕ = m g$ $2 m_{1} g \frac{1}{\sqrt{1 + μ^{2}}} = m g$ $\Rightarrow \frac{m_{1}}{m} = \frac{\sqrt{1 + μ^{2}}}{2}$ $\frac{L}{2} \cos θ = h \tan ϕ = μ h$ $\Rightarrow \cos θ = \frac{L}{2 μ h}$
	Commented by ajfour last updated on 01/Aug/22

	$g i v e n m_{1} = m$ $T = m g$ $N + T \cos ϕ = m g$ $N = m g (1 - \cos ϕ)$ $f = μ N = μ m g (1 - \cos ϕ) = T \sin ϕ$ $\Rightarrow μ (1 - \cos ϕ) = \sin ϕ$ $μ = \cot \frac{ϕ}{2}$ $m g (\frac{L \cos θ}{2}) = (T \cos ϕ) (L \cos θ)$ $+ (T \sin ϕ) (L \sin θ)$ $\Rightarrow \tan θ = \frac{1}{2} - \cos ϕ$ $h = L \cos ϕ + L \sin θ$ $\frac{h}{L} = \cos ϕ + \sin θ$ $= \cos ({2 \cot}^{- 1} μ)$ $+ \frac{1}{\sqrt{1 + \frac{4}{{[1 - 2 \cos ({2 \cot}^{- 1} μ)]}^{2}}}}$
	Commented by ajfour last updated on 01/Aug/22

	$S i r, f o r y o u r s o l u t i o n :$ $l i n e o f m g p a s s e s t h r o u g h t h e$ $p u l l e y n e e d s j u s t i f i c a t i o n! ? s i r .$
	Commented by Tawa11 last updated on 01/Aug/22

	$Great sirs$
	Commented by mr W last updated on 02/Aug/22

	$t h e t h r e e f o r c e s a c t i n g o n t h e r o d :$ $m g, R, T$ $t h e y m u s t i n t e r s e c t a t o n p o i n t .$ $m_{1} m u s t b e l e s s t h a n m, o t h e r w i s e$ $t h e r o d c a n o n l y h a v e o n e s i n g l e$ $p o s i t i o n : v e r t i c a l p o s i t i o n .$
	Commented by ajfour last updated on 02/Aug/22

	$s a y f o r e x a m p l e (e x t r e m e c a s e)$ $μ = \cot \frac{ϕ}{2} = \sqrt{3} \Rightarrow ϕ = 60 °$ $\tan θ = \frac{1}{2} - \frac{1}{2} = 0$ $\frac{h}{L} = \frac{1}{2} \Rightarrow h = \frac{L}{2}$ $N + T \cos ϕ = m g$ $N = \frac{m g}{2}$ $f = T \sin ϕ = \frac{m g \sqrt{3}}{2} = μ N$
	Commented by ajfour last updated on 02/Aug/22

	Commented by mr W last updated on 02/Aug/22

	$y e s s i r .$
	Commented by ajfour last updated on 02/Aug/22

	$i n y o u r s o l u t i o n$ $s i r, h o w 2 T \cos ϕ = m g$ $t h e r e i s e v e n t o r q u e d u e t o$ $o t h e r c o m p o n e n t o f t e n s i o n a s$ $w e l l .$
	Commented by mr W last updated on 03/Aug/22

	$t h a t m i g h t b e i n c o r r e c t .$
	Commented by mr W last updated on 03/Aug/22

	Commented by mr W last updated on 03/Aug/22

	$T = m g$ $φ = π - 2 ϕ$
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