Math Question and Answers


Question and Answers Forum

All Questions Topic List
Mechanics Questions
Previous in All Question Next in All Question
Previous in Mechanics Next in Mechanics
Question Number 44951 by rahul 19 last updated on 06/Oct/18

Commented by rahul 19 last updated on 07/Oct/18

Commented by rahul 19 last updated on 07/Oct/18

$p l s e x p l a i n t h e s o l u t i o n! (1^{s t}, 2^{n d} l i n e)$
Commented by MrW3 last updated on 07/Oct/18

$t h e s o l u t i o n g i v e n i s n o t c o r r e c t, I t h i n k .$ $i t i g n o r e s t h e f r i c t i o n c a u s e d t h r o u g h$ $t h e t e n s i o n i n t h e r o p e .$
	Answered by MrW3 last updated on 06/Oct/18

	Commented by rahul 19 last updated on 07/Oct/18
	sir, Ans. is AD I feel the solution is too advance for me . Anyways Thanks!!☺️��
	Commented by MrW3 last updated on 07/Oct/18

	Commented by MrW3 last updated on 07/Oct/18
	$G=λgds=λgRdθ N=(T/R)ds+G sin θ=(T+λgR sin θ)dθ f=μN=μ(T+λgR sin θ)dθ T+dT+f=T+G cos θ dT+μ(T+λgR sin θ)dθ=λgR cos θ dθ ⇒(dT/dθ)+μT=λgR(cos θ−μ sin θ) let T=e^(−μθ) S (dT/dθ)=−μe^(−μθ) S+e^(−μθ) (dS/dθ) −μe^(−μθ) S+e^(−μθ) (dS/dθ)+μe^(−μθ) S=λgR(cos θ−μ sin θ) e^(−μθ) (dS/dθ)=λgR(cos θ−μ sin θ) ⇒(dS/dθ)=λgRe^(μθ) (cos θ−μ sin θ) S=λgR∫e^(μθ) (cos θ−μ sin θ)dθ =((λgRe^(μθ) {2μ cos θ+(1−μ^2 )sin θ})/(1+μ^2 ))+C e^(μθ) T=((λgRe^(μθ) {2μ cos θ+(1−μ^2 )sin θ})/(1+μ^2 ))+C ⇒T=((λgR{2μ cos θ+(1−μ^2 )sin θ})/(1+μ^2 ))+Ce^(−μθ) at θ=0: T=0 ((2λgRμ)/(1+μ^2 ))+C=0 C=−((2λgRμ)/(1+μ^2 )) ⇒T=((λgR)/(1+μ^2 ))[(1−μ^2 ) sin θ+2μ(cos θ−e^(−μθ) )] the friction coeifficient is big enough if no tension force is needed to hold the rope at its topmost point. at θ=(π/2): T=^(!) 0 ((λgR)/(1+μ^2 ))(1−μ^2 −2μe^(−((μπ)/2)) )=0 ⇒1−μ^2 −2μe^(−((μπ)/2)) =0 ⇒μ≈0.7324 [Option B gives (1/(√2))=0.707] (dT/dθ)=...[(1−μ^2 )cos θ+2μ(−sin θ+μe^(−μθ) )]=0 ⇒((sin θ−μe^(−μθ) )/(cos θ))=((1−μ^2 )/(2μ)) ⇒θ in terms of μ with μ=0.7324⇒θ=41.74° ⇒T_(max) =0.353λRg [Option D gives ((√2)−1)λRg=0.414λRg]$
	$G = λ g d s = λ g R d θ$ $N = \frac{T}{R} d s + G \sin θ = (T + λ g R \sin θ) d θ$ $f = μ N = μ (T + λ g R \sin θ) d θ$ $T + d T + f = T + G \cos θ$ $d T + μ (T + λ g R \sin θ) d θ = λ g R \cos θ d θ$ $\Rightarrow \frac{d T}{d θ} + μ T = λ g R (\cos θ - μ \sin θ)$ $l e t T = e^{- μ θ} S$ $\frac{d T}{d θ} = - μ e^{- μ θ} S + e^{- μ θ} \frac{d S}{d θ}$ $- μ e^{- μ θ} S + e^{- μ θ} \frac{d S}{d θ} + μ e^{- μ θ} S = λ g R (\cos θ - μ \sin θ)$ $e^{- μ θ} \frac{d S}{d θ} = λ g R (\cos θ - μ \sin θ)$ $\Rightarrow \frac{d S}{d θ} = λ {g R e}^{μ θ} (\cos θ - μ \sin θ)$ $S = λ g R \int e^{μ θ} (\cos θ - μ \sin θ) d θ$ $= \frac{λ {g R e}^{μ θ} {2 μ \cos θ + (1 - μ^{2}) \sin θ}}{1 + μ^{2}} + C$ $e^{μ θ} T = \frac{λ {g R e}^{μ θ} {2 μ \cos θ + (1 - μ^{2}) \sin θ}}{1 + μ^{2}} + C$ $\Rightarrow T = \frac{λ g R {2 μ \cos θ + (1 - μ^{2}) \sin θ}}{1 + μ^{2}} + {C e}^{- μ θ}$ $a t θ = 0 : T = 0$ $\frac{2 λ g R μ}{1 + μ^{2}} + C = 0$ $C = - \frac{2 λ g R μ}{1 + μ^{2}}$ $\Rightarrow T = \frac{λ g R}{1 + μ^{2}} [(1 - μ^{2}) \sin θ + 2 μ (\cos θ - e^{- μ θ})]$ $t h e f r i c t i o n c o e i f f i c i e n t i s b i g e n o u g h$ $i f n o t e n s i o n f o r c e i s n e e d e d t o h o l d$ $t h e r o p e a t i t s t o p m o s t p o i n t .$ $a t θ = \frac{π}{2} : T \overset{!}{=} 0$ $\frac{λ g R}{1 + μ^{2}} (1 - μ^{2} - 2 μ e^{- \frac{μ π}{2}}) = 0$ $\Rightarrow 1 - μ^{2} - 2 μ e^{- \frac{μ π}{2}} = 0$ $\Rightarrow μ \approx 0.7324$ $[O p t i o n B g i v e s \frac{1}{\sqrt{2}} = 0.707]$ $\frac{d T}{d θ} = . . . [(1 - μ^{2}) \cos θ + 2 μ (- \sin θ + μ e^{- μ θ})] = 0$ $\Rightarrow \frac{\sin θ - μ e^{- μ θ}}{\cos θ} = \frac{1 - μ^{2}}{2 μ}$ $\Rightarrow θ i n t e r m s o f μ$ $w i t h μ = 0.7324 \Rightarrow θ = 41.74 °$ $\Rightarrow T_{m a x} = 0.353 λ R g$ $[O p t i o n D g i v e s (\sqrt{2} - 1) λ R g = 0.414 λ R g]$
	Commented by ajfour last updated on 07/Oct/18

	$y e s S i r, t r u e .$
	Commented by MrW3 last updated on 07/Oct/18
	$If we ignore the friction caused by the tension force in rope, i.e. if we ignore the term μT in the eqn. (dT/dθ)+μT=λgR(cos θ−μ sin θ), then we will get (dT/dθ)=λgR(cos θ−μ sin θ) ⇒T=λgR(sin θ+μ cos θ)+C at θ=0, T=0⇒C=−μλRg ⇒T=λgR{sin θ+μ(cos θ−1)} at θ=(π/2),T=0: λgR{1−μ}=0⇒μ=1 (dT/dθ)=λRg{cos θ−μ sin θ}=0 ⇒cos θ−μ sin θ=0 ⇒tan θ=(1/μ)=1⇒θ=(π/4) T_(max) =λRg{((√2)/2)+1×(((√2)/2)−1)}=λRg((√2)−1) you see if we ignore the friction caused by tension in rope, we will get the same result as given in book. but why should we ignore it? that′s why I say that the solution given in book is not correct (enough). certainly, as we can see, ignoring that term makes the solution much easier than considering that term.$
	$I f w e i g n o r e t h e f r i c t i o n c a u s e d b y t h e$ $t e n s i o n f o r c e i n r o p e, i . e . i f w e i g n o r e$ $t h e t e r m μ T i n t h e e q n .$ $\frac{d T}{d θ} + μ T = λ g R (\cos θ - μ \sin θ),$ $t h e n w e w i l l g e t$ $\frac{d T}{d θ} = λ g R (\cos θ - μ \sin θ)$ $\Rightarrow T = λ g R (\sin θ + μ \cos θ) + C$ $a t θ = 0, T = 0 \Rightarrow C = - μ λ R g$ $\Rightarrow T = λ g R {\sin θ + μ (\cos θ - 1)}$ $a t θ = \frac{π}{2}, T = 0 :$ $λ g R {1 - μ} = 0 \Rightarrow μ = 1$ $\frac{d T}{d θ} = λ R g {\cos θ - μ \sin θ} = 0$ $\Rightarrow \cos θ - μ \sin θ = 0$ $\Rightarrow \tan θ = \frac{1}{μ} = 1 \Rightarrow θ = \frac{π}{4}$ $T_{m a x} = λ R g {\frac{\sqrt{2}}{2} + 1 \times (\frac{\sqrt{2}}{2} - 1)} = λ R g (\sqrt{2} - 1)$ $y o u s e e i f w e i g n o r e t h e f r i c t i o n$ $c a u s e d b y t e n s i o n i n r o p e, w e w i l l$ $g e t t h e s a m e r e s u l t a s g i v e n i n b o o k .$ $b u t w h y s h o u l d w e i g n o r e i t ?$ ${t h a t}^{'} s w h y I s a y t h a t t h e s o l u t i o n$ $g i v e n i n b o o k i s n o t c o r r e c t (e n o u g h) .$ $c e r t a i n l y, a s w e c a n s e e, i g n o r i n g t h a t$ $t e r m m a k e s t h e s o l u t i o n m u c h e a s i e r$ $t h a n c o n s i d e r i n g t h a t t e r m .$
	Commented by rahul 19 last updated on 07/Oct/18

	$y e s s i r! D e f i n i t e l y p r o b l e m i s n o t c l e a r .$ $N o w {i t}^{'} s u n d e r s t o o d a b l e :)$
	Commented by MrW3 last updated on 07/Oct/18

	Commented by MrW3 last updated on 07/Oct/18

	$D i a g r a m a b o v e s h o w s t h e c a s e μ = 1 .$ $U p o n θ = 74 ° t h e t e n s i o n i n r o p e i s z e r o .$ $T h e m a x i m a l t e n s i o n i n r o p e i s$ $0.28 λ R g w h i c h o c c u r s a t θ = 33.7 ° .$
	Commented by MrW3 last updated on 07/Oct/18

	$T h i s d i a g r a m s h o w s t h e c a s e μ = 0.5 .$ $T o k e e p t h e r o p e f r o m s l i d i n g, a t e n s i o n$ $f o r c e o f 0.23 λ R g i s n e e d e d t o h o l d t h e$ $r o p e a t i t s t o p m o s t p o i n t .$ $T h e m a x . t e n s i o n i n r o p e i s 0.457 λ R g$ $w h i c h o c c u r s a t θ = 51.6 ° .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum