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Question Number 202816 by ajfour last updated on 03/Jan/24

Commented by ajfour last updated on 03/Jan/24

With what constant force should trolley be pushed towards left such that inclination of plank dont change and system moves at uniform speed?

$${With}\:{what}\:{constant}\:{force}\:{should} \\ $$$${trolley}\:{be}\:{pushed}\:{towards}\:{left}\:{such} \\ $$$${that}\:{inclination}\:{of}\:{plank}\:{dont}\:{change} \\ $$$${and}\:{system}\:{moves}\:{at}\:{uniform}\:{speed}? \\ $$

Commented by mr W last updated on 04/Jan/24

is the trolley fixed with the plank and therefore the inclination of the plank given, e.g. with inclanation angle θ?

$${is}\:{the}\:{trolley}\:{fixed}\:{with}\:{the}\:{plank} \\ $$$${and}\:{therefore}\:{the}\:{inclination}\:{of}\:{the}\: \\ $$$${plank}\:{given},\:{e}.{g}.\:{with}\:{inclanation} \\ $$$${angle}\:\theta? \\ $$

Commented by mr W last updated on 04/Jan/24

Commented by mr W last updated on 04/Jan/24

or the trolley is not fixed with the plank, but their contact is frictionless?

$${or}\:{the}\:{trolley}\:{is}\:{not}\:{fixed}\:{with}\:{the} \\ $$$${plank},\:{but}\:{their}\:{contact}\:{is}\:{frictionless}? \\ $$

Commented by mr W last updated on 04/Jan/24

i have solved both cases.

$${i}\:{have}\:{solved}\:{both}\:{cases}. \\ $$

Answered by a.lgnaoui last updated on 03/Jan/24

F_x −F_f =constante M(dv/dt)−𝛍mvx=0 v_x =vcos 𝛂=v(√(1−(b^2 /L^2 ))) ⇒ M(dv/dt)=mv((√(L^2 −b^2 ))/L) donc F=((mv(√(L^2 −b^2 )))/L) { ((m ; mssse de[la tige ( L: longeur))),((v=vitesse[ de[deplacement )) :}

$$\mathrm{F}_{\boldsymbol{\mathrm{x}}} −\boldsymbol{\mathrm{F}}_{\boldsymbol{\mathrm{f}}} =\boldsymbol{\mathrm{constante}} \\ $$$$\boldsymbol{\mathrm{M}}\frac{\boldsymbol{\mathrm{dv}}}{\boldsymbol{\mathrm{dt}}}−\boldsymbol{\mu\mathrm{m}}\mathrm{v}\boldsymbol{\mathrm{x}}=\mathrm{0} \\ $$$$\boldsymbol{\mathrm{v}}_{\boldsymbol{\mathrm{x}}} =\boldsymbol{\mathrm{v}}\mathrm{cos}\:\boldsymbol{\alpha}=\boldsymbol{\mathrm{v}}\sqrt{\mathrm{1}−\frac{\boldsymbol{\mathrm{b}}^{\mathrm{2}} }{\boldsymbol{\mathrm{L}}^{\mathrm{2}} }} \\ $$$$\Rightarrow\:\boldsymbol{\mathrm{M}}\frac{\boldsymbol{\mathrm{dv}}}{\boldsymbol{\mathrm{dt}}}=\boldsymbol{\mathrm{mv}}\frac{\sqrt{\mathrm{L}^{\mathrm{2}} −\boldsymbol{\mathrm{b}}^{\mathrm{2}} }}{\boldsymbol{\mathrm{L}}} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{donc}}\:\:\:\:\:\:\boldsymbol{\mathrm{F}}=\frac{\boldsymbol{\mathrm{mv}}\sqrt{\boldsymbol{\mathrm{L}}^{\mathrm{2}} −\boldsymbol{\mathrm{b}}^{\mathrm{2}} }}{\boldsymbol{\mathrm{L}}} \\ $$$$\begin{cases}{\boldsymbol{\mathrm{m}}\:;\:\:\boldsymbol{\mathrm{mssse}}\:\boldsymbol{\mathrm{de}}\left[\boldsymbol{\mathrm{la}}\:\boldsymbol{\mathrm{tige}}\:\left(\:\boldsymbol{\mathrm{L}}:\:\mathrm{longeur}\right)\right.}\\{\boldsymbol{\mathrm{v}}=\boldsymbol{\mathrm{vitesse}}\left[\:\boldsymbol{\mathrm{de}}\left[\boldsymbol{\mathrm{deplacement}}\:\right.\right.}\end{cases} \\ $$

Answered by mr W last updated on 04/Jan/24

CASE 1: trolley and plank are connected with a hinge and the inclanation of the plank is known (θ).

$${CASE}\:\mathrm{1}: \\ $$$${trolley}\:{and}\:{plank}\:{are}\:{connected}\:{with} \\ $$$${a}\:{hinge}\:{and}\:{the}\:{inclanation}\:{of}\:{the} \\ $$$${plank}\:{is}\:{known}\:\left(\theta\right). \\ $$

Commented by mr W last updated on 04/Jan/24

Commented by mr W last updated on 04/Jan/24

tan φ=μ ((CD)/(sin ((π/2)−φ−θ)))=((AC)/(sin φ)) ⇒CD=((L cos (φ+θ))/(2 sin φ)) =(L/2)(((cos θ)/(tan φ))−sin θ)=(L/2)(((cos θ)/μ)−sin θ) ((CD)/(sin ((π/2)−ϕ+θ)))=((CB)/(sin ϕ)) CD=((((b/(sin θ))−(L/2))cos (ϕ−θ))/(sin ϕ)) =((b/(sin θ))−(L/2))(((cos θ)/(tan ϕ))+sin θ) (L/2)(((cos θ)/μ)−sin θ)=((b/(sin θ))−(L/2))(((cos θ)/(tan ϕ))+sin θ) (1/μ)−tan θ=(((2b)/(L sin θ))−1)((1/(tan ϕ))+tan θ) ⇒(1/(tan ϕ))=(((1/μ)−tan θ)/(((2b)/(L sin θ))−1))−tan θ (R/(sin ϕ))=((mg)/(sin (φ+ϕ))) R=((mg sin ϕ)/(sin (φ+ϕ)))=((mg)/(((sin φ)/(tan ϕ))+cos φ)) F=R sin φ=((mg)/((1/(tan ϕ))+(1/(tan φ)))) =((mg)/((((1/μ)−tan θ)/(((2b)/(L sin θ))−1))−tan θ+(1/μ))) ⇒(F/(mg))=(1/(((1/μ)−tan θ)((1/(((2b)/(L sin θ))−1))+1)))

$$\mathrm{tan}\:\phi=\mu \\ $$$$\frac{{CD}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\phi−\theta\right)}=\frac{{AC}}{\mathrm{sin}\:\phi} \\ $$$$\Rightarrow{CD}=\frac{{L}\:\mathrm{cos}\:\left(\phi+\theta\right)}{\mathrm{2}\:\mathrm{sin}\:\phi} \\ $$$$\:\:\:\:\:\:=\frac{{L}}{\mathrm{2}}\left(\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\phi}−\mathrm{sin}\:\theta\right)=\frac{{L}}{\mathrm{2}}\left(\frac{\mathrm{cos}\:\theta}{\mu}−\mathrm{sin}\:\theta\right) \\ $$$$\frac{{CD}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\varphi+\theta\right)}=\frac{{CB}}{\mathrm{sin}\:\varphi} \\ $$$${CD}=\frac{\left(\frac{{b}}{\mathrm{sin}\:\theta}−\frac{{L}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\varphi−\theta\right)}{\mathrm{sin}\:\varphi} \\ $$$$\:\:\:\:\:\:\:=\left(\frac{{b}}{\mathrm{sin}\:\theta}−\frac{{L}}{\mathrm{2}}\right)\left(\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}+\mathrm{sin}\:\theta\right) \\ $$$$\frac{{L}}{\mathrm{2}}\left(\frac{\mathrm{cos}\:\theta}{\mu}−\mathrm{sin}\:\theta\right)=\left(\frac{{b}}{\mathrm{sin}\:\theta}−\frac{{L}}{\mathrm{2}}\right)\left(\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}+\mathrm{sin}\:\theta\right) \\ $$$$\frac{\mathrm{1}}{\mu}−\mathrm{tan}\:\theta=\left(\frac{\mathrm{2}{b}}{{L}\:\mathrm{sin}\:\theta}−\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{tan}\:\varphi}+\mathrm{tan}\:\theta\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{tan}\:\varphi}=\frac{\frac{\mathrm{1}}{\mu}−\mathrm{tan}\:\theta}{\frac{\mathrm{2}{b}}{{L}\:\mathrm{sin}\:\theta}−\mathrm{1}}−\mathrm{tan}\:\theta \\ $$$$\frac{{R}}{\mathrm{sin}\:\varphi}=\frac{{mg}}{\mathrm{sin}\:\left(\phi+\varphi\right)} \\ $$$${R}=\frac{{mg}\:\mathrm{sin}\:\varphi}{\mathrm{sin}\:\left(\phi+\varphi\right)}=\frac{{mg}}{\frac{\mathrm{sin}\:\phi}{\mathrm{tan}\:\varphi}+\mathrm{cos}\:\phi} \\ $$$${F}={R}\:\mathrm{sin}\:\phi=\frac{{mg}}{\frac{\mathrm{1}}{\mathrm{tan}\:\varphi}+\frac{\mathrm{1}}{\mathrm{tan}\:\phi}} \\ $$$$\:\:\:=\frac{{mg}}{\frac{\frac{\mathrm{1}}{\mu}−\mathrm{tan}\:\theta}{\frac{\mathrm{2}{b}}{{L}\:\mathrm{sin}\:\theta}−\mathrm{1}}−\mathrm{tan}\:\theta+\frac{\mathrm{1}}{\mu}} \\ $$$$\Rightarrow\frac{{F}}{{mg}}=\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{\mu}−\mathrm{tan}\:\theta\right)\left(\frac{\mathrm{1}}{\frac{\mathrm{2}{b}}{{L}\:\mathrm{sin}\:\theta}−\mathrm{1}}+\mathrm{1}\right)} \\ $$

Answered by mr W last updated on 04/Jan/24

CASE 2: trolley and plank are not fixed connected, but their contact is frictionless.

$${CASE}\:\mathrm{2}: \\ $$$${trolley}\:{and}\:{plank}\:{are}\:{not}\:{fixed} \\ $$$${connected},\:{but}\:{their}\:{contact}\:{is} \\ $$$${frictionless}. \\ $$

Commented by mr W last updated on 04/Jan/24

Commented by mr W last updated on 04/Jan/24

tan φ=μ tan φ_1 =μ_1 =0 ⇒φ_1 =0 ⇒ϕ=θ θ≥sin^(−1) (b/L) ((CD)/(sin ((π/2)−φ−θ)))=((AC)/(sin φ)) ⇒CD=((L cos (φ+θ))/(2 sin φ)) =(L/2)(((cos θ)/(tan φ))−sin θ)=(L/2)(((cos θ)/μ)−sin θ) ((CD)/(sin ((π/2)−ϕ+θ)))=((CB)/(sin ϕ)) CD=((((b/(sin θ))−(L/2))cos (ϕ−θ))/(sin ϕ)) =((b/(sin θ))−(L/2))(((cos θ)/(tan ϕ))+sin θ) (L/2)(((cos θ)/μ)−sin θ)=((b/(sin θ))−(L/2))(((cos θ)/(tan ϕ))+sin θ) (1/μ)−tan θ=(((2b)/(L sin θ))−1)((1/(tan ϕ))+tan θ) (1/μ)−tan θ=(((2b)/(L sin θ))−1)((1/(tan θ))+tan θ) ⇒((2b)/(L sin θ))((1/(tan θ))+tan θ)−(1/(tan θ))=(1/μ) ⇒sin θ cos θ (((sin θ)/μ)+cos θ)=((2b)/L) ⇒θ=.... (R/(sin ϕ))=((mg)/(sin (φ+ϕ))) R=((mg sin ϕ)/(sin (φ+ϕ)))=((mg)/(((sin φ)/(tan ϕ))+cos φ)) F=R sin φ=((mg)/((1/(tan ϕ))+(1/(tan φ))))=((mg)/((1/(tan θ))+(1/μ))) ⇒(F/(mg))=(1/((1/(tan θ))+(1/μ))) example: L=5, b=2, μ=0.5 ⇒θ≈0.5182 (29.69°) ⇒(F/(mg))≈0.2664 ⇒θ≈1.1710 (67.09°) ⇒(F/(mg))≈0.4128

$$\mathrm{tan}\:\phi=\mu \\ $$$$\mathrm{tan}\:\phi_{\mathrm{1}} =\mu_{\mathrm{1}} =\mathrm{0}\: \\ $$$$\Rightarrow\phi_{\mathrm{1}} =\mathrm{0}\:\Rightarrow\varphi=\theta \\ $$$$\theta\geqslant\mathrm{sin}^{−\mathrm{1}} \frac{{b}}{{L}} \\ $$$$\frac{{CD}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\phi−\theta\right)}=\frac{{AC}}{\mathrm{sin}\:\phi} \\ $$$$\Rightarrow{CD}=\frac{{L}\:\mathrm{cos}\:\left(\phi+\theta\right)}{\mathrm{2}\:\mathrm{sin}\:\phi} \\ $$$$\:\:\:\:\:\:=\frac{{L}}{\mathrm{2}}\left(\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\phi}−\mathrm{sin}\:\theta\right)=\frac{{L}}{\mathrm{2}}\left(\frac{\mathrm{cos}\:\theta}{\mu}−\mathrm{sin}\:\theta\right) \\ $$$$\frac{{CD}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\varphi+\theta\right)}=\frac{{CB}}{\mathrm{sin}\:\varphi} \\ $$$${CD}=\frac{\left(\frac{{b}}{\mathrm{sin}\:\theta}−\frac{{L}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\varphi−\theta\right)}{\mathrm{sin}\:\varphi} \\ $$$$\:\:\:\:\:\:\:=\left(\frac{{b}}{\mathrm{sin}\:\theta}−\frac{{L}}{\mathrm{2}}\right)\left(\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}+\mathrm{sin}\:\theta\right) \\ $$$$\frac{{L}}{\mathrm{2}}\left(\frac{\mathrm{cos}\:\theta}{\mu}−\mathrm{sin}\:\theta\right)=\left(\frac{{b}}{\mathrm{sin}\:\theta}−\frac{{L}}{\mathrm{2}}\right)\left(\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}+\mathrm{sin}\:\theta\right) \\ $$$$\frac{\mathrm{1}}{\mu}−\mathrm{tan}\:\theta=\left(\frac{\mathrm{2}{b}}{{L}\:\mathrm{sin}\:\theta}−\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{tan}\:\varphi}+\mathrm{tan}\:\theta\right) \\ $$$$\frac{\mathrm{1}}{\mu}−\mathrm{tan}\:\theta=\left(\frac{\mathrm{2}{b}}{{L}\:\mathrm{sin}\:\theta}−\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{tan}\:\theta}+\mathrm{tan}\:\theta\right) \\ $$$$\Rightarrow\frac{\mathrm{2}{b}}{{L}\:\mathrm{sin}\:\theta}\left(\frac{\mathrm{1}}{\mathrm{tan}\:\theta}+\mathrm{tan}\:\theta\right)−\frac{\mathrm{1}}{\mathrm{tan}\:\theta}=\frac{\mathrm{1}}{\mu} \\ $$$$\Rightarrow\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\:\left(\frac{\mathrm{sin}\:\theta}{\mu}+\mathrm{cos}\:\theta\right)=\frac{\mathrm{2}{b}}{{L}} \\ $$$$\Rightarrow\theta=.... \\ $$$$ \\ $$$$\frac{{R}}{\mathrm{sin}\:\varphi}=\frac{{mg}}{\mathrm{sin}\:\left(\phi+\varphi\right)} \\ $$$${R}=\frac{{mg}\:\mathrm{sin}\:\varphi}{\mathrm{sin}\:\left(\phi+\varphi\right)}=\frac{{mg}}{\frac{\mathrm{sin}\:\phi}{\mathrm{tan}\:\varphi}+\mathrm{cos}\:\phi} \\ $$$${F}={R}\:\mathrm{sin}\:\phi=\frac{{mg}}{\frac{\mathrm{1}}{\mathrm{tan}\:\varphi}+\frac{\mathrm{1}}{\mathrm{tan}\:\phi}}=\frac{{mg}}{\frac{\mathrm{1}}{\mathrm{tan}\:\theta}+\frac{\mathrm{1}}{\mu}} \\ $$$$\Rightarrow\frac{{F}}{{mg}}=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{tan}\:\theta}+\frac{\mathrm{1}}{\mu}} \\ $$$$ \\ $$$${example}: \\ $$$${L}=\mathrm{5},\:{b}=\mathrm{2},\:\mu=\mathrm{0}.\mathrm{5} \\ $$$$\Rightarrow\theta\approx\mathrm{0}.\mathrm{5182}\:\left(\mathrm{29}.\mathrm{69}°\right)\:\Rightarrow\frac{{F}}{{mg}}\approx\mathrm{0}.\mathrm{2664} \\ $$$$\Rightarrow\theta\approx\mathrm{1}.\mathrm{1710}\:\left(\mathrm{67}.\mathrm{09}°\right)\:\Rightarrow\frac{{F}}{{mg}}\approx\mathrm{0}.\mathrm{4128} \\ $$

Answered by mr W last updated on 04/Jan/24

CASE 3: as case 2, but with friction coefficient μ_1 ≠0 between the plank and trolley. tan φ=μ tan φ_1 =μ_1 ϕ=θ+φ_1 θ≥sin^(−1) (b/L) ((CD)/(sin ((π/2)−φ−θ)))=((AC)/(sin φ)) ⇒CD=((L cos (φ+θ))/(2 sin φ)) =(L/2)(((cos θ)/(tan φ))−sin θ)=(L/2)(((cos θ)/μ)−sin θ) ((CD)/(sin ((π/2)−ϕ+θ)))=((CB)/(sin ϕ)) CD=((((b/(sin θ))−(L/2))cos (ϕ−θ))/(sin ϕ)) =((b/(sin θ))−(L/2))(((cos θ)/(tan ϕ))+sin θ) (L/2)(((cos θ)/μ)−sin θ)=((b/(sin θ))−(L/2))(((cos θ)/(tan ϕ))+sin θ) (1/μ)−tan θ=(((2b)/(L sin θ))−1)((1/(tan ϕ))+tan θ) (1/μ)−tan θ=(((2b)/(L sin θ))−1)(((1−μ_1 tan θ)/(tan θ+μ_1 ))+tan θ) ⇒(((2b)/(L sin θ))−1)(((1−μ_1 tan θ)/(tan θ+μ_1 ))+tan θ)+tan θ=(1/μ) ⇒θ=.... (R/(sin ϕ))=((mg)/(sin (φ+ϕ))) R=((mg sin ϕ)/(sin (φ+ϕ)))=((mg)/(((sin φ)/(tan ϕ))+cos φ)) F=R sin φ=((mg)/((1/(tan ϕ))+(1/(tan φ))))=((mg)/(((1−μ_1 tan θ)/(μ_1 +tan θ))+(1/μ))) ⇒(F/(mg))=(1/(((1−μ_1 tan θ)/(μ_1 +tan θ))+(1/μ)))

$${CASE}\:\mathrm{3}: \\ $$$${as}\:{case}\:\mathrm{2},\:{but}\:{with}\:{friction}\: \\ $$$${coefficient}\:\mu_{\mathrm{1}} \neq\mathrm{0}\:{between}\:{the}\:{plank}\: \\ $$$${and}\:{trolley}.\: \\ $$$$ \\ $$$$\mathrm{tan}\:\phi=\mu \\ $$$$\mathrm{tan}\:\phi_{\mathrm{1}} =\mu_{\mathrm{1}} \\ $$$$\varphi=\theta+\phi_{\mathrm{1}} \\ $$$$\theta\geqslant\mathrm{sin}^{−\mathrm{1}} \frac{{b}}{{L}} \\ $$$$\frac{{CD}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\phi−\theta\right)}=\frac{{AC}}{\mathrm{sin}\:\phi} \\ $$$$\Rightarrow{CD}=\frac{{L}\:\mathrm{cos}\:\left(\phi+\theta\right)}{\mathrm{2}\:\mathrm{sin}\:\phi} \\ $$$$\:\:\:\:\:\:=\frac{{L}}{\mathrm{2}}\left(\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\phi}−\mathrm{sin}\:\theta\right)=\frac{{L}}{\mathrm{2}}\left(\frac{\mathrm{cos}\:\theta}{\mu}−\mathrm{sin}\:\theta\right) \\ $$$$\frac{{CD}}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{2}}−\varphi+\theta\right)}=\frac{{CB}}{\mathrm{sin}\:\varphi} \\ $$$${CD}=\frac{\left(\frac{{b}}{\mathrm{sin}\:\theta}−\frac{{L}}{\mathrm{2}}\right)\mathrm{cos}\:\left(\varphi−\theta\right)}{\mathrm{sin}\:\varphi} \\ $$$$\:\:\:\:\:\:\:=\left(\frac{{b}}{\mathrm{sin}\:\theta}−\frac{{L}}{\mathrm{2}}\right)\left(\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}+\mathrm{sin}\:\theta\right) \\ $$$$\frac{{L}}{\mathrm{2}}\left(\frac{\mathrm{cos}\:\theta}{\mu}−\mathrm{sin}\:\theta\right)=\left(\frac{{b}}{\mathrm{sin}\:\theta}−\frac{{L}}{\mathrm{2}}\right)\left(\frac{\mathrm{cos}\:\theta}{\mathrm{tan}\:\varphi}+\mathrm{sin}\:\theta\right) \\ $$$$\frac{\mathrm{1}}{\mu}−\mathrm{tan}\:\theta=\left(\frac{\mathrm{2}{b}}{{L}\:\mathrm{sin}\:\theta}−\mathrm{1}\right)\left(\frac{\mathrm{1}}{\mathrm{tan}\:\varphi}+\mathrm{tan}\:\theta\right) \\ $$$$\frac{\mathrm{1}}{\mu}−\mathrm{tan}\:\theta=\left(\frac{\mathrm{2}{b}}{{L}\:\mathrm{sin}\:\theta}−\mathrm{1}\right)\left(\frac{\mathrm{1}−\mu_{\mathrm{1}} \:\mathrm{tan}\:\theta}{\mathrm{tan}\:\theta+\mu_{\mathrm{1}} }+\mathrm{tan}\:\theta\right) \\ $$$$\Rightarrow\left(\frac{\mathrm{2}{b}}{{L}\:\mathrm{sin}\:\theta}−\mathrm{1}\right)\left(\frac{\mathrm{1}−\mu_{\mathrm{1}} \:\mathrm{tan}\:\theta}{\mathrm{tan}\:\theta+\mu_{\mathrm{1}} }+\mathrm{tan}\:\theta\right)+\mathrm{tan}\:\theta=\frac{\mathrm{1}}{\mu} \\ $$$$\Rightarrow\theta=.... \\ $$$$ \\ $$$$\frac{{R}}{\mathrm{sin}\:\varphi}=\frac{{mg}}{\mathrm{sin}\:\left(\phi+\varphi\right)} \\ $$$${R}=\frac{{mg}\:\mathrm{sin}\:\varphi}{\mathrm{sin}\:\left(\phi+\varphi\right)}=\frac{{mg}}{\frac{\mathrm{sin}\:\phi}{\mathrm{tan}\:\varphi}+\mathrm{cos}\:\phi} \\ $$$${F}={R}\:\mathrm{sin}\:\phi=\frac{{mg}}{\frac{\mathrm{1}}{\mathrm{tan}\:\varphi}+\frac{\mathrm{1}}{\mathrm{tan}\:\phi}}=\frac{{mg}}{\frac{\mathrm{1}−\mu_{\mathrm{1}} \:\mathrm{tan}\:\theta}{\mu_{\mathrm{1}} +\mathrm{tan}\:\theta}+\frac{\mathrm{1}}{\mu}} \\ $$$$\Rightarrow\frac{{F}}{{mg}}=\frac{\mathrm{1}}{\frac{\mathrm{1}−\mu_{\mathrm{1}} \:\mathrm{tan}\:\theta}{\mu_{\mathrm{1}} +\mathrm{tan}\:\theta}+\frac{\mathrm{1}}{\mu}} \\ $$

Answered by ajfour last updated on 04/Jan/24

Commented by ajfour last updated on 04/Jan/24

Case: free roller vontact between plank and trolley. μ(mg−Ncos θ)=Nsin θ ⇒ N=((μmg)/(sin θ+μcos θ)) =((μmg(1+t^2 ))/(2t+μ(1−t^2 ))) & N((b/(sin θ)))=((mgLcos θ)/2) ⇒ N=((mgL)/(4b))sin 2θ=((mgL()/(4b))((4t)(1−t^2 ))/((1+t^2 )^2 )) If t=tan (θ/2) ⇒ μ=(sin θ+μcos θ)(((Lsin θcos θ)/(2b))) or ((2t+μ(1−t^2 ))/((1+t^2 )))=((μ(b/L)(1+t^2 )^2 )/(t(1−t^2 ))) t(1−t^2 ){2t+μ(1−t^2 )}=kμ(1+t^2 )^3 matter of degree six.. a=(sin θ+ccos θ)sin θcos

$${Case}:\:{free}\:{roller}\:{vontact}\:{between} \\ $$$${plank}\:{and}\:{trolley}. \\ $$$$\mu\left({mg}−{N}\mathrm{cos}\:\theta\right)={N}\mathrm{sin}\:\theta\: \\ $$$$\Rightarrow\:\:{N}=\frac{\mu{mg}}{\mathrm{sin}\:\theta+\mu\mathrm{cos}\:\theta} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mu{mg}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}{\mathrm{2}{t}+\mu\left(\mathrm{1}−{t}^{\mathrm{2}} \right)} \\ $$$$\&\:\:\:\:{N}\left(\frac{{b}}{\mathrm{sin}\:\theta}\right)=\frac{{mgL}\mathrm{cos}\:\theta}{\mathrm{2}} \\ $$$$\Rightarrow\:{N}=\frac{{mgL}}{\mathrm{4}{b}}\mathrm{sin}\:\mathrm{2}\theta=\frac{{mgL}\left(\right.}{\mathrm{4}{b}}\frac{\left.\mathrm{4}{t}\right)\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${If}\:{t}=\mathrm{tan}\:\frac{\theta}{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mu=\left(\mathrm{sin}\:\theta+\mu\mathrm{cos}\:\theta\right)\left(\frac{{L}\mathrm{sin}\:\theta\mathrm{cos}\:\theta}{\mathrm{2}{b}}\right) \\ $$$${or} \\ $$$$\frac{\mathrm{2}{t}+\mu\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}=\frac{\mu\left({b}/{L}\right)\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{{t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)} \\ $$$${t}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\left\{\mathrm{2}{t}+\mu\left(\mathrm{1}−{t}^{\mathrm{2}} \right)\right\}={k}\mu\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{3}} \\ $$$${matter}\:{of}\:{degree}\:{six}.. \\ $$$${a}=\left(\mathrm{sin}\:\theta+{c}\mathrm{cos}\:\theta\right)\mathrm{sin}\:\theta\mathrm{cos}\: \\ $$

Commented by mr W last updated on 04/Jan/24

μ(mg−Ncos θ)=F+Nsin θ F=N sin θ

$$\mu\left({mg}−{N}\mathrm{cos}\:\theta\right)=\cancel{{F}+}{N}\mathrm{sin}\:\theta \\ $$$${F}={N}\:\mathrm{sin}\:\theta \\ $$

Commented by mr W last updated on 04/Jan/24

θ must fulfill sin θ cos θ(((sin θ)/μ)+cos θ)=((2b)/L)

$$\theta\:{must}\:{fulfill} \\ $$$$\mathrm{sin}\:\theta\:\mathrm{cos}\:\theta\left(\frac{\mathrm{sin}\:\theta}{\mu}+\mathrm{cos}\:\theta\right)=\frac{\mathrm{2}{b}}{{L}} \\ $$

Commented by mr W last updated on 04/Jan/24

we get the same for (F/(mg)): N=((μmg)/(sin θ+μ cos θ)) F=N sin θ=((μmg sin θ)/(sin θ+μ cos θ))=((mg)/((1/(tan θ))+(1/μ))) ⇒(F/(mg))=(1/((1/μ)+(1/(tan θ))))

$${we}\:{get}\:{the}\:{same}\:{for}\:\frac{{F}}{{mg}}: \\ $$$${N}=\frac{\mu{mg}}{\mathrm{sin}\:\theta+\mu\:\mathrm{cos}\:\theta} \\ $$$${F}={N}\:\mathrm{sin}\:\theta=\frac{\mu{mg}\:\mathrm{sin}\:\theta}{\mathrm{sin}\:\theta+\mu\:\mathrm{cos}\:\theta}=\frac{{mg}}{\frac{\mathrm{1}}{\mathrm{tan}\:\theta}+\frac{\mathrm{1}}{\mu}} \\ $$$$\Rightarrow\frac{{F}}{{mg}}=\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mu}+\frac{\mathrm{1}}{\mathrm{tan}\:\theta}} \\ $$

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