Question Number 225338 by mr W last updated on 22/Oct/25
Commented by mr W last updated on 24/Oct/25
$${as}\:{Q}\mathrm{225146},\:{but}\:{the}\:{wedge}\:{can} \\ $$$${slide}\:{on}\:{the}\:{ground}\:{frictionlessly}. \\ $$
Answered by mr W last updated on 25/Oct/25
![V, A = velocity and acceleration of wedge (→) v, a = relative velocity and acceleration of block along the wedge (↙) let λ=(M/m) A=(dV/dt) v=(dx/dt) a=(dv/dt)=v(dv/dx) conservation of momentum: MV=m(v cos α−V) V=((mv cos α)/(M+m))=((v cos α)/(λ+1)) ⇒A=((a cos α)/(λ+1)) mA sin α=mg cos α−N ⇒N=m(g cos α−A sin α) m(a−A cos α)=mg sin α−f f=μN=kxN m(a−A cos α)=mg sin α−kxm(g cos α−A sin α) a−A(cos α+kx sin α)=g(sin α−kx cos α) a−((a cos α)/(λ+1))(cos α+kx sin α)=g(sin α−kx cos α) [λ+sin α(sin α−kx cos α)]a=(λ+1)g(sin α−kx cos α) ⇒a=(((λ+1)g(sin α−kx cos α))/(λ+sin α(sin α−kx cos α))) ⇒A=((g cos α (sin α−kx cos α))/(λ+sin α(sin α−kx cos α))) let δ=k cos α sin α v(dv/dx)=(((λ+1)g(sin α−kx cos α))/(λ+sin α(sin α−kx cos α))) vdv=(((λ+1)g(sin α−kx cos α)dx)/(λ+sin α(sin α−kx cos α))) ∫_0 ^v vdv=(((λ+1)g)/(sin α))∫_0 ^x (((sin α−kx cos α)dx)/((λ/(sin α))+(sin α−kx cos α))) ∫_0 ^v vdv=−(((λ+1)g)/δ)∫_0 ^x (((sin α−kx cos α)d(sin α−kx cos α))/((λ/(sin α))+(sin α−kx cos α))) (v^2 /2)=−(((λ+1)g)/δ)∫_(sin α) ^(sin α−kx cos α) ((udu)/((λ/(sin α))+u)) (v^2 /2)=(((λ+1)g)/δ)[u−(λ/(sin α)) ln ((λ/(sin α))+u)]_(sin α−kx cos α) ^(sin α) (v^2 /2)=(((λ+1)g)/δ)[kx cos α+(λ/(sin α)) ln (((λ/(sin α))+sin α−kx cos α)/((λ/(sin α))+sin α))] (v^2 /2)=(((λ+1)g)/(δ sin α))[δx+λ ln (1−((δx)/(λ+sin^2 α)))] ⇒v=(√((2(λ+1)g)/(δ sin α))) (√(δx+λ ln (1−((δx)/(λ+sin^2 α))))) (dx/dt)=(√((2(λ+1)g)/(δ sin α))) (√(δx+λ ln (1−((δx)/(λ+sin^2 α))))) t=(√((δ sin α)/(2(λ+1)g)))∫_0 ^x (dx/( (√(δx+λ ln (1−((δx)/(λ+sin^2 α))))))) t=(√((sin α)/(2δ(λ+1)g)))∫_0 ^x ((d(δx))/( (√(δx+λ ln (1−((δx)/(λ+sin^2 α))))))) ⇒t=(√((sin α)/(2δ(λ+1)g)))∫_0 ^(δx) (du/( (√(u+λ ln (1−(u/(λ+sin^2 α))))))) this integral can only be solved nummerically. at x_(max) : v=0 δx_(max) +λ ln (1−((δx_(max) )/(λ+sin^2 α)))=0 with ξ=((δx_(max) )/λ) ln (1−((ξλ)/(λ+sin^2 α)))=−ξ 1−(ξ/(1+((sin^2 α)/λ)))=e^(−ξ) (ξ−1−((sin^2 α)/λ))e^ξ =−(1+((sin^2 α)/λ)) (ξ−1−((sin^2 α)/λ))e^(ξ−1−((sin^2 α)/λ)) =−((1+((sin^2 α)/λ))/e^(1+((sin^2 α)/λ)) ) using lambert W function we get ⇒ξ−1−((sin^2 α)/λ)=W(−((1+((sin^2 α)/λ))/e^(1+((sin^2 α)/λ)) )) ⇒ξ=1+((sin^2 α)/λ)+W(−((1+((sin^2 α)/λ))/e^(1+((sin^2 α)/λ)) )) ⇒((δx_(max) )/λ)=1+((sin^2 α)/λ)+W(−((1+((sin^2 α)/λ))/e^(1+((sin^2 α)/λ)) )) ⇒x_(max) =(λ/δ)[1+((sin^2 α)/λ)+W(−((1+((sin^2 α)/λ))/e^(1+((sin^2 α)/λ)) ))] W(−((1+((sin^2 α)/λ))/e^(1+((sin^2 α)/λ)) )) has two values. one is −(1+((sin^2 α)/λ)), which leads to x=0, the other value delivers the x_(max) .](https://www.tinkutara.com/question/Q225339.png)
$${V},\:{A}\:=\:{velocity}\:{and}\:{acceleration}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{of}\:{wedge}\:\:\left(\rightarrow\right) \\ $$$${v},\:{a}\:=\:{relative}\:{velocity}\:{and}\:{acceleration}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{of}\:{block}\:{along}\:{the}\:{wedge}\:\:\left(\swarrow\right) \\ $$$${let}\:\lambda=\frac{{M}}{{m}} \\ $$$${A}=\frac{{dV}}{{dt}} \\ $$$${v}=\frac{{dx}}{{dt}} \\ $$$${a}=\frac{{dv}}{{dt}}={v}\frac{{dv}}{{dx}} \\ $$$${conservation}\:{of}\:{momentum}: \\ $$$${MV}={m}\left({v}\:\mathrm{cos}\:\alpha−{V}\right) \\ $$$${V}=\frac{{mv}\:\mathrm{cos}\:\alpha}{{M}+{m}}=\frac{{v}\:\mathrm{cos}\:\alpha}{\lambda+\mathrm{1}} \\ $$$$\Rightarrow{A}=\frac{{a}\:\mathrm{cos}\:\alpha}{\lambda+\mathrm{1}} \\ $$$${mA}\:\mathrm{sin}\:\alpha={mg}\:\mathrm{cos}\:\alpha−{N} \\ $$$$\Rightarrow{N}={m}\left({g}\:\mathrm{cos}\:\alpha−{A}\:\mathrm{sin}\:\alpha\right) \\ $$$${m}\left({a}−{A}\:\mathrm{cos}\:\alpha\right)={mg}\:\mathrm{sin}\:\alpha−{f} \\ $$$${f}=\mu{N}={kxN} \\ $$$${m}\left({a}−{A}\:\mathrm{cos}\:\alpha\right)={mg}\:\mathrm{sin}\:\alpha−{kxm}\left({g}\:\mathrm{cos}\:\alpha−{A}\:\mathrm{sin}\:\alpha\right) \\ $$$${a}−{A}\left(\mathrm{cos}\:\alpha+{kx}\:\mathrm{sin}\:\alpha\right)={g}\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right) \\ $$$${a}−\frac{{a}\:\mathrm{cos}\:\alpha}{\lambda+\mathrm{1}}\left(\mathrm{cos}\:\alpha+{kx}\:\mathrm{sin}\:\alpha\right)={g}\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right) \\ $$$$\left[\lambda+\mathrm{sin}\:\alpha\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right)\right]{a}=\left(\lambda+\mathrm{1}\right){g}\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right) \\ $$$$\Rightarrow{a}=\frac{\left(\lambda+\mathrm{1}\right){g}\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right)}{\lambda+\mathrm{sin}\:\alpha\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right)} \\ $$$$\Rightarrow{A}=\frac{{g}\:\mathrm{cos}\:\alpha\:\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right)}{\lambda+\mathrm{sin}\:\alpha\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right)} \\ $$$${let}\:\delta={k}\:\mathrm{cos}\:\alpha\:\mathrm{sin}\:\alpha \\ $$$${v}\frac{{dv}}{{dx}}=\frac{\left(\lambda+\mathrm{1}\right){g}\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right)}{\lambda+\mathrm{sin}\:\alpha\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right)} \\ $$$${vdv}=\frac{\left(\lambda+\mathrm{1}\right){g}\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right){dx}}{\lambda+\mathrm{sin}\:\alpha\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right)} \\ $$$$\int_{\mathrm{0}} ^{{v}} {vdv}=\frac{\left(\lambda+\mathrm{1}\right){g}}{\mathrm{sin}\:\alpha}\int_{\mathrm{0}} ^{{x}} \frac{\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right){dx}}{\frac{\lambda}{\mathrm{sin}\:\alpha}+\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right)} \\ $$$$\int_{\mathrm{0}} ^{{v}} {vdv}=−\frac{\left(\lambda+\mathrm{1}\right){g}}{\delta}\int_{\mathrm{0}} ^{{x}} \frac{\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right){d}\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right)}{\frac{\lambda}{\mathrm{sin}\:\alpha}+\left(\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha\right)} \\ $$$$\frac{{v}^{\mathrm{2}} }{\mathrm{2}}=−\frac{\left(\lambda+\mathrm{1}\right){g}}{\delta}\int_{\mathrm{sin}\:\alpha} ^{\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha} \frac{{udu}}{\frac{\lambda}{\mathrm{sin}\:\alpha}+{u}} \\ $$$$\frac{{v}^{\mathrm{2}} }{\mathrm{2}}=\frac{\left(\lambda+\mathrm{1}\right){g}}{\delta}\left[{u}−\frac{\lambda}{\mathrm{sin}\:\alpha}\:\mathrm{ln}\:\left(\frac{\lambda}{\mathrm{sin}\:\alpha}+{u}\right)\right]_{\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha} ^{\mathrm{sin}\:\alpha} \\ $$$$\frac{{v}^{\mathrm{2}} }{\mathrm{2}}=\frac{\left(\lambda+\mathrm{1}\right){g}}{\delta}\left[{kx}\:\mathrm{cos}\:\alpha+\frac{\lambda}{\mathrm{sin}\:\alpha}\:\mathrm{ln}\:\frac{\frac{\lambda}{\mathrm{sin}\:\alpha}+\mathrm{sin}\:\alpha−{kx}\:\mathrm{cos}\:\alpha}{\frac{\lambda}{\mathrm{sin}\:\alpha}+\mathrm{sin}\:\alpha}\right] \\ $$$$\frac{{v}^{\mathrm{2}} }{\mathrm{2}}=\frac{\left(\lambda+\mathrm{1}\right){g}}{\delta\:\mathrm{sin}\:\alpha}\left[\delta{x}+\lambda\:\mathrm{ln}\:\left(\mathrm{1}−\frac{\delta{x}}{\lambda+\mathrm{sin}^{\mathrm{2}} \:\alpha}\right)\right] \\ $$$$\Rightarrow{v}=\sqrt{\frac{\mathrm{2}\left(\lambda+\mathrm{1}\right){g}}{\delta\:\mathrm{sin}\:\alpha}}\:\sqrt{\delta{x}+\lambda\:\mathrm{ln}\:\left(\mathrm{1}−\frac{\delta{x}}{\lambda+\mathrm{sin}^{\mathrm{2}} \:\alpha}\right)} \\ $$$$\frac{{dx}}{{dt}}=\sqrt{\frac{\mathrm{2}\left(\lambda+\mathrm{1}\right){g}}{\delta\:\mathrm{sin}\:\alpha}}\:\sqrt{\delta{x}+\lambda\:\mathrm{ln}\:\left(\mathrm{1}−\frac{\delta{x}}{\lambda+\mathrm{sin}^{\mathrm{2}} \:\alpha}\right)} \\ $$$${t}=\sqrt{\frac{\delta\:\mathrm{sin}\:\alpha}{\mathrm{2}\left(\lambda+\mathrm{1}\right){g}}}\int_{\mathrm{0}} ^{{x}} \frac{{dx}}{\:\sqrt{\delta{x}+\lambda\:\mathrm{ln}\:\left(\mathrm{1}−\frac{\delta{x}}{\lambda+\mathrm{sin}^{\mathrm{2}} \:\alpha}\right)}} \\ $$$${t}=\sqrt{\frac{\mathrm{sin}\:\alpha}{\mathrm{2}\delta\left(\lambda+\mathrm{1}\right){g}}}\int_{\mathrm{0}} ^{{x}} \frac{{d}\left(\delta{x}\right)}{\:\sqrt{\delta{x}+\lambda\:\mathrm{ln}\:\left(\mathrm{1}−\frac{\delta{x}}{\lambda+\mathrm{sin}^{\mathrm{2}} \:\alpha}\right)}} \\ $$$$\Rightarrow{t}=\sqrt{\frac{\mathrm{sin}\:\alpha}{\mathrm{2}\delta\left(\lambda+\mathrm{1}\right){g}}}\int_{\mathrm{0}} ^{\delta{x}} \frac{{du}}{\:\sqrt{{u}+\lambda\:\mathrm{ln}\:\left(\mathrm{1}−\frac{{u}}{\lambda+\mathrm{sin}^{\mathrm{2}} \:\alpha}\right)}} \\ $$$${this}\:{integral}\:{can}\:{only}\:{be}\:{solved}\: \\ $$$${nummerically}. \\ $$$$ \\ $$$${at}\:{x}_{{max}} :\:{v}=\mathrm{0} \\ $$$$\delta{x}_{{max}} +\lambda\:\mathrm{ln}\:\left(\mathrm{1}−\frac{\delta{x}_{{max}} }{\lambda+\mathrm{sin}^{\mathrm{2}} \:\alpha}\right)=\mathrm{0} \\ $$$${with}\:\xi=\frac{\delta{x}_{{max}} }{\lambda} \\ $$$$\mathrm{ln}\:\left(\mathrm{1}−\frac{\xi\lambda}{\lambda+\mathrm{sin}^{\mathrm{2}} \:\alpha}\right)=−\xi \\ $$$$\mathrm{1}−\frac{\xi}{\mathrm{1}+\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\lambda}}={e}^{−\xi} \\ $$$$\left(\xi−\mathrm{1}−\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\lambda}\right){e}^{\xi} =−\left(\mathrm{1}+\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\lambda}\right) \\ $$$$\left(\xi−\mathrm{1}−\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\lambda}\right){e}^{\xi−\mathrm{1}−\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\lambda}} =−\frac{\mathrm{1}+\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\lambda}}{{e}^{\mathrm{1}+\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\lambda}} } \\ $$$${using}\:{lambert}\:{W}\:{function}\:{we}\:{get} \\ $$$$\Rightarrow\xi−\mathrm{1}−\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\lambda}={W}\left(−\frac{\mathrm{1}+\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\lambda}}{{e}^{\mathrm{1}+\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\lambda}} }\right) \\ $$$$\Rightarrow\xi=\mathrm{1}+\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\lambda}+{W}\left(−\frac{\mathrm{1}+\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\lambda}}{{e}^{\mathrm{1}+\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\lambda}} }\right) \\ $$$$\Rightarrow\frac{\delta{x}_{{max}} }{\lambda}=\mathrm{1}+\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\lambda}+{W}\left(−\frac{\mathrm{1}+\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\lambda}}{{e}^{\mathrm{1}+\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\lambda}} }\right) \\ $$$$\Rightarrow{x}_{{max}} =\frac{\lambda}{\delta}\left[\mathrm{1}+\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\lambda}+{W}\left(−\frac{\mathrm{1}+\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\lambda}}{{e}^{\mathrm{1}+\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\lambda}} }\right)\right] \\ $$$${W}\left(−\frac{\mathrm{1}+\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\lambda}}{{e}^{\mathrm{1}+\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\lambda}} }\right)\:{has}\:{two}\:{values}.\:{one} \\ $$$${is}\:−\left(\mathrm{1}+\frac{\mathrm{sin}^{\mathrm{2}} \:\alpha}{\lambda}\right),\:{which}\:{leads}\:{to}\:{x}=\mathrm{0}, \\ $$$${the}\:{other}\:{value}\:{delivers}\:{the}\:{x}_{{max}} . \\ $$
Commented by mr W last updated on 22/Oct/25
$${examples}\:{for}\:\alpha=\mathrm{30}°,\:{k}=\mathrm{0}.\mathrm{5},\:\lambda=\frac{{M}}{{m}} \\ $$$${green}:\:{v}\left({x}\right)\:{with}\:{sliding}\:{wedge}\: \\ $$$${pink}:\:{v}\left({x}\right)\:{with}\:{fixed}\:{wedge}\: \\ $$
Commented by mr W last updated on 22/Oct/25
Commented by mr W last updated on 22/Oct/25
Commented by mr W last updated on 22/Oct/25
Commented by mr W last updated on 24/Oct/25
$${we}\:{see}\:{in}\:{case}\:{of}\:{sliding}\:{wedge}\:{on} \\ $$$${the}\:{ground}\:{the}\:{block}\:{moves}\:{faster} \\ $$$${than}\:{when}\:{the}\:{wedge}\:{is}\:{fixed}\:{on}\:{the} \\ $$$${ground}. \\ $$$${but}\:{with}\:{increasing}\:{mass}\:{of}\:{the}\:{wedge} \\ $$$${both}\:{cases}\:{approach}\:{to}\:{being}\:{the} \\ $$$${same}. \\ $$$${in}\:{both}\:{cases}\:{there}\:{exists}\:{a}\:{maximum} \\ $$$${velocity}\:{for}\:{the}\:{block}\:{and}\:{it}\:{will}\:{stop} \\ $$$${after}\:{it}\:{has}\:{moved}\:{a}\:{certain}\:{distance}. \\ $$
Commented by fantastic last updated on 22/Oct/25
$${WOW}\:{sir} \\ $$