Q225146-here-if-the-wedge-and-the-ground-had-no-friction-it-would-start-going-what-is-acc-at-time-t-kx-

Question Number 225282 by fantastic last updated on 20/Oct/25

Q225146 here if the wedge and the ground had no friction it would start going → what is acc. at time t? μ=kx

$${Q}\mathrm{225146} \\ $$$${here}\:{if}\:{the}\:{wedge}\:{and}\:{the} \\ $$$${ground}\:\:{had}\:{no}\:{friction} \\ $$$${it}\:{would}\:{start}\:{going}\:\rightarrow \\ $$$${what}\:{is}\:{acc}.\:{at}\:{time}\:{t}? \\ $$$$\mu={kx} \\ $$

Commented by mr W last updated on 20/Oct/25

$${nice}\:{question}! \\ $$$${in}\:{this}\:{case}\:{we}\:{need}\:{to}\:{know}\:{the} \\ $$$${masses}\:{from}\:{block}\:{and}\:{wedge}. \\ $$

Commented by fantastic last updated on 20/Oct/25

$${Take}\:{m}\:{and}\:{M} \\ $$

Commented by fantastic last updated on 20/Oct/25

$${Also}\:{find}\:{the}\:{acc}.{of}\:{the}\:{block} \\ $$

Commented by fantastic last updated on 21/Oct/25

pseudo force=mA a=(mgsin α+mAcos α−μn)/m n+mAsin α=mgcos α ⇒n=mgcos α−mAsin α μ=ξx a=(gsin α+Acos α−ξxgcos α+ξxAsin α) (dv/dt)=(gsin α+Acos α−ξxgcos α+ξxAsin α) dv=gsin αdt+Acos αdt−ξxgcos αdt+ξxAsin αdt ∫_0 ^v dv=∫_0 ^t gsin αdt+∫_0 ^t Acos αdt−∫_0 ^t ξxgcos αdt+∫_0 ^t ξxAsin αdt (dx/dt)=mgsin αt+mAcos αt−ξxmgcos αt+ξxmAsin αt ∫_0 ^x dx=∫_0 ^t gsin αtdt+∫_0 ^t Acos αtdt−∫_0 ^t ξxgcos αtdt+∫_0 ^t ξxAsin αtdt x=(1/2)gt^2 sin α+(1/2)At^2 cos α−(1/2)ξgxt^2 cos α+(1/2)ξAxt^2 sin α x(2+ξgt^2 cos α−ξAt^2 sin α)=t^2 (gsin α+Acos α) ∴x=((t^2 (gsin α+Acos α))/((2+ξt^2 (gcos α−Asin α))) = So at time t μ=ξx=ξ f=μn=ξ m(gcos α−Asin α) mgsin α+mAcos α−ξ m(gcos α−Asin α)=ma a=gsin α+Acos α−ξ (gcos α−Asin α).....(i) now if we draw the FBD of the wedge normal force will act on its surface perpendicularly one component will be horizontal ncos α one will be perpendicular nsin α so ncos α=MA ⇒mgcos^2 α−mAsin αcos α=MA ⇒mgcos^2 α=A(M+msin αcos α) determinant (((∴A=((mgcos^2 α)/((M+msin αcos α)))))) a=gsin α+Acos α−ξ (gcos α−Asin α) a=(gsin α+(((mgcos^2 α)/((M+msin αcos α))))cos α−ξ(((t^2 (gsin α+Acos α))/((2+ξt^2 (gcos α−Asin α))))(gcos α−((mgcos^2 α)/((M+msin αcos α)))sin α))

$$ \\ $$$${pseudo}\:{force}={mA} \\ $$$${a}=\left({mg}\mathrm{sin}\:\alpha+{mA}\mathrm{cos}\:\alpha−\mu{n}\right)/{m} \\ $$$${n}+{mA}\mathrm{sin}\:\alpha={mg}\mathrm{cos}\:\alpha \\ $$$$\Rightarrow{n}={mg}\mathrm{cos}\:\alpha−{mA}\mathrm{sin}\:\alpha \\ $$$$\mu=\xi{x} \\ $$$${a}=\left({g}\mathrm{sin}\:\alpha+{A}\mathrm{cos}\:\alpha−\xi{xg}\mathrm{cos}\:\alpha+\xi{xA}\mathrm{sin}\:\alpha\right) \\ $$$$\frac{{dv}}{{dt}}=\left({g}\mathrm{sin}\:\alpha+{A}\mathrm{cos}\:\alpha−\xi{xg}\mathrm{cos}\:\alpha+\xi{xA}\mathrm{sin}\:\alpha\right) \\ $$$${dv}={g}\mathrm{sin}\:\alpha{dt}+{A}\mathrm{cos}\:\alpha{dt}−\xi{xg}\mathrm{cos}\:\alpha{dt}+\xi{xA}\mathrm{sin}\:\alpha{dt} \\ $$$$\int_{\mathrm{0}} ^{{v}} {dv}=\int_{\mathrm{0}} ^{{t}} \:{g}\mathrm{sin}\:\alpha{dt}+\int_{\mathrm{0}} ^{{t}} {A}\mathrm{cos}\:\alpha{dt}−\int_{\mathrm{0}} ^{{t}} \xi{xg}\mathrm{cos}\:\alpha{dt}+\int_{\mathrm{0}} ^{{t}} \xi{xA}\mathrm{sin}\:\alpha{dt} \\ $$$$\frac{{dx}}{{dt}}={mg}\mathrm{sin}\:\alpha{t}+{mA}\mathrm{cos}\:\alpha{t}−\xi{xmg}\mathrm{cos}\:\alpha{t}+\xi{xmA}\mathrm{sin}\:\alpha{t} \\ $$$$\int_{\mathrm{0}} ^{{x}} {dx}=\int_{\mathrm{0}} ^{{t}} {g}\mathrm{sin}\:\alpha{tdt}+\int_{\mathrm{0}} ^{{t}} {A}\mathrm{cos}\:\alpha{tdt}−\int_{\mathrm{0}} ^{{t}} \xi{xg}\mathrm{cos}\:\alpha{tdt}+\int_{\mathrm{0}} ^{{t}} \xi{xA}\mathrm{sin}\:\alpha{tdt} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \mathrm{sin}\:\alpha+\frac{\mathrm{1}}{\mathrm{2}}{At}^{\mathrm{2}} \mathrm{cos}\:\alpha−\frac{\mathrm{1}}{\mathrm{2}}\xi{gxt}^{\mathrm{2}} \mathrm{cos}\:\alpha+\frac{\mathrm{1}}{\mathrm{2}}\xi{Axt}^{\mathrm{2}} \mathrm{sin}\:\alpha \\ $$$${x}\left(\mathrm{2}+\xi{gt}^{\mathrm{2}} \mathrm{cos}\:\alpha−\xi{At}^{\mathrm{2}} \mathrm{sin}\:\alpha\right)={t}^{\mathrm{2}} \left({g}\mathrm{sin}\:\alpha+{A}\mathrm{cos}\:\alpha\right) \\ $$$$\therefore{x}=\frac{{t}^{\mathrm{2}} \left({g}\mathrm{sin}\:\alpha+{A}\mathrm{cos}\:\alpha\right)}{\left(\mathrm{2}+\xi{t}^{\mathrm{2}} \left({g}\mathrm{cos}\:\alpha−{A}\mathrm{sin}\:\alpha\right)\right.}\:=\: \\ $$$${So}\:{at}\:{time}\:{t}\:\mu=\xi{x}=\xi\: \\ $$$${f}=\mu{n}=\xi\:{m}\left({g}\mathrm{cos}\:\alpha−{A}\mathrm{sin}\:\alpha\right) \\ $$$${mg}\mathrm{sin}\:\alpha+{mA}\mathrm{cos}\:\alpha−\xi\:{m}\left({g}\mathrm{cos}\:\alpha−{A}\mathrm{sin}\:\alpha\right)={ma} \\ $$$${a}={g}\mathrm{sin}\:\alpha+{A}\mathrm{cos}\:\alpha−\xi\:\left({g}\mathrm{cos}\:\alpha−{A}\mathrm{sin}\:\alpha\right)…..\left({i}\right) \\ $$$${now}\:{if}\:{we}\:{draw}\:{the}\:{FBD}\:{of}\:{the}\:{wedge}\: \\ $$$${normal}\:{force}\:{will}\:{act}\:{on}\:{its}\:{surface}\:{perpendicularly} \\ $$$${one}\:{component}\:{will}\:{be}\:{horizontal}\:{n}\mathrm{cos}\:\alpha \\ $$$${one}\:{will}\:{be}\:{perpendicular}\:{n}\mathrm{sin}\:\alpha \\ $$$${so} \\ $$$${n}\mathrm{cos}\:\alpha={MA} \\ $$$$\Rightarrow{mg}\mathrm{cos}^{\mathrm{2}} \:\alpha−{mA}\mathrm{sin}\:\alpha\mathrm{cos}\:\alpha={MA} \\ $$$$\Rightarrow{mg}\mathrm{cos}\:^{\mathrm{2}} \alpha={A}\left({M}+{m}\mathrm{sin}\:\alpha\mathrm{cos}\:\:\alpha\right) \\ $$$$\begin{array}{|c|}{\therefore{A}=\frac{{mg}\mathrm{cos}\:^{\mathrm{2}} \alpha}{\left({M}+{m}\mathrm{sin}\:\alpha\mathrm{cos}\:\alpha\right)}}\\\hline\end{array} \\ $$$${a}={g}\mathrm{sin}\:\alpha+{A}\mathrm{cos}\:\alpha−\xi\:\left({g}\mathrm{cos}\:\alpha−{A}\mathrm{sin}\:\alpha\right) \\ $$$${a}=\left({g}\mathrm{sin}\:\alpha+\left(\frac{{mg}\mathrm{cos}\:^{\mathrm{2}} \alpha}{\left({M}+{m}\mathrm{sin}\:\alpha\mathrm{cos}\:\alpha\right)}\right)\mathrm{cos}\:\alpha−\xi\left(\frac{{t}^{\mathrm{2}} \left({g}\mathrm{sin}\:\alpha+{A}\mathrm{cos}\:\alpha\right)}{\left(\mathrm{2}+\xi{t}^{\mathrm{2}} \left({g}\mathrm{cos}\:\alpha−{A}\mathrm{sin}\:\alpha\right)\right.}\right)\left({g}\mathrm{cos}\:\alpha−\frac{{mg}\mathrm{cos}\:^{\mathrm{2}} \alpha}{\left({M}+{m}\mathrm{sin}\:\alpha\mathrm{cos}\:\alpha\right)}\mathrm{sin}\:\alpha\right)\right) \\ $$

Answered by mr W last updated on 20/Oct/25

Commented by fantastic last updated on 20/Oct/25

sir i think A=(((mgcos^2 α)/(M+msin αcos α))) and a=(((mgcos^2 α)/(M+msin αcos α)))(cos α+sin α)+g(sin α−ϕ(((1/2)gt^2 sin α)/(1+gϕ(1/2)t^2 cos α))cos α)

$${sir}\:{i}\:{think} \\ $$$${A}=\left(\frac{{mg}\mathrm{cos}\:^{\mathrm{2}} \alpha}{{M}+{m}\mathrm{sin}\:\alpha\mathrm{cos}\:\alpha}\right) \\ $$$${and}\: \\ $$$${a}=\left(\frac{{mg}\mathrm{cos}\:^{\mathrm{2}} \alpha}{{M}+{m}\mathrm{sin}\:\alpha\mathrm{cos}\:\alpha}\right)\left(\mathrm{cos}\:\alpha+\mathrm{sin}\:\alpha\right)+{g}\left(\mathrm{sin}\:\alpha−\varphi\frac{\frac{\mathrm{1}}{\mathrm{2}}{gt}^{\mathrm{2}} \mathrm{sin}\:\alpha}{\mathrm{1}+{g}\varphi\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} \mathrm{cos}\:\alpha}\mathrm{cos}\:\alpha\right) \\ $$

Commented by mr W last updated on 21/Oct/25

how can the acc. of the wedge be constant? it must be a function in terms of x or t, i think. at x=0: μ=0, there is no friction, the A should be maximum. due to the increasing of friction, A is getting smaller. anyway A≠constant.

$${how}\:{can}\:{the}\:{acc}.\:\:{of}\:{the}\:{wedge}\:{be} \\ $$$${constant}?\:{it}\:{must}\:{be}\:{a}\:{function} \\ $$$${in}\:{terms}\:{of}\:{x}\:{or}\:{t},\:{i}\:{think}. \\ $$$${at}\:{x}=\mathrm{0}:\:\mu=\mathrm{0},\:\:{there}\:{is}\:{no}\:{friction}, \\ $$$${the}\:{A}\:{should}\:{be}\:{maximum}.\:{due}\:{to} \\ $$$${the}\:{increasing}\:{of}\:{friction},\:{A}\:{is} \\ $$$${getting}\:{smaller}.\:{anyway}\:{A}\neq{constant}. \\ $$

Commented by fantastic last updated on 21/Oct/25