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 214248 by mr W last updated on 03/Dec/24

Commented by mr W last updated on 02/Dec/24

$$amanwithmassMisstandingon$$$$thetopofatwinstepladder.each$$$$ladderwithlengthlhasamassm.$$$$bothladdersformanangleof45°.$$$$suddentlythespreadersarebrocken$$$$andtheladdersbegintosliponthe$$$$smoothsurfaceofthefloor.$$$$afterwhattimeandwithwhat$$$$speedwillthemanhitthefloor?$$

Answered by a.lgnaoui last updated on 02/Dec/24

$$\mathrm{hauteur}\mathrm{h}=\mathrm{Lcos}\left(\frac{45}{2}\right)$$$$\mathrm{h}=\mathrm{L}\frac{\sqrt{2+\sqrt{2}}}{2}$$$${\mathrm{a}}_{\mathrm{y}}=\mathrm{g}\mathrm{v}=\mathrm{gt}+{\mathrm{v}}_0$$$${\mathrm{v}}_0=0\mathbf{v}=\mathbf{gt}$$$$\mathbf{h}=\frac{1}{2}{\mathbf{gt}}^2\Rightarrow {\mathbf{v}}^2=2\mathbf{gh}$$$$\mathbf{v}=\sqrt{2\mathbf{gh}}=\sqrt{\mathbf{gL}\sqrt{2+\sqrt{2}}}$$$$\mathrm{t}=\frac{\sqrt{\mathbf{Lg}\sqrt{2+\sqrt{2}})}}{\mathbf{g}}$$$$$$

Commented by a.lgnaoui last updated on 02/Dec/24

Commented by mr W last updated on 03/Dec/24

$$youseemnottounderstandwhat$$$$thequestionisabout.$$$$themanfallswiththeladder.he$$$${doesn}^{\prime }tfallfromtheladder!$$$$besidesthetotalmassisM+2m.$$$$thecenterofmassofeachladderis$$$$atitsmiddlepoint,notonitstop.$$

Commented by mr W last updated on 03/Dec/24

Commented by mr W last updated on 05/Dec/24

$$theladderorthisdrawing?$$

Commented by 0202391905769 last updated on 05/Dec/24

$$\mathrm{comment}\mathrm{as}\mathrm{tu}\mathrm{construire}\mathrm{sa}?$$

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

Commented by mr W last updated on 05/Dec/24

$$h=l\sin \theta$$$$let\omega =-\frac{d\theta }{dt}$$$$u_{My}=-\frac{dh}{dt}=\omega l\cos \theta$$$$a_{My}={\omega }^{2}l\sin \theta$$$$u_{mx}=\frac{d(0.5l\cos \theta )}{dt}=\frac{\omega l\sin \theta }{2}$$$$u_{my}=-\frac{d(0.5l\sin \theta )}{dt}=\frac{\omega l\cos \theta }{2}$$$$h_0=l\cos 22.5°$$$$Mg(h_0-h)+2mg(\frac{h_0}{2}-\frac{h}{2})$$$$=\frac{M{\omega }^2{l}^2{\cos }^2\theta }{2}+2\times \frac{m{\omega }^2{l}^2}{2}\left(\frac{{\sin }^2\theta +{\cos }^2\theta }{4}\right)+2\times \frac{{ml}^2}{12}\times {\omega }^2$$$$12(M+m)g(\cos 22.5°-\sin \theta )=(6M{\cos }^2\theta +5m){\omega }^{2}l$$$$with\mu =\frac{m}{M}$$$$\Rightarrow {\omega }^2=\frac{12(1+\mu )g(\cos \frac{3\pi }{8}-\sin \theta )}{(6{\cos }^2\theta +5\mu )l}$$$$T=\sqrt{\frac{2h_0}{g}}{\int }_0^{\frac{3\pi }{8}}\sqrt{\frac{6{\cos }^2\theta +5\mu }{24(1+\mu )\sin \frac{3\pi }{8}(\cos \frac{3\pi }{8}-\sin \theta )}}d\theta$$$$u_P=\omega l\cos \theta =\cos \theta \sqrt{\frac{12(1+\mu )gl(\cos \frac{3\pi }{8}-\sin \theta )}{6{\cos }^2\theta +5\mu }}$$$$example:$$$$\mu =0.1$$$$T\approx 1.055212\sqrt{\frac{2h_0}{g}}$$$$u_P=\sqrt{\frac{1+\mu }{1+\frac{5\mu }{6}}}\sqrt{2{gh}_0}\approx 1.008\sqrt{2{gh}_0}$$$$$$$$Mg-N=M{\omega }^{2}l\sin \theta$$$$\frac{N}{Mg}=1-\frac{{\omega }^{2}l\sin \theta }{g}=1-\frac{12(1+\mu )\sin \theta (\sin \frac{3\pi }{8}-\sin \theta )}{6{\cos }^2\theta +5\mu }$$

Answered by a.lgnaoui last updated on 04/Dec/24

$$\mathrm{la}\mathrm{situation}$$$$\{\begin{array}{l}{\mathbf{x}}_0=\mathbf{L}\sin \frac{\pi }{8}\mathrm{t}=0\\ {\mathbf{y}}_0=\mathbf{L}\cos \frac{\pi }{8}\end{array}$$$$\{\begin{array}{l}\mathbf{x}=\mathbf{L}(1-\sin \frac{\pi }{8})({\mathbf{t}}_0+\mathbf{t})\\ \mathbf{y}=\mathrm{o}\end{array}$$$$\{\begin{array}{l}\mathrm{ax}=\mathrm{o}\mathbf{x}=\mathbf{vt}\\ \mathbf{ay}=+\mathbf{g}\mathbf{v}=\mathbf{gt}\mathbf{y}=\frac{1}{2}{\mathbf{gt}}^2+\mathbf{L}\sin \frac{\pi }{8}\end{array}$$$$\mathrm{a}\mathbf{t}\mathbf{t}=\frac{\mathbf{x}}{\mathbf{v}}=\frac{\mathbf{L}(1-\sin \frac{\pi }{8})}{\mathrm{v}}=\frac{\mathbf{v}}{\mathbf{g}}$$$$\mathbf{v}=\sqrt{\mathbf{gL}(1-\sin \frac{\pi }{8})}$$$$\mathbf{y}=\frac{{\mathbf{x}}^2}{2{\mathbf{v}}^2}\mathbf{gL}(1-\sin \frac{\pi }{8})+\mathbf{L}\sin \frac{\pi }{8}$$$$\{\begin{array}{l}\mathrm{x}=\mathrm{L}(1-\sin \frac{\pi }{8});\mathbf{y}=0\\ \frac{{\mathbf{x}}^2}{2{\mathbf{v}}^2}\mathbf{g}(1-\sin \frac{\mathit{\pi }}{8})+\sin \frac{\pi }{8}=0\end{array}$$$$\mathrm{g}(1-\sin ){\mathrm{x}}^2=-{2\mathrm{v}}^2\sin \left(\frac{\pi }{8}\right)$$$$\mid \mathrm{v}\mid =\mathrm{x}\sqrt{\frac{\mathrm{g}(1-\sin \frac{\pi }{8})}{2\sin \frac{\pi }{8}}}$$$$\mathbf{v}=\mathbf{L}\sqrt{\frac{\mathrm{g}{(1-\sin \frac{\pi }{8})}^3}{2\sin \frac{\pi }{8}}}$$$$\mathbf{t}=\frac{\mathbf{x}}{\mathbf{v}}\Rightarrow \mathbf{x}=\mathbf{L}(1-\sin \frac{\mathit{\pi }}{8})$$$$\mathbf{t}=\sqrt{\frac{2\sin \frac{\pi }{8}}{\mathrm{g}(1-\sin \frac{\pi }{8})}}$$$$$$$$$$$$\mathbf{soit}\mathrm{v}=\mathrm{gt}$$$$\mathbf{v}=\sqrt{\frac{2\mathbf{g}\sin \frac{\mathit{\pi }}{8}}{1-\sin \frac{\mathit{\pi }}{8}}}$$$$$$

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

$$stillwrong!$$$$laddersrotatealso!$$$$theunitoftimeissecond,not$$$$second/\sqrt{m}!$$

Commented by a.lgnaoui last updated on 04/Dec/24

Terms of Service

Privacy Policy

Contact: info@tinkutara.com