Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 216587 by Tawa11 last updated on 11/Feb/25

Commented by Tawa11 last updated on 11/Feb/25

In the mechanical system represented in Figure 16, the bar, whose mass is "M", can slide along the rails without friction. At the initial moment the load suspended in wire deviates from the vertical to the angle "α" and is released. What is the mass "m" of this load, if the angle formed by the thread with the vertical does not vary during the movement of the system?*

Commented by mr W last updated on 12/Feb/25

$$\frac{m}{M}=\frac{\sin \alpha }{{(1-\sin \alpha )}^2}$$$$T=\frac{Mg\tan \alpha }{1-\sin \alpha }=tensioninwire$$$$doyouknowthecorrectanswer?$$

Commented by Tawa11 last updated on 12/Feb/25

$$\mathrm{No}\mathrm{sir}.$$$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{the}\mathrm{correct}\mathrm{answer}.$$

Commented by Tawa11 last updated on 12/Feb/25

$$\mathrm{Please}\mathrm{sir},\mathrm{show}\mathrm{me}\mathrm{the}\mathrm{full}\mathrm{working}.$$$$\mathrm{Thanks}\mathrm{sir}.$$

Answered by mr W last updated on 12/Feb/25

Commented by mr W last updated on 12/Feb/25

$$U=velocityofM(\leftarrow )$$$$A=accelerationofM(\leftarrow )$$$$u_x,u_y=velocityofm(\to ,\downarrow )$$$$a_x,a_y=accelerationofm(\to ,\downarrow )$$$$U=-\frac{db}{dt}$$$$A=\frac{dU}{dt}$$$$totallengthofwireL=b+l=constant$$$$\Rightarrow \frac{db}{dt}+\frac{dl}{dt}=0\Rightarrow \frac{dl}{dt}=-\frac{db}{dt}=U$$$$$$$$x_m=b+l\sin \alpha$$$$y_m=l\cos \alpha$$$$\alpha remainsconstant\Rightarrow \frac{d\alpha }{dt}=0$$$$u_x=\frac{{dx}_m}{dt}=\frac{db}{dt}+\frac{dl}{dt}\sin \alpha =-U+U\sin \alpha$$$$a_x=\frac{{du}_x}{dt}=-\frac{dU}{dt}+\frac{dU}{dt}\sin \alpha =-A(1-\sin \alpha )$$$$u_y=\frac{{dy}_m}{dt}=\frac{dl}{dt}\cos \alpha =U\cos \alpha$$$$a_y=\frac{{du}_y}{dt}=\frac{dU}{dt}\cos \alpha =A\cos \alpha$$$$MA=T-T\sin \alpha$$$$\Rightarrow A=\frac{T(1-\sin \alpha )}{M}$$$${ma}_x=-T\sin \alpha$$$$\Rightarrow -m\frac{T(1-\sin \alpha )}{M}(1-\sin \alpha )=-T\sin \alpha$$$$\Rightarrow \frac{m}{M}=\frac{\sin \alpha }{{(1-\sin \alpha )}^2}$$$${ma}_y=mg-T\cos \alpha$$$$m\frac{T(1-\sin \alpha )}{M}\cos \alpha =mg-T\cos \alpha$$$$\frac{m(1-\sin \alpha )\cos \alpha }{M}=\frac{mg}{T}-\cos \alpha$$$$\frac{\sin \alpha (1-\sin \alpha )\cos \alpha }{{(1-\sin \alpha )}^2}=\frac{mg}{T}-\cos \alpha$$$$(\frac{\sin \alpha }{1-\sin \alpha }+1)\cos \alpha =\frac{mg}{T}$$$$\Rightarrow T=\frac{mg(1-\sin \alpha )}{\cos \alpha }=\frac{Mg\tan \alpha }{1-\sin \alpha }$$

Commented by Tawa11 last updated on 12/Feb/25

$$\mathrm{Thanks}\mathrm{for}\mathrm{your}\mathrm{time}\mathrm{sir},\mathrm{I}\mathrm{appreciate}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com