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Question-216587

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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
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
(m/M)=((sin α)/((1−sin α)^2 )) T=((Mg tan α)/(1−sin α))=tension in wire do you know the correct answer?
$$\frac{{m}}{{M}}=\frac{\mathrm{sin}\:\alpha}{\left(\mathrm{1}−\mathrm{sin}\:\alpha\right)^{\mathrm{2}} } \\ $$$${T}=\frac{{Mg}\:\mathrm{tan}\:\alpha}{\mathrm{1}−\mathrm{sin}\:\alpha}={tension}\:{in}\:{wire} \\ $$$${do}\:{you}\:{know}\:{the}\:{correct}\:{answer}? \\ $$
Commented by Tawa11 last updated on 12/Feb/25
No sir. I don′t know the correct answer.
$$\mathrm{No}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{know}\:\mathrm{the}\:\mathrm{correct}\:\mathrm{answer}. \\ $$
Commented by Tawa11 last updated on 12/Feb/25
Please sir, show me the full working. Thanks sir.
$$\mathrm{Please}\:\mathrm{sir},\:\mathrm{show}\:\mathrm{me}\:\mathrm{the}\:\mathrm{full}\:\mathrm{working}. \\ $$$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$
Answered by mr W last updated on 12/Feb/25
Commented by mr W last updated on 12/Feb/25
U=velocity of M (←) A=acceleration of M (←) u_x , u_y =velocity of m (→, ↓) a_x , a_y =acceleration of m (→, ↓) U=−(db/dt) A=(dU/dt) total length of wire L=b+l=constant ⇒(db/dt)+(dl/dt)=0 ⇒(dl/dt)=−(db/dt)=U x_m =b+l sin α y_m =l cos α α remains constant ⇒(dα/dt)=0 u_x =(dx_m /dt)=(db/dt)+(dl/dt) sin α=−U+U sin α a_x =(du_x /dt)=−(dU/dt)+(dU/dt) sin α=−A(1−sin α) u_y =(dy_m /dt)=(dl/dt) cos α=U cos α a_y =(du_y /dt)=(dU/dt) cos α=A cos α MA=T−T sin α ⇒A=((T(1−sin α))/M) ma_x =−T sin α ⇒−m((T(1−sin α))/M)(1−sin α)=−T sin α ⇒(m/M)=((sin α)/((1−sin α)^2 )) ma_y =mg−T cos α m((T(1−sin α))/M) cos α=mg−T cos α ((m(1−sin α)cos α)/M) =((mg)/T)−cos α ((sin α(1−sin α)cos α)/((1−sin α)^2 )) =((mg)/T)−cos α (((sin α)/(1−sin α))+1)cos α =((mg)/T) ⇒T =((mg(1−sin α))/(cos α))=((Mg tan α)/(1−sin α))
$${U}={velocity}\:{of}\:{M}\:\left(\leftarrow\right) \\ $$$${A}={acceleration}\:{of}\:{M}\:\left(\leftarrow\right) \\ $$$${u}_{{x}} ,\:{u}_{{y}} ={velocity}\:{of}\:{m}\:\left(\rightarrow,\:\downarrow\right) \\ $$$${a}_{{x}} ,\:{a}_{{y}} ={acceleration}\:{of}\:{m}\:\left(\rightarrow,\:\downarrow\right) \\ $$$${U}=−\frac{{db}}{{dt}} \\ $$$${A}=\frac{{dU}}{{dt}} \\ $$$${total}\:{length}\:{of}\:{wire}\:{L}={b}+{l}={constant} \\ $$$$\Rightarrow\frac{{db}}{{dt}}+\frac{{dl}}{{dt}}=\mathrm{0}\:\Rightarrow\frac{{dl}}{{dt}}=−\frac{{db}}{{dt}}={U} \\ $$$$ \\ $$$${x}_{{m}} ={b}+{l}\:\mathrm{sin}\:\alpha \\ $$$${y}_{{m}} ={l}\:\mathrm{cos}\:\alpha \\ $$$$\alpha\:{remains}\:{constant}\:\Rightarrow\frac{{d}\alpha}{{dt}}=\mathrm{0} \\ $$$${u}_{{x}} =\frac{{dx}_{{m}} }{{dt}}=\frac{{db}}{{dt}}+\frac{{dl}}{{dt}}\:\mathrm{sin}\:\alpha=−{U}+{U}\:\mathrm{sin}\:\alpha \\ $$$${a}_{{x}} =\frac{{du}_{{x}} }{{dt}}=−\frac{{dU}}{{dt}}+\frac{{dU}}{{dt}}\:\mathrm{sin}\:\alpha=−{A}\left(\mathrm{1}−\mathrm{sin}\:\alpha\right) \\ $$$${u}_{{y}} =\frac{{dy}_{{m}} }{{dt}}=\frac{{dl}}{{dt}}\:\mathrm{cos}\:\alpha={U}\:\mathrm{cos}\:\alpha \\ $$$${a}_{{y}} =\frac{{du}_{{y}} }{{dt}}=\frac{{dU}}{{dt}}\:\mathrm{cos}\:\alpha={A}\:\mathrm{cos}\:\alpha \\ $$$${MA}={T}−{T}\:\mathrm{sin}\:\alpha \\ $$$$\Rightarrow{A}=\frac{{T}\left(\mathrm{1}−\mathrm{sin}\:\alpha\right)}{{M}} \\ $$$${ma}_{{x}} =−{T}\:\mathrm{sin}\:\alpha \\ $$$$\Rightarrow−{m}\frac{{T}\left(\mathrm{1}−\mathrm{sin}\:\alpha\right)}{{M}}\left(\mathrm{1}−\mathrm{sin}\:\alpha\right)=−{T}\:\mathrm{sin}\:\alpha \\ $$$$\Rightarrow\frac{{m}}{{M}}=\frac{\mathrm{sin}\:\alpha}{\left(\mathrm{1}−\mathrm{sin}\:\alpha\right)^{\mathrm{2}} } \\ $$$${ma}_{{y}} ={mg}−{T}\:\mathrm{cos}\:\alpha \\ $$$${m}\frac{{T}\left(\mathrm{1}−\mathrm{sin}\:\alpha\right)}{{M}}\:\mathrm{cos}\:\alpha={mg}−{T}\:\mathrm{cos}\:\alpha \\ $$$$\frac{{m}\left(\mathrm{1}−\mathrm{sin}\:\alpha\right)\mathrm{cos}\:\alpha}{{M}}\:=\frac{{mg}}{{T}}−\mathrm{cos}\:\alpha \\ $$$$\frac{\mathrm{sin}\:\alpha\left(\mathrm{1}−\mathrm{sin}\:\alpha\right)\mathrm{cos}\:\alpha}{\left(\mathrm{1}−\mathrm{sin}\:\alpha\right)^{\mathrm{2}} }\:=\frac{{mg}}{{T}}−\mathrm{cos}\:\alpha \\ $$$$\left(\frac{\mathrm{sin}\:\alpha}{\mathrm{1}−\mathrm{sin}\:\alpha}+\mathrm{1}\right)\mathrm{cos}\:\alpha\:=\frac{{mg}}{{T}} \\ $$$$\Rightarrow{T}\:=\frac{{mg}\left(\mathrm{1}−\mathrm{sin}\:\alpha\right)}{\mathrm{cos}\:\alpha}=\frac{{Mg}\:\mathrm{tan}\:\alpha}{\mathrm{1}−\mathrm{sin}\:\alpha} \\ $$
Commented by Tawa11 last updated on 12/Feb/25
Thanks for your time sir, I appreciate.
$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir},\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

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