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u4905047
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9.5
« on: September 13, 2007, 09:17:32 PM »

9.5 Figure 9.13 shows two blocks of masses M and m that slide on smooth planes inclined at angles \alpha and \beta to the horizontal.The blocks are connected by a light inextensible string that passes over a light frictionaless pulley. Find the acceleration of the block of mass m up the plane, and deduce the tension in the string.

Sol-n Let the system consists of two blocks of mass M and m.
Initially,The system is at rest.
At time t = t, two blocks have
i)the same distance ( which is parallel to the incline) S
ii)the same velocity v = \dot{S}
iii)the same acceleration a = \dot{v}
because they are fixed with the same string,and this string is light and inextensible.Thus tension T is constant.

1.Find the acceleration of mass m

                     readingMethod 1 Newton 's law of motion

From the newton 's second law (in the direction which is parallel to the incline) of mass

m;                     T\;-\;mg\sin\beta             =                            ma........(1)
   
M;                     Mg\sin\alpha\;-T\;             =                            Ma........(2)

(1)+(2);           (Mg\sin \alpha\;-\;mg\sin \beta)             =         (m+M)a

                       a                                        =            \dfrac{(M\sin \alpha\;-\;m\sin \beta)g}{m+M}


                  readingMethod 2 Energy conservation
From energy conservation principle which can simplify as


                      the decreasing of potential energy       =         the increasing of kinetic energy.

                                      -\Delta V                          =                         +\Delta K

                      -(mgh_{1} - Mgh_{2})                     =         +(\frac{1}{2}mv^{2} + \frac{1}{2}Mv^{2})

                      -(mgS\sin \beta - MgS\sin \alpha)    =         \frac{1}{2}(m+M)v^{2}

                      (M\sin \alpha\;-\;m\sin \beta)gS         =         \frac{1}{2}(m+M)v^{2}

                     differentiate with respect to t all the equation,and use v\;=\;\dot{S}=\frac{dS}{dt} and a\;=\;\dot{v}=\frac{dv}{dt}.then

                     \frac{d}{dt}(M\sin \alpha\;-\;m\sin \beta)gS      =         \frac{d}{dt}\frac{1}{2}(m+M)v^{2}

                    (M\sin \alpha\;-\;m\sin \beta)g\frac{dS}{dt}          =         \frac{1}{2}(m+M)\frac{d}{dt}v^{2}

                   (M\sin \alpha\;-\;m\sin \beta)gv             =         \frac{1}{2}(m+M)2va

                  and divided by v,thus

                   (M\sin \alpha\;-\;m\sin \beta)g             =         (m+M)a


\therefore a\;=\;\dfrac{(M\sin \alpha\;-\;m\sin \beta)g}{m+M}


2.Find the tension T.
From the newton 's second law of mass m.

                     T\;-\;mg\sin\beta             =                            ma

                     T                                   =                            m(a\;+\;g\sin\beta)

substitute a\;=\;\dfrac{(M\sin \alpha\;-\;m\sin \beta)g}{m+M} then


                                T                          =                            m(\dfrac{(M\sin \alpha\;-\;m\sin \beta)g}{m+M}\;+\;g\sin\beta)


                                                               =                            mg(\dfrac{M\sin \alpha\;-\;m\sin \beta+\;m\sin \beta+\;M\sin \beta}{m+M})

                                                                =                            mMg\dfrac{(sin \alpha\;+\sin \beta)}{m+M}

\therefore T\;=\;mMg\dfrac{(sin \alpha\;+\sin \beta)}{m+M}



7.19 A spacecraft travelling with speed V approaches a planet of mass M along straight lin whos perpendicular distance from the center of the planet is p. When the spacecraft is at a distance c from the planet, it fires its engines so as to multiply its current speed by a factor k( 0 < k < 1 ), its direction of motion being unaffected. [ You may neglect the time taken for this operation.] Find the condition that the spacecraft should go into orbit around the planet.





\oint_{C}Mdx + Ndy        =  \iint \left( \frac{\partial N}{\partial x}-\frac{\partial M}{\partial y}\right)dA

\frac{\partial N}{\partial x}
« Last Edit: September 22, 2007, 09:10:51 PM by u4905047 » Logged
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