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Towmassesm_(1) and m_(2)(m_(1)gtm_(2)) are incontact with each other on a smooth horizontal surface calculate the magnitude of contact force between them |
Answer» Solution :Let usconsidertwoblocksof masses `m_(1)`and`m_(2) (m_(1) gt m_(2))` . Keptin contactwitheach otheron asmoothhorizontalfrictionalsurfaceasshownin the figure  Bythe application of ahorizontalforceF boththe blocksaresetintomotionwithacceleration asimultaneouslyin thedirectionof theforce F . Tofind theacceleration `VEC(a)` Newton'ssecondlawhas tobeappliedto thesystemmass `m=m_(1)+ m_(2)` If weselectthe motionof the twomassesalongthe positivex direction`F hat(j)= MAHAT(i)` Theaccelerationof thesystemis given by `:.,a= (F )/( m_(1) + m_(2))` Theforce exerted by teh block`m_(1)` on `m_(2)` dueto itsmotionis calledforceof contact `(f_(n))`Accordingto Newton's thirdlaw the block`m_(2)` willexert anequivalentoppositereaction Figureshowsthe freebodydiagramof block`m_(1)`  `Fhat(i) - f_(12) hat(i) = m_(1)a hat(i)` Bycomparingthe componentsof theaboveequationwe get `F-f_(12)= m_(1) a` `f_(12) = F - m_(1) a` Substitutingthe value of accelerationfromequation(1) in (2)we get ` f_(12) = f- m_(1) (F)/(m_(1)+m_(2))` `F_(12)= (Fm_(2))/(m_(1) +m_(2))` EQUATION (3) showsthat themagnitudeofcontactforcedependson mass`m_(2)`whichprovides the reactionforce. Notethat thisforceis actingalongthe negativex direction`m_(1)` isgiven by `f_(12)= - (Fm_(2))/( m_(1) + m_(2))` For `m_(2)` thereis onlyone forceforceactingon itthe xdirectionand it isdenoted by`f_(21)` . thisforceis EXERTEDBY mass `m_(1)` .THEFREE bodydiagramfor mass `m_(2)` is shown in the figure .  ApplyingNewton'secondlaw for mass `m_(2)` `f_(21) hat(i) = m_(2) hat(i)` Bycomparingthe componentson bothsidesof the aboveequation `f_(21) = m_(2) a` Substitutingforaccelerationfromequation (1)in equation(4) we get `f_(21) = (Fm_(2))/( m_(1) +m_(2))` In thiscase the magnitude of the contactforce is `f_(21) = (Fm)/(m_(1) + m_(2))` |
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