N° 11Complex motion (2)1) The inextensible string: i.c if M moves a distance x then m moves the same distance x and thereforeM and m have the same speed and the same linear acceleration.R2) The mass M is under the action of its weight Mg, the reaction of the supportM4.R and the tension of the string T.Applying Newton's second law on M:EFeu =Ma = Mg+T+R =MaMgProjecting this relation on a horizontal axis parallel and in the same direction ofthe motion of M: 0+T+0= Ma T=MaThe mass m is under the action of its weight mg and the tension of the string t1.mApplying Newton's second law on m:ext = ma = mg+t = m amgProjecting this relation on the vertically downward axis :mg -t = ma =t =m(g-a)|.3) The pulley is under the action of its weight u g, the reaction R' of the axis ofrotation (A) and of tensions T' and t' of the string on the pulley.Applying Newton's second law for rotation on the pulley:= 10"Fat/(A)+ M,= 10"u g/(A)R/(A)T'/(A)t'/(A)-T'.rt'r=0 since u g intersects with (A)=0 since R'intersects with (A)10"=-T'r+t'r = 10" > t'-T' =-But the string is of negligible mass, hence the tension of the string is constant. :t' =t and T'= T

1) The inextensible string: i.c if M moves a distance x then m moves the same distance x and therefore M and m have the same speed and the same lincar acceleration. R 2) The mass M is under the action of its weight Mg, the reaction of the support M T. 4. R and the tension of the string Applying Newton's second law on M:Fet =Ma Mg+T+R =Ma Mg Projecting this relation on a horizontal axis parallel and in the same direction of the motion of M: 0+T+0= Ma T= Ma The mass m is under the action of its weight mg and the tension of the string t. Applying Newton's second law on m : Fext = ma =mg+ t ma -> mg Projecting this relation on the vertically downward axis : mg -t = ma t =m(g-a)| 3) The pulley is under the action of its weight ug, the reaction R' of the axis of R' rotation (A) and of tensions T' and t' of the string on the pulley. Applying Newton's second law for rotation on the pulley: = 10" Fat/(A) + M. + M, = 10" u g/(A) R/ T/(A) t'/(A) -T'r t'r =0 since u g intersects with (A) =0 since R' intersects with (A) 10" =-T'r+t'r = 10" => t'-T' =- But the string is of negligible mass, hence the tension of the string is constant. : t' =t and T'= T

1) The inextensible string: i.c if M moves a distance x then m moves the same distance x and therefore M and m have the same speed and the same lincar acceleration. R 2) The mass M is under the action of its weight Mg, the reaction of the support M T. 4. R and the tension of the string Applying Newton's second law on M:Fet =Ma Mg+T+R =Ma Mg Projecting this relation on a horizontal axis parallel and in the same direction of the motion of M: 0+T+0= Ma T= Ma The mass m is under the action of its weight mg and the tension of the string t. Applying Newton's second law on m : Fext = ma =mg+ t ma -> mg Projecting this relation on the vertically downward axis : mg -t = ma t =m(g-a)| 3) The pulley is under the action of its weight ug, the reaction R' of the axis of R' rotation (A) and of tensions T' and t' of the string on the pulley. Applying Newton's second law for rotation on the pulley: = 10" Fat/(A) + M. + M, = 10" u g/(A) R/ T/(A) t'/(A) -T'r t'r =0 since u g intersects with (A) =0 since R' intersects with (A) 10" =-T'r+t'r = 10" => t'-T' =- But the string is of negligible mass, hence the tension of the string is constant. : t' =t and T'= T

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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