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MECHANICS
لKEITH R. SYMON

رد: ممكن تساعدوني ضروري جدا جدا
 

رد: ممكن تساعدوني ضروري جدا جدا
 

ممكن تساعدوني في حل هذا السؤال من الوحدة الثالثه:
A particle in the xy-plane is attracted toward the origin by a force F =k/y, inversely proportional to its distance from the x-axis. Calculate the work done by the force when the particle moves from the point x =0, y =a to the point x =2a, y=0 along a path which follows the side of rectangle consisting of a segment parallel to the x-axis from x =0, y=a to x=2a, y=a, and a vertical segment from a latter point to the x-axis.
B) Calculate the work done by the same force when the particle moves a long an ellipse of semiaxis a, 2a. (Hint: set x= 2a sin , y= a cos)

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I'll try to give some hints to you>>>
the problem seems simple...
just you have a force that is dependent on y ...
to give the work done , you have to integrate along the path given...
when on x axis direction you just integrate for dx or (Fdx = k/y dx) and x has the two values as lower and upper bounds...here y is constant.
when on vertical axis , you just integrate with respect to y , i.e. Fdy = k/y dy which gives you the ln y for the extremes... you substitute for this...
for the inclined path(if any exists) it is not more than you have now x and y changing so that you are to integrate with respect to them both ... but since you have a line so you can give a relation between x and y and then integrate F.dr rather than Fdx or Fdy ...
the remainder is up to you now,,,
ok???

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as for the second part:
you are given a hint of giving the polar coordinates so that you give r= rcos theta i + r sin theta j and then give F.dr to integrate...
that is it >>>
ok???

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sorry ... the hint says it is 2a cos and 2 a sin...
sorry you use this relation not rcos or r sin ...
that is r = xi + y j and substitute...
thanks

رد: ممكن تساعدوني ضروري جدا جدا
 

رد: ممكن تساعدوني ضروري جدا جدا
 

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