ch11-p087 - 87. This problem involves the vector cross...

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which yields 32 ˆ (30 k) kg m /s. t =⋅ G A (f) Again, the acceleration vector is given by G at =− 60 2 . # i in SI units, and the net force on the car is ma G . In a similar argument to that given in the previous part, we have G G G G G τ = ×= mr a t b g b g b g b gb g b g ch e j o k 30 20 0 50 2 .. . . # which yields 2 ˆ (90 k) N m. t G 87. This problem involves the vector cross product of vectors lying in the xy plane. For such vectors, if we write G ′ = ′′ rx y ## i + j , then (using Eq. 3-30) we find G G rv x vy v yx di # k. (a) Here, G r points in either the + # i or the # i direction (since the particle moves along the x axis). It has no y or z components, and neither does G v , so it is clear from the above expression (or, more simply, from the fact that ii = 0 × ) that G A GG = mr v b g 0 in this case. (b) The net force is in the # i direction (as one finds from differentiating the velocity expression, yielding the acceleration), so, similar to what we found in part (a), we obtain = G G rF 0.
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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