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Unformatted text preview: Midterm 1, Math 20C (Lecture C) November 2nd, 2007 This is the solution set for one of the types of the exam. The solutions to the other type are straightforward modifications of the solutions below. 1. An object slides 5 meters over the inclined surface with slope 60 due to the vertical gravitational force F g with magnitude | F g | = 50 Newton as illustrated in the figure below. vector j 60 F g d vector i a) (1 point) Express the displacement vector d (note that | d | = 5 meters) in terms of the standard basis vectors vector i and vector j . Solution: d = (-| d | cos(60)) vector i + (-| d | sin(60)) vector j =- 5 2 vector i- 5 3 2 vector j The signs of both of the components are negative, since the horizontal and vertical compo- nents of d point in the directions opposite to vector i and vector j , respectively. b) (2 points) Find the projection of F g onto d . Solution: Projection of F g onto d is the vector pointing in the direction of d with magnitude | F g | cos( ) where = 30 is the angle between F g and d . proj d F g = | F g | cos(30) d | d | = 25 3- 2 . 5 vector i- 2 . 5 3 vector j 5 =- 12 . 5 3 vector i- 37 . 5 vector j c) (2 points) Find the work done by the gravitational force in Newton meters. Solution: By definition the work done is the dot product of the force and the displacement vectors. W = F g d =- 50 vector j (- 5 2 vector i- 5 3 2 vector j ) =- 50 - 5 3 2 = 125 3 2. The position of a particle in 3D space as a function of time t is given by r ( t ) = t vector i + t 2 vector j + t 3 vector k, t ....
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This note was uploaded on 04/29/2008 for the course MATH 20C taught by Professor Helton during the Fall '08 term at UCSD.
- Fall '08