ENGG 202 Lecture Notes - Lecture 1: Unit Vector, Cross Product, Harmonic Oscillator
26 Jan 2015
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97. 7 n-m cw: fb = 48. 8 n left. Solution strategy: part a) 1 couple is shown as a pair of forces, the other is represented by its moment. To find the single equivalent couple (or moment) acting on the plate, we simply need the total moment acting on the plate since the sum of the forces will be zero (only couples acting on it). Part b) draw a horizontal force at b, (assume the direction right or left as you wish). Then sum the moments about point d to find the magnitude of this force. Solution strategy: you can use vectors to solve: |rab x f| = 600 nm, or use the scalar definition of the moment for the couple. The diagonal distance ab is the distance that is perpendicular to both lines of action (vertical lines) so. 600 = fd, calc the diagonal distance d and you can solve for f.
