367_Physics ProblemsTechnical Physics

367_Physics ProblemsTechnical Physics - Chapter 12 P12.49...

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Chapter 12 369 P12.49 From the free-body diagram, the angle T makes with the rod is θ + ° 60 0 20 0 80 0 .. . and the perpendicular component of T is T sin . 80 0 ° . Summing torques around the base of the rod, τ = 0: −° + ° = 400 10000 60 80 0 .c o s . s i n m N m af bg T T = ° ° 10 000 60 0 80 0 508 10 3 N N cos . sin . . F x = FT H = cos . 20 0 0 H = × cos . . 20 0 4 77 10 3 N F y = V −= sin . 20 0 10 000 0 N and V =− ° = × 10 000 20 0 8 26 10 3 N N sin . . F V T 60 ° 20 ° F H 10 000 N FIG. P12.49 P12.50 Choosing the origin at R , (1) FR T x =+ °− = sin . sin 15 0 0 (2) T y ° + = 700 15 0 0 cos . cos (3) τθ + = 700 0 180 0 070 0 0 cos . . T Solve the equations for from (3), T = 1800cos from (1), R = ° 1800 15 0 sin cos sin . θθ Then (2) gives 700 150 15 0 0 2 ° ° += sin cos cos . sin . cos or cos . . sin cos 2 03889 3732 0 +− = Squaring, cos . cos . 42 08809 001013 0 −+ = Let u = cos 2 then using the quadratic equation, u = 001165
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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