355_Physics ProblemsTechnical Physics

355_Physics ProblemsTechnical Physics - Chapter 12 P12.18...

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Chapter 12 357 P12.18 FFF xbt =−+ = 550 0 . N (1) Fnm g y =− = 0 Summing torques about point O, τ O m m m =− = F t 1 50 5 50 10 0 0 .. . af which yields F t = 36 7 . N to the left Then, from Equation (1), F b =−= 36 7 5 50 31 2 . N N N to the right 10.0 m 5.50 N 1.50 m m g F t F b O n FIG. P12.18 P12.19 (a) T e sin . . 42 0 20 0 °= N T e = 29 9 . N (b) TT em cos . 42 0 T m = 22 2 P12.20 Relative to the hinge end of the bridge, the cable is attached horizontally out a distance x = 500 200 470 .c o s . . m and vertically down a distance y = 171 .s i n . . m . The cable then makes the following angle with the horizontal: θ = + L N M O Q P tan . 1 12 0 1 71 711 m 4.70 m . (a) Take torques about the hinge end of the bridge: RR xy 00 1 9 6 2 0 0 71 1 1 71 71 1 4 70 980 0 a f +− °
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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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