353_Physics ProblemsTechnical Physics

353_Physics ProblemsTechnical Physics - 355 Chapter 12...

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Chapter 12 355 P12.13 (a) Ff n xw =− = 0 Fn yg =− = 800 500 0 N N Taking torques about an axis at the foot of the ladder, 800 4 00 30 0 500 7 50 30 0 15 0 30 0 0 m m cm af a f a f .s i n . i n . .c o s . °+ ° −° = n w Solving the torque equation, n w = = 4 0 08 0 0 7 5 05 0 0 3 0 0 15 0 268 .. t a n . . m N N m N a f a f . Next substitute this value into the F x equation to find n g f n w 500 N 800 N A FIG. P12.13 fn w == 268 N in the positive x direction. Solving the equation F y = 0, n g = 1300 N in the positive y direction. (b) In this case, the torque equation τ A = 0 gives: 9 00 800 30 0 7 50 500 30 0 15 0 60 0 0 . sin . . sin . . sin . N N m a fa f a fa f a f bg °+ °− °= n w or n w = 421 N . Since w 421 N and ff n g max µ , we find = f n g max . 421 0324 N . P12.14 (a) n 0 (1) Fnm gm g = 12 0 (2) τθ θ Aw mg L x nL F H G I K J −+ = 2
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