357_Physics ProblemsTechnical Physics

357_Physics ProblemsTechnical Physics - Chapter 12 P12.22...

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Chapter 12 359 P12.22 Call the required force F , with components FF x cos . 15 0 and y =− ° sin . 15 0 , transmitted to the center of the wheel by the handles. Just as the wheel leaves the ground, the ground exerts no force on it. F x = 0 : Fn x cos . 15 0 °− (1) F y = 0 : −° + = y sin . 15 0 400 0 N (2) R n x n y F x F y 400 N b 8.00 cm distances forces a b a FIG. P12.22 Take torques about its contact point with the brick. The needed distances are seen to be: bR aR b = = = 800 200 800 120 16 0 22 .. . cm cm cm cm af (a) τ = 0 : −++ = Fb Fa a xy 400 0 N a f , or F + = 12 0 15 0 16 0 15 0 400 16 0 0 .c o s . .s i n . . cm cm N cm a f so F = = 6400 859 Ncm 7.45 cm N (b) Then, using Equations (1) and (2), n x = 859 15 0 830 N N cos . and n y =+ ° = 400 859 15 0 622 N N a f sin . nn n n n y x = = F H G I K J == ° −− 11 104 0749 369 . tan tan . . kN to the left and upward θ *P12.23 When xx = min , the rod is on the verge of slipping, so
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