hw4 solution

hw4 solution - Hw #4 solution Chapter 6 No.5 (a) In the...

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Hw #4 solution Chapter 6 No.5 (a) In the horizontal plane, the only force that can keep the coin in circular motion is friction. And because the coin is stationary, so the friction is static friction. (b) 22 2 30.0 50.0 / (50.0 / ) 83.3 / 30.0 r rr s s R cm vc m s m s ac Rc m Fm a f m g μ = = == = = m s m Cancel on both sides 2 2 83.3 / 1 / 0.0850 9.8 / 100 rs s r cm s m ag a g ms cm μμ =⇒ = = = No.9 (a) In vertical direction, the net force on the bob should be zero, because the bob won’t move in the vertical direction. So we have
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2 cos 0 cos / cos 80.0 9.8 / / cos5.00 787 y FT m g Tm g T m g k g m s θ θθ =− = =⇒ = = × = N N j So we can write the tension in vector way: 787sin 5.00 787cos5.00 (68.6 ) (784 ) Ti j N i =+ = + JG G G G G the vertical component :(784 ) Nj G the horizontal component: (68.6 ) Ni G (b) Just consider the horizontal plane. The bob is in uniform circular motion, so the horizontal component works as the radial force: sin5.00 sin5.00/ rr a a T = m s 2 68.6 /80.0 0.857 / r aN k g m == No.19 (a) At point A, gravity and normal force cause the radial force. And the radial force should be upward. Since the gravity is downward, so the normal
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hw4 solution - Hw #4 solution Chapter 6 No.5 (a) In the...

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