HW_10_solutions

# HW_10_solutions - b) dL dt = 2 kg ( ) 9.8 N kg ( )...

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Problem Set 10 10.3: Taking positive torques to be counterclockwise (out of the page), m, N 2.34 N) m)(26.0 (0.09 m, N 1.62 N) (180.0 m) 090 . 0 ( 2 1 ! = = ! " = # " = ô ô ( ) m, N 78 . 1 N) (14.0 m) 090 . 0 ( 2 3 ! = = " so the net torque is m, N 50 . 2 ! with the direction counterclockwise (out of the page). Note that for 3 ! the applied force is perpendicular to the lever arm. 10.13: ( ) n t MR Rn I n R n f 2 0 k # μ = = = = ( )( ) ( ) ( ) ( )( ) . 482 . 0 N 160 s 50 . 7 2 min rev 850 m 260 . 0 kg 0 . 50 min rev s rad 30 = = 10.35: a) mvr sin(36.9deg) = 115 kg " m 2 /s, From the right hand rule: direction is into the page.
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Unformatted text preview: b) dL dt = 2 kg ( ) 9.8 N kg ( ) &quot; 8 m ( ) &quot; sin 90 # 36.9 ( ) = 125 N &quot; m = 125 kg &quot; m 2 s 2 , out of the page. 11.16: (a) Free body diagram of wheelbarrow: N 1200 ) m 70 . ( ) m 70 . )( N 80 ( ) m . 2 )( N 450 ( L wheel = = + + ! = &quot; W W L (b) The extra force comes from the ground....
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