Transcription of PHYSICS 111 HOMEWORK SOLUTION #10 - New Jersey …
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PHYSICS 111 HOMEWORK SOLUTION #10 - New Jersey …
PHYSICS 111 HOMEWORK . SOLUTION #10. April 10, 2013. Given M~ = 4~i + ~j 3~k and N. ~ = ~i 2~j 5~k, calculate the vector ~ ~. product M N . By simply following the rules of the cross product: ~i ~i = ~j ~j = ~k ~k = ~0. ~i ~j = ~k = ~j ~i ~j ~k = ~i = ~k ~j ~k ~i = ~j = ~i ~k ~ N. M ~ = (4~i + ~j 3~k) (~i 2~j 5~k). = 8~k + 20~j ~k 5~i 3~j 6~i = 11~i + 17~j 9~k Calculate the net torque (magnitude and direction) on the beam in the figure below about the following axes. 2. We will choose clockwise as our positive direction and apply the formula for a torque: X. ~ net = F~i ~ri X. net = Fi ri sin i a) About the O-axis: net = 25 2 sin 60 + 10 4 sin 20 + 0. = This net torque is counterclockwise b) About the C-axis: net = 0 + 10 2 sin 20 30 2 sin 45.
= mr2 r g r = m p gr2 = 5:35 104 r 9:81 130 = 2:48 108 kg:m2=s b) If the two rockets supply a thrust of F=110 N each, we can write Newton 2nd law as follows: 2Fr = I = mr2 = 2F mr The average time interval is then calculated from and != p g r: t =! = mr 2F r g r = m p rg 2F = 8680 s 0.8 A playground merry-go-round of radius R = 1.60 m has a moment
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