Transcription of PHYSICS 111 HOMEWORK SOLUTION #10
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PHYSICS 111 HOMEWORK SOLUTION #10
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.
X F~ i ~r i ˝ net = X F ir i sin i a) About the O-axis: ˝ net = 25 2 sin60 + 10 4 sin20 + 0 = 29:6 N:m This net torque is counterclockwise b) About the C-axis: ˝ net = 0 + 10 2 sin20 30 2 sin45 = 35:6 N:m This net torque is again counterclockwise 0.3 A light, rigid rod of length l = 1.00 m joins two particles, with masses m 1 = 4.00 kg and m ...
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