ch05-p101

# ch05-p101 - 101 We first analyze the forces on m1=1.0 kg...

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Unformatted text preview: 101. We first analyze the forces on m1=1.0 kg. The +x direction is “downhill” (parallel to T ). With the acceleration (5.5 m/s2) in the positive x direction for m1, then Newton’s second law, applied to the x axis, becomes T + m1 g sin β = m1 ( 5.5m/s 2 ) But for m2=2.0 kg, using the more familiar vertical y axis (with up as the positive direction), we have the acceleration in the negative direction: F + T − m2 g = m2 ( −5.5m/s 2 ) where the tension comes in as an upward force (the cord can pull, not push). (a) From the equation for m2, with F = 6.0 N, we find the tension T = 2.6 N. (b) From the equation for m, using the result from part (a), we obtain the angle β = 17° . ...
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## This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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