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Unformatted text preview: / CHAPTER 3 3.18 (a) F, =120 N 1;, = 60.0 N Pl, = 104 N P2 = 80.0 N P2,, =2o.7 N 13y = 77.3 N R: (213,)2 +(2:1;,)2 = ‘/(39.3 N)2 +(181 N)2 = 185 N )=tan"(4.61)=77.8° The resultant is R = 185 N at 77.8° from the xaxis (b) To have zero net force on‘the mule, the resultant above must be Cancelled by a force  . ual in ma 'tude and o  oositely directed. Thus, the required force is
185 N at 258° from the xaxis CHAPTER 4 4.2 From I: = '0, + at, the acceleration given to the football is _ 12—0! 10 m/s—O 2
=—=———=50 .
t 0.20s m/s Then, from Newton’s 2"cl law, we find (2F) = mi = (0.50 kg)(50 m/sz) = 4448N)——
..(38...)( —m 2
_ 3 172—12,? (320 m/s) ‘0
4.8 'i‘he acceleration of the bullet IS given by a = ~——2 (Ax) = m 2
Then, nememxmr kg)[(:_:‘i_8I;L/§1))_]= 4.12 (a) Choose the positive yaxis in the forward direction. We
resolve the forces into their components as The magnitude and direction of the resultant force is [EE1=,/(i:1—;)2 +(1—‘y)2 = 799 N , t9=tan‘1 ($) = 8.77° to right of y—axis.
1/ Thus, SE = 799 N at 8.77° to the right of the forward direction [2F]: 180N)2+(390N2=430N, at a=tan’[39°N)=65.2°Nom. lTh I =IE=430N= 2 0
us a m 270kg 1.59m/s at65.2 NofE 180N ...
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 Spring '11
 raul

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