ksaimouli
A block is pulled by a string that makes an angle of 25º to the horizontal. If the mass of the block is 12.0 kg and the coefficient of friction is 0.25, what force would keep the block moving at a constant velocity?
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henpen
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|dw:1349633059599:dw|
ksaimouli
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|dw:1349633142335:dw|
ksaimouli
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what to do after this i am totally confused
ksaimouli
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i know that ff=u*fn
ksaimouli
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we need to find fn
henpen
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\[F_n=mg-F_y=12*9.8-Fsin(25)\]
ksaimouli
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i got this but how do u find the F
henpen
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\[F_f= \mu F_n= 0.25(117.6-Fsin(25))=F_x=Fcos(25)\]
ksaimouli
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how did u get 117.6
henpen
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mg
henpen
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12*9.8
henpen
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\[29.4-0.25\sin(25)F=Fcos(25)\]
\[\frac{29.4}{\cos(25)+0.25\sin(25)}=F\]
ksaimouli
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thx
ksaimouli
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why =fcos(25)
henpen
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|dw:1349634909926:dw|Basic trig
henpen
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\[F_f=F_x=Fcos(25)\]
henpen
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No acceleration => no net force in x-direction