ksaimouli
  • 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?
Physics
katieb
  • katieb
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anonymous
  • anonymous
|dw:1349633059599:dw|
ksaimouli
  • ksaimouli
|dw:1349633142335:dw|
ksaimouli
  • ksaimouli
what to do after this i am totally confused

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ksaimouli
  • ksaimouli
i know that ff=u*fn
ksaimouli
  • ksaimouli
we need to find fn
anonymous
  • anonymous
\[F_n=mg-F_y=12*9.8-Fsin(25)\]
ksaimouli
  • ksaimouli
i got this but how do u find the F
anonymous
  • anonymous
\[F_f= \mu F_n= 0.25(117.6-Fsin(25))=F_x=Fcos(25)\]
ksaimouli
  • ksaimouli
how did u get 117.6
anonymous
  • anonymous
mg
anonymous
  • anonymous
12*9.8
anonymous
  • anonymous
\[29.4-0.25\sin(25)F=Fcos(25)\] \[\frac{29.4}{\cos(25)+0.25\sin(25)}=F\]
ksaimouli
  • ksaimouli
thx
ksaimouli
  • ksaimouli
why =fcos(25)
anonymous
  • anonymous
|dw:1349634909926:dw|Basic trig
anonymous
  • anonymous
\[F_f=F_x=Fcos(25)\]
anonymous
  • anonymous
No acceleration => no net force in x-direction

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