anonymous
  • anonymous
Physics help with friction! the problem is "You push a 25kg wooden box across a wooden floor at a constant velocity of 1.0 m/s. If the coefficient of friction is .20, how much force do you exert on the box?"
Physics
schrodinger
  • schrodinger
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anonymous
  • anonymous
how do you set up the question?
anonymous
  • anonymous
I know that m= 25kg
anonymous
  • anonymous
F=\[nuN\]

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anonymous
  • anonymous
First, you draw the free body diagram. And note that the question tells you: there is no acceleration(constant velocity) Then you apply Newton's second law to it.
anonymous
  • anonymous
ok so newtons second law you would use f=ma?
anonymous
  • anonymous
yes. \(\Sigma F = m\Sigma a\)
anonymous
  • anonymous
oh ok a
anonymous
  • anonymous
so we know that m=25 kg and what is the a?
anonymous
  • anonymous
|dw:1352779234119:dw|
anonymous
  • anonymous
first your normal force equals your mass times gravity (245N) and then plug them into your equation to find a. a=.20x245/25
anonymous
  • anonymous
well, since it is constant velocity, a=0.
anonymous
  • anonymous
for your a you should get 1.96 m/s^2
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
so how do you get 1.96 m/s^2
anonymous
  • anonymous
Hmm... @Sea15era why is there acceleration from the total force?
anonymous
  • anonymous
|dw:1352779486046:dw|
anonymous
  • anonymous
but the box is moving with constant velocity, i.e. no acceleration, right?
anonymous
  • anonymous
thats just a velocity acceleration is in m/s^2 not m/s i believe.
anonymous
  • anonymous
F=25x1.96 F=49N should be the answer if i did my equations and sum of forces statement wrong.
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
I hope this helps you , good luck and if by chance it's wrong I'm sorry!
anonymous
  • anonymous
Yes. The acceleration is not given, but since you're pushing it in constant velocity, there should not be any acceleration.
anonymous
  • anonymous
so how do you find the force that is being exerted?
anonymous
  • anonymous
\(\Sigma F=ma\) Start with this picture. |dw:1352781084312:dw| As shown, the forces are F_x and F_f, of different directions. Applying the second law, \(F_x - F_f =ma=0\) as it is constant velocity.
anonymous
  • anonymous
Now, sea told you \(F_f=\mu F_N\), where \(F_N=F_g\) in this case.|dw:1352781249983:dw| Do you follow?
anonymous
  • anonymous
kinda
anonymous
  • anonymous
im still a bit confused about the Fn= Fg
anonymous
  • anonymous
Now, as \(F_g = m g\), where m=25kg and g=9.8, so you get \(F_g\) Now after you get \(F_g\) you can get \(F_N\) After you get \(F_N\), You get \(F_f\) Then since \( F_x-F_f=0\), \(F_x=F_f\).
anonymous
  • anonymous
On the y-axis, v=0,(it doesn't move vertically), so a=0 too. Since \(\Sigma F_y=F_N-F_g= 0\),it follows that \(F_N=F_g\)
anonymous
  • anonymous
oh ok i got you so far
anonymous
  • anonymous
Step by step, have you got the value of \(F_N\)?
anonymous
  • anonymous
im still confused on how to get Fn
anonymous
  • anonymous
i got Fg= 245
anonymous
  • anonymous
yup, and that's the value of Fn. Fn is the normal force of the ground acting on the box.
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
yeah, so next you will get the value of friction force...
anonymous
  • anonymous
oh ok so how do you find the friction force
anonymous
  • anonymous
\(F_f = \mu F_N\) where \(\mu\) is the coefficient of kinetic friction.
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
(.20)(25) ?
anonymous
  • anonymous
nvm i mean (245)(.2)
anonymous
  • anonymous
yup. Then after you get friction force, as you see from posts above, the value of friction force is the force exerted.

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