abannavong
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?"
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abannavong
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how do you set up the question?
abannavong
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I know that m= 25kg
Sea15era
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F=\[nuN\]
Shadowys
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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.
abannavong
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ok
so newtons second law you would use f=ma?
Shadowys
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yes. \(\Sigma F = m\Sigma a\)
abannavong
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oh ok a
abannavong
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so we know that m=25 kg and what is the a?
Sea15era
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|dw:1352779234119:dw|
Sea15era
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first your normal force equals your mass times gravity (245N) and then plug them into your equation to find a. a=.20x245/25
Shadowys
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well, since it is constant velocity, a=0.
Sea15era
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for your a you should get 1.96 m/s^2
abannavong
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oh ok
abannavong
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so how do you get 1.96 m/s^2
Shadowys
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Hmm... @Sea15era why is there acceleration from the total force?
Sea15era
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|dw:1352779486046:dw|
Shadowys
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but the box is moving with constant velocity, i.e. no acceleration, right?
Sea15era
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thats just a velocity acceleration is in m/s^2 not m/s i believe.
Sea15era
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F=25x1.96 F=49N should be the answer if i did my equations and sum of forces statement wrong.
abannavong
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oh ok
Sea15era
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I hope this helps you , good luck and if by chance it's wrong I'm sorry!
Shadowys
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Yes. The acceleration is not given, but since you're pushing it in constant velocity, there should not be any acceleration.
abannavong
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so how do you find the force that is being exerted?
Shadowys
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\(\Sigma F=ma\)
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.
Shadowys
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Now, sea told you \(F_f=\mu F_N\), where \(F_N=F_g\) in this case.|dw:1352781249983:dw|
Do you follow?
abannavong
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kinda
abannavong
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im still a bit confused about the Fn= Fg
Shadowys
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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 \( F_x-F_f=0\), \(F_x=F_f\).
Shadowys
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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 \(F_N=F_g\)
abannavong
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oh ok i got you so far
Shadowys
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Step by step, have you got the value of \(F_N\)?
abannavong
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im still confused on how to get Fn
abannavong
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i got Fg= 245
Shadowys
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yup, and that's the value of Fn. Fn is the normal force of the ground acting on the box.
abannavong
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oh ok
Shadowys
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yeah, so next you will get the value of friction force...
abannavong
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oh ok so how do you find the friction force
Shadowys
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\(F_f = \mu F_N\) where \(\mu\) is the coefficient of kinetic friction.
abannavong
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oh ok
abannavong
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(.20)(25) ?
abannavong
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nvm i mean (245)(.2)
Shadowys
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yup. Then after you get friction force, as you see from posts above, the value of friction force is the force exerted.