Physics Help Forum Friction Problem

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 Jan 16th 2010, 02:44 AM #1 Junior Member   Join Date: Nov 2009 Posts: 7 Friction Problem I need to check my answer ^^ An 85-N box of orange is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and floor.
 Jan 16th 2010, 08:28 AM #2 Physics Team   Join Date: Feb 2009 Posts: 1,425 Why dont you state your answer and let us check if it is ok ?
Jan 17th 2010, 06:22 AM   #3
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 Originally Posted by jasonlewiz I need to check my answer ^^ An 85-N box of orange is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and floor.

Solution:

coefficient of kinetic friction = kinetic friction/normal force

since W=N therefore N= 85 N

I don't know what will be my next solution,.,

Jan 17th 2010, 10:24 AM   #4
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OK

 Originally Posted by jasonlewiz I need to check my answer ^^
 Originally Posted by jasonlewiz An 85-N box of orange is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20N and a vertical component of 25 N downward. Calculate the coefficient of kinetic friction between the box and floor.

I would start by summing all the forces acting on the box and setting them equal to the net Force acting upon the box. All we really care about are the forces that tend to move the box, in other words we care only about the x components of force acting upon the box.

Now friction is due to the amount of force acting downward, including the box's weight, but it acts to counter the x component of force that is pushing the box (the 20N of force that acts to keep the box moving). So let's first find the force from friction as this is the only other x component of force other than the 20N that is acting to keep the box moving. The force from friction is:

(weight + 25N)Kf where Kf is the kinetic coefficient of friction and the 25N comes from the downward component of force that is acting along with the 20N of x directed force that is being applied upon the box. So let's put the weight in there so we are working with only one unknown, Kf. So the force from friction is (85N +25N)Kf = 110Kf

Now this force acts to slow the box down so let's call this a negative force and the 20N that acts to keep the box moving we will keep positive, so summing the x directed forces upon the box we get:
20 - 110Kf = ma = 8.673a = 8.673*(-0.9m/s^2)
where 8.673 is the mass of an 85 N box,
mass = weight/g = 85N/(9.8m/s^2)

That is we are given it is accelerating at a rate of -0.9 m/s every second. Actually we are told it is slowing at a rate of 0.9 m/s every second which is the same as it negatively accelerating at -0.9 m/s^2

Now the only unknown in the following equation is the kinetic coefficient of friction, so let's solve for it:

(20-110Kf) = -7.806

rearranging and solving for Kf yields:

(-7.806 - 20)/-110 = -27.8061/-110 = Kf

Kf = 0.2528

Let's check the answer by placing it back into the sum of the x directed forces acting on the box and see if we find the box accelerating at:
-0.9m/s^2

20 -110*0.2528 = 8.673a

a = (20-27.80612)/8.673 = -7.8061/8.673 = -0.9 m/s^2

Yup, that seems to work. I noticed you posted something like:

"Solution:

coefficient of kinetic friction = kinetic friction/normal force

since W=N therefore N= 85 N"

I don't know what you mean by this statement, the first part I think I understand what you are saying that the force countering the motion of the box that is due to kinetic friction divided by the force normal to the box equals the coefficient of kinetic friction, OK, but then you say W = N therefore N = 85 N''. The weight does not equal the normal force applied by the box onto the surface it is sliding on, you have to also add in the additional 25N of downward force upon the box to find the force normal to the surface that is = 85N + 25N = 110N which does not equal the weight of the box, so no W = N is not true (assuming N is the force normal to the surface) then N = W + 25N, and even if the normal force was only due to the weight of the box what do you mean by N = 85 N''? Are you saying the normal force is equal to 85 Newtons and are using N'' to convey Newtons because you used N as a variable for the normal force?

Anyway, those are more like questions you should ask yourself, and remember that the force normal to the surface is not always the weight of "thing" sitting upon that surface, but it is the sum of all y directed forces acting upon the "thing", including its weight, that sum to the force seen normal to the surface on which this "thing" is on.

In your case you have a box that weighs 85N with a downward force of 25N acting upon it making it "appear" to be heavier as far as the floor is concerned, so while in motion the box offers a counter force to being pushed of 110N*0.2528 = 27.8061N and is being pushed with a force of only 20N so the net force acting upon the box is 7.8061 which acts counter to the force that is trying to keep the box moving in the same direction it is now sliding in, so it must be slowing at a rate of:

a = F/m = 7.8061/8.673 = 0.9 m/s^2

Long story short the kinetic coefficient of friction is 0.25279 and this value is always smaller than the static coefficient of friction, which eventually when that box comes to a stop is what you would need to know in order to figure out how much force is needed to get it moving again.

Many Smiles, Craig

Last edited by clombard1973; Jan 17th 2010 at 10:29 AM.

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