Negative Values of the Frictional Force?

[AilingLore21]

Homework Statement

Ignore B & C for now

The block in the figure reaches a velocity of 40 m/sec in 100 m, starting from rest. Compute the coefficient of kinetic friction between the block and the ground.

W = 100N

F = 40 N

2. Homework Equations
∑Fy = 0 so Normal Force = Weight
∑Fx = Force - Inertia - Frictional Force - Weight = 0

The Attempt at a Solution

0.5 (V^2 -Vo^2) = as

0.5 (40^2 - 0) = a(100m)

a = 8 m/s^2

So I got this equation:

40 N = 100 N ( cos90) + 100N (8m/s^2 / 9.806 ) + 100 N ( μ)

I ended up getting a negative coefficient of friction which is

-0.415827. Is there something wrong with my equation?

 

[Comeback City]

You cannot have a negative coefficient of friction.
Also, write out your equation in variable form.

 

[John Park]

Is there something wrong with my equation?

Think about the directions of the forces, how they add or subtract.

 

[AilingLore21]

You cannot have a negative coefficient of friction.
Also, write out your equation in variable form.

0.5 (V² -V_o²) = 2 * a * d

∑Fy = 0

W = 100N
N - W = 0
so W =100 N∑Fx = 0

F = 40 N

F - Ff - W ( a / g ) - W = 0

 

[AilingLore21]

Think about the directions of the forces, how they add or subtract.

The inertia naturally resists the motion so it's negative, frictional force too. Weight is multiplied by the cosine of 90 so it's technically zero. Force appears to be going to the right so it's positive. I'm still having trouble why it's a negative value when computed

 

[John Park]

F - Ff - W ( a / g ) - W = 0

If Ff is the frictional force, what is W?

 

[AilingLore21]

If Ff is the frictional force, what is W?

Weight, so are you saying Weight does not exist on a horizontal movement ?

 

[John Park]

The inertia naturally resists the motion so it's negative

But the acceleration must have the same direction and the same sign as the net force.

 

[John Park]

Weight, so are you saying Weight does not exist on a horizontal movement ?

It has no effect that isn't already accounted for: weight by definition acts only vertically, and in this case is balanced by the normal force. You have the coresponding mass in the term that includes g.

 

[kuruman]

∑Fx = Force - Inertia - Frictional Force - Weight = 0

I see a couple of problems with this equation.
1. You say that the net force (sum of all the forces) is zero. That is true only if the acceleration is zero. This is not true here because the block starts from zero velocity and reaches 40 m/s some time later. Therefore the acceleration cannot possibly be zero.
2. You show 4 forces acting in the horizontal direction. I can only see 2 that make sense, "Force" (pushing the block) and "Frictional force" (from the ground contact). Where does "Inertia" come from? What about "Weight"? Is that in the horizontal direction?

 

[TomHart]

-0.415827. Is there something wrong with my equation?

I got the same answer. Maybe we are both crazy.

 

[AilingLore21]

It has no effect that isn't already accounted for: weight by definition acts only vertically, and in this case is balanced by the normal force. You have the coresponding mass in the term that includes g.

Ok , so this gave me an idea

∑Fx = 0

F = 40 N

F - Ff + W ( a / g ) = 0

So if I compute again

40 - Ff + 100 ( 8m/s² / 9.806)

I would get a Ff of 121.5827 N

divide that by 100 and I get a coefficient of 1.22

Seems pretty large for a coefficient of friction

 

[Comeback City]

∑Fx = 0

This is wrong.

Read what @kuruman said.

 

[TomHart]

If a force of 40 N is applied to a mass of 10 kg, you get an acceleration of 4 m/s^2. So how can you get an acceleration of 8 m/s^2 unless you have another force acting in the same direction. Thus, a negative μ.

Maybe it's too earl for me and I'm not fully awake yet.
Edit: See ^. I can't even type "early".

 

[Comeback City]

Interestingly enough, I also got that same negative answer. I'm not sure what could've gone wrong here unless the data from the original question was copied down incorrectly, since a negative coefficient of friction is physically impossible.

 

[TomHart]

Interestingly enough, I also got that same negative answer. I'm not sure what could've gone wrong here unless the data from the original question was copied down incorrectly, since a negative coefficient of friction is physically impossible.

Either written down wrong, or a poorly-thought-out problem.

 

[Comeback City]

Either written down wrong, or a poorly-thought-out problem.

Yeah both are very possible

 

[TomHart]

@AilingLore21 Can you check if all of the values in the problem are written down correctly.

 

[John Park]

I think you have the wrong sign for the acceleration; but I agree something is strange here. The block was accelerated at almost g, which as far as I can see is incompatible with a 40 N force and a 100/g kg mass. Have you checked the given data, including units?

 

[AilingLore21]

I somewhat figured out that

@AilingLore21 Can you check if all of the values in the problem are written down correctly.

They are apparently. I'm used to solving the inclined versions of these. I wonder why I'm having this much trouble on a horizontal one

 

[TomHart]

They are apparently. I'm used to solving the inclined versions of these. I wonder why I'm having this much trouble on a horizontal one

Actually, your first answer of μ = -0.416 looks like it was correct. But I could not understand your method. It seemed like there were some problems.
For example, if you could explain how this equation came about.

40 N = 100 N ( cos90) + 100N (8m/s^2 / 9.806 ) + 100 N ( μ)

 

[kuruman]

Look, if the contact were frictionless, the acceleration would be
$$a=\frac{F}{W/g}=\frac{40}{100/9.8} = 3.92~m/s^2$$
How can the acceleration with friction be about twice a much? If the numbers are copied correctly, they are not thought out correctly.

 

[TomHart]

40 N = 100 N ( cos90) + 100N (8m/s^2 / 9.806 ) + 100 N ( μ)

Actually, the more I look at your equation, the more it makes sense. The only thing that is really questionable in that equation is the 100cos90 term. Where did that come from?

One other tip: It is best to leave out units in these equations. Just include them in the final result.

If I throw out the 100cos90 term in your equation, I am left with:
40 = (100)(8/9.8) + 100μ, which is derived from:
ΣF = ma
F - f = ma (where F is the 40 N applied force and f is the friction force)
40 - μ(100) = (100/9.8)(8)
Solving gives μ = -0.416.

So I think you were pretty much on the right track, except for that 100cos90 term.

 

[kuruman]

If I throw out the 100cos90 term in your equation

You can safely do that, $\cos 90^o =0$.

 

[TomHart]

You can safely do that, cos90=0.

For sure. I just think it shows a lack of understanding that it was there in the first place.

 

[John Park]

I don't know if you'd find it helpful, but I like to start from Newton's law for this type of problem.

Net Force = Mass x Acceleration.

You've found the acceleration. The mass is 100/g kg.

The net force is the applied force less the frictional force = 40 - 100.μ N (where μ is the frictional coefficient to be found).

I get the same negative answer you did originally, too.

 

[kuruman]

I just think it shows a lack of understanding that it was there in the first place.

Perhaps. Although not as simple as zero, 100 N cos90^o is, nevertheless, the horizontal component of the weight.

 

[TomHart]

Perhaps. Although not as simple as zero, 100 N cos90o is, nevertheless, the horizontal component of the weight.

Usually as folks progress on in the study of physics, they learn that they can usually omit the horizontal component of weight. :)

 

[TomHart]

Perhaps. Although not as simple as zero, 100 N cos90o is, nevertheless, the horizontal component of the weight.

Actually, the more I think about it, the more I think I agree with you - that it might not show a lack of understanding. It could actually be an indication of a fairly thorough understanding of the fundamental concept.

 

[kuruman]

Usually as folks progress on in the study of physics, they learn that they can usually omit the horizontal component of weight. :)

Indeed, but I think this is not a good problem for honing one's physical understanding as I indicated in post #22.