Calculating acceleration from the coefficient of kinetic friction

AI Thread Summary
The discussion focuses on calculating the acceleration of two boxes being raised on a plank, given their weights and coefficients of friction. For part (a), the angle at which each box begins to slide is determined to be 21.8 degrees, as the mass cancels out in the friction equations. In part (b), participants discuss using the net force equation, F = ma, to find acceleration, emphasizing that the gravitational force component and frictional force change when the boxes start sliding. The normal force remains relatively constant, while the frictional force shifts from static to kinetic, affecting the net force and thus the acceleration. Understanding these force dynamics is crucial for accurately calculating the boxes' acceleration down the ramp.
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Homework Statement


At a construction side, a small crane is raising two boxes of nails on a plank to the roof. One box has already been opened and is half full, while the other box is new. The boxes, including the nails, weight 10 kg and 20 kg, respectively, and are the same size.

Diagram:
BLCZgpB.png


a) If the coefficient of static friction is 0.4, draw an FBD for each box of nails and use it to calculate the angle at which each box beings to slide.
b) If the coefficient of kinetic friction is 0.3, how fast will the boxes accelerate along the plank, once they start to slide?

Homework Equations

The Attempt at a Solution


[/B]
a)

Box 1 = 10 kg, box 2 = 20 kg. Both FBDs are the same except for the mass of the box. For my FBD (if it's correct) should I include mgcosθ and mgsinθ?
hr6DMXh.png


Using the FBD:

\begin{array}{l}
μF_N=F_g \\
μF_N=mg\sinθ \\
(0.4)(mg\cosθ)=mg\sinθ \\
0.4=\frac {\sin}{\cos} \\
θ=tan^{-1}(0.4) \\
θ=21.8°
\end{array}

This is the answer for both boxes because mass and gravity cancel out in the equation, meaning that the mass of the box does not effect the angle at which each box begins to slide but the coefficient of friction does.

b) I don't know where to start for this question. I thought of using F=ma but I need to use the angle at which they begin to slide somewhere so F=ma won't work.
 

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Your diagram and your equations are not consistent. But your equation is OK. (By the way, put a backslash \ before "sin", "cos" etc and the function name will appear in a non-italic font).
1. ##F_g## according to your equation is the component of gravity parallel to the ramp. It's the force that would cause the object to move, with friction opposing it. So the direction on the diagram is incorrect.

2. ##F_n##, according to your equation, is the other component of gravity, the one normal to the surface. So it should be normal to the surface.

3. And since the two forces in that equation are not parallel / anti-parallel, they shouldn't look that way in your diagram.

b) F = ma is correct. That would be the net force, the force which pulls it down the ramp (which depends on the angle in the way you showed in part a), minus the frictional force (which also depends on the angle, in the way you showed in part a).
 
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RPinPA said:
Your diagram and your equations are not consistent. But your equation is OK. (By the way, put a backslash \ before "sin", "cos" etc and the function name will appear in a non-italic font).
1. ##F_g## according to your equation is the component of gravity parallel to the ramp. It's the force that would cause the object to move, with friction opposing it. So the direction on the diagram is incorrect.

2. ##F_n##, according to your equation, is the other component of gravity, the one normal to the surface. So it should be normal to the surface.

3. And since the two forces in that equation are not parallel / anti-parallel, they shouldn't look that way in your diagram.

b) F = ma is correct. That would be the net force, the force which pulls it down the ramp (which depends on the angle in the way you showed in part a), minus the frictional force (which also depends on the angle, in the way you showed in part a).
Yeah I am really confused. I edited my equations right as you posted this because something looked wrong. Also I am sort of confused about why mg sin and mg cos are actually doing in this problem and why they are where they are.

Is this FBD better?

hv7HlsU.png
 

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Yes, if ##F_g## means ##mg##. That is the weight vector, and it points vertically downward. Although as I noted that's not the way you used ##F_g## when you wrote your equation.

Specter said:
Also I am sort of confused about why mg sin and mg cos are actually doing in this problem and why they are where they are.

Look at the triangle you have drawn, that has ##F_g## as hypotenuse and two dashed lines as legs. Those are the components of ##F_g## which are perpendicular and parallel to the ramp.

The angle between the normal dashed line and ##F_g## is the same ##\theta## as between the ramp and the horizontal. It might take some trig to convince yourself of that, but it's true. One argument that might help you think about it is to think what happens as the ramp angle goes to 0. When the ramp is horizontal, that component is the whole weight. The angle between ##mg## and the normal component is 0.

So the normal component is the adjacent of ##\theta##, meaning the normal component is ##F_g \cos \theta##. That's why ##F_N## is ##mg \cos \theta##. And the other component, the one parallel to the ramp, the one that pulls the object down the ramp, is the opposite of angle ##\theta## and so the component is ##F_g \sin \theta ##.

Again, think about what happens when ##\theta## goes to 0 and the sine goes to 0. When the ramp is horizontal, there's no gravity acting in the parallel direction. So something that goes to 0 is the right trig function to use here.

The equation you write is that static friction ##\mu F_N## is equal and opposite to the component of gravity pulling down the ramp, ##mg \sin \theta##. And ##F_N## as we said is ##mg \cos \theta##.
 
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RPinPA said:
Yes, if ##F_g## means ##mg##. That is the weight vector, and it points vertically downward. Although as I noted that's not the way you used ##F_g## when you wrote your equation.
Look at the triangle you have drawn, that has ##F_g## as hypotenuse and two dashed lines as legs. Those are the components of ##F_g## which are perpendicular and parallel to the ramp.

The angle between the normal dashed line and ##F_g## is the same ##\theta## as between the ramp and the horizontal. It might take some trig to convince yourself of that, but it's true. One argument that might help you think about it is to think what happens as the ramp angle goes to 0. When the ramp is horizontal, that component is the whole weight. The angle between ##mg## and the normal component is 0.

So the normal component is the adjacent of ##\theta##, meaning the normal component is ##F_g \cos \theta##. That's why ##F_N## is ##mg \cos \theta##. And the other component, the one parallel to the ramp, the one that pulls the object down the ramp, is the opposite of angle ##\theta## and so the component is ##F_g \sin \theta ##.

Again, think about what happens when ##\theta## goes to 0 and the sine goes to 0. When the ramp is horizontal, there's no gravity acting in the parallel direction. So something that goes to 0 is the right trig function to use here.

The equation you write is that static friction ##\mu F_N## is equal and opposite to the component of gravity pulling down the ramp, ##mg \sin \theta##. And ##F_N## as we said is ##mg \cos \theta##.

Wow that makes perfect sense. Much easier to understand than how my notes explain it. Thanks so much!

For part B you said I can use ##F=ma## . I know it can be rearranged to ##\displaystyle a=\frac {F_{net}} {m}## . I don't really understand how I am supposed to use it for this question. I know that the boxes start to slide at 21.8 degrees, and that the coefficient of kinetic friction is 0.3.
 
Specter said:
Wow that makes perfect sense. Much easier to understand than how my notes explain it. Thanks so much!

For part B you said I can use ##F=ma## . I know it can be rearranged to ##\displaystyle a=\frac {F_{net}} {m}## . I don't really understand how I am supposed to use it for this question. I know that the boxes start to slide at 21.8 degrees, and that the coefficient of kinetic friction is 0.3.
Just before it started to slide, everything was in balance. The static friction was equal to the downslope component of the gravitational force.
As it starts to slide, what forces change and by how much? What does that tell you about the net force parallel to the slope now?
 
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haruspex said:
Just before it started to slide, everything was in balance. The static friction was equal to the downslope component of the gravitational force.
As it starts to slide, what forces change and by how much? What does that tell you about the net force parallel to the slope now?
I'm not sure. I don't really understand this. Maybe the net force is equal to the normal force minus the force of kinetic friction? I'm lost. I think that if I find the net force I can use F=ma to find acceleration.
 
Please try again to answer this question.
haruspex said:
As it starts to slide, what forces change and by how much?
Let's go through the list. We have
  • Gravitational force
  • Normal force
  • Frictional force
Which of those might change when it goes from stationary to sliding?
Specter said:
Maybe the net force is equal to the normal force minus the force of kinetic friction?
There are three forces. The net force comes from adding all three as vectors.

It may help to go back to the equations you used to solve the first part but make them more general.
You wrote ##\mu F_N=mg\sin(\theta)##.
We can generalise that by allowing for acceleration and not assuming maximum friction: ##ma=\Sigma F=mg\sin(\theta)-F_{friction}##.
For the first part, you could plug in a=0 and ##F_{friction}=\mu_sF_N##. In the second part we need different substitutions.
 
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haruspex said:
Please try again to answer this question.

Let's go through the list. We have
  • Gravitational force
  • Normal force
  • Frictional force
Which of those might change when it goes from stationary to sliding?

There are three forces. The net force comes from adding all three as vectors.

It may help to go back to the equations you used to solve the first part but make them more general.
You wrote ##\mu F_N=mg\sin(\theta)##.
We can generalise that by allowing for acceleration and not assuming maximum friction: ##ma=\Sigma F=mg\sin(\theta)-F_{friction}##.
For the first part, you could plug in a=0 and ##F_{friction}=\mu_sF_N##. In the second part we need different substitutions.

Which of those might change when it goes from stationary to sliding?
Normal force and frictional force. Gravitational force would not change.

It may help to go back to the equations you used to solve the first part but make them more general.
You wrote ##\mu F_N=mg\sin(\theta)##.
We can generalise that by allowing for acceleration and not assuming maximum friction: ##ma=\Sigma F=mg\sin(\theta)-F_{friction}##.
For the first part, you could plug in a=0 and ##F_{friction}=\mu_sF_N##. In the second part we need different substitutions.

I'm still confused.. sorry. What does this part mean? ##ma=\Sigma F=mg\sin(\theta)-F_{friction}##, I haven't seen learned (##\Sigma##) this yet. Does it mean ##ma=mg\sin(\theta)-F_{friction}##? If so, then you said to let ##a=0## and ##F_{friction}=\mu_sF_N##? If this is the correct way to write this, does that mean that mass cancels out again just like in part a?
 
  • #10
##\Sigma## is a capital "sigma". It means "sum." ##ma = \Sigma F## means "##ma## is the sum of all the forces."
 
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  • #11
RPinPA said:
##\Sigma## is a capital "sigma". It means "sum." ##ma = \Sigma F## means "##ma## is the sum of all the forces."
Thank you.
 
  • #12
haruspex said:
Which of those might change when it goes from stationary to sliding?
Specter said:
Normal force and frictional force. Gravitational force would not change.
Would the normal force change very much in a transition from just-before slipping to just-after slipping?
 
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  • #13
jbriggs444 said:
Would the normal force change very much in a transition from just-before slipping to just-after slipping?
I guess it wouldn't. Would friction be the only force that changes then?
 
  • #14
Yes, That is what @haruspex was trying to get you to realize.
 
  • #15
jbriggs444 said:
Yes, That is what @haruspex was trying to get you to realize.
Okay. If only friction changes how can I use F=ma to solve for acceleration? I'm lost.
 
  • #16
Before it broke loose, it was stationary. Acceleration was what?
After it broke loose, friction changed. By how much did it change?
As a result of that change, how much should acceleration change?
So how much acceleration will result?

That is an outline of the strategy I would use.
 
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  • #17
jbriggs444 said:
Before it broke loose, it was stationary. Acceleration was what?
After it broke loose, friction changed. By how much did it change?
As a result of that change, how much should acceleration change?
So how much acceleration will result?

That is an outline of the strategy I would use.

Before it broke loose the acceleration was ##a=0.0m/s^2##
After it broke loose, the friction changed from ##\mu_s=0.4## to ##\mu_k=0.3##. I guess this would result in a change of 0.1 in friction? I haven't had to do that yet.
How can I find how the acceleration will change from the change in friction?
 
  • #18
Specter said:
Before it broke loose the acceleration was ##a=0.0m/s^2##
Yep.
After it broke loose, the friction changed from ##\mu_s=0.4## to ##\mu_k=0.3##. I guess this would result in a change of 0.1 in friction? I haven't had to do that yet.
The relevant coefficient of friction changed by 0.1. If we just knew the normal force, we could translate that into a change in frictional force. Had we calculated the normal force yet? If not, we can write a force balance equation in the normal direction to figure it out.
 
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  • #19
jbriggs444 said:
Yep.

The relevant coefficient of friction changed by 0.1. If we just knew the normal force, we could translate that into a change in frictional force. Had we calculated the normal force yet? If not, we can write a force balance equation in the normal direction to figure it out.
Alright. This would be the equation to calculate the normal force. I would have to do this for each box because of the different masses right? 10kg and 20kg.

Hopefully this is correct. I just have to plug everything in.

\begin{array}{l}
F_{net}=F_N-F_g \\
F_{net}=F_N-mg \\
0 N=F_N-mg \cos \theta \\
F_N=mg\cos \theta
\end{array}
 
  • #20
Good. Now with that in hand, can you take the next steps?
 
  • #21
jbriggs444 said:
Good. Now with that in hand, can you take the next steps?
In my text this is how a similar example had solved the problem. Have I done this correctly?The normal force of box a (10 kg) is 91 N
The normal force of box b (20 kg) is 182 N

Box A:
##F_f=\mu F_N##
##F_f=(0.1)(90.00)##
##F_f=9.01 N##

Box B:
##F_f=\mu F_N##
##F_f=(0.1)(182.00)##
##F_f=18.2 N##
 
  • #22
Specter said:
a similar example
It is not similar enough to help. There you had two boxes but only one coefficient of friction. In the present problem you have two coefficients, and although there are two boxes we are just dealing with one at this point. (Maybe there are other parts to the question where two boxes becomes relevant.)

You have established that the normal force does not change when the box starts to slide. Write an expression for the normal force.
You have established that the coefficient of friction changes from 0.4 to 0.3. What is the change in maximum frictional force?
 
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  • #23
haruspex said:
It is not similar enough to help. There you had two boxes but only one coefficient of friction. In the present problem you have two coefficients, and although there are two boxes we are just dealing with one at this point. (Maybe there are other parts to the question where two boxes becomes relevant.)

You have established that the normal force does not change when the box starts to slide. Write an expression for the normal force.
You have established that the coefficient of friction changes from 0.4 to 0.3. What is the change in maximum frictional force?

The change in the frictional force would be 0.1.

The motion is only occurring along the x-axis, right? So the net force in the y-direction would be 0? Which expression should I use? I feel like the second one I wrote is the correct one because I tried solving with the first one and it works out to an angle.

Previously, I had written this as the expression for normal force:

\begin{array}{l}
F_{net}=F_N-F_g \\
F_{net}=F_N-mg \\
0 N=F_N-mg \cos \theta \\
F_N=mg\cos \theta
\end{array}

This is what I have done now:

\begin{array}{l}
F_{net}=F_N-mg \\
F_N=mg \\
F_N=(10)(9.8) \\
F_N=98 N
\end{array}

Now I could use F=ma to solve for a, if the above is correct:

\begin{array}{l}
\displaystyle F=ma \\
\displaystyle a=\frac {f}{m} \\
\displaystyle a=\frac {98}{10} \\
\displaystyle a=9.8 m/s^2
\end{array}

The only thing is that I haven't used the coefficient of friction in my work so I don't know if I did it correctly.
 
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  • #24
Specter said:
The change in the frictional force would be 0.1.
You keep confusing coefficient of friction with frictional force. @jbriggs444 already corrected you on this in post #18.
If the coefficient changes by 0.1 and the normal force is N, how much does the frictional force change by?
 
  • #25
Specter said:
Previously, I had written this as the expression for normal force:

\begin{array}{l}
F_{net}=F_N-F_g \\
F_{net}=F_N-mg \\
0 N=F_N-mg \cos \theta \\
F_N=mg\cos \theta
\end{array}

This is what I have done now:

\begin{array}{l}
F_{net}=F_N-mg \\
F_N=mg \\
F_N=(10)(9.8) \\
F_N=98 N
\end{array}
How exactly are you defining Fg? You seem to be confused whether it is the whole gravitational force, mg, or just the component normal to the surface, mg cos(θ).
Fnet=FN-mg makes no sense because FN and mg act in different directions.
 
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  • #26
haruspex said:
How exactly are you defining Fg? You seem to be confused whether it is the whole gravitational force, mg, or just the component normal to the surface, mg cos(θ).
Fnet=FN-mg makes no sense because FN and mg act in different directions.

So how about this? ##F_N## and ##mg\cos (\theta)## both act in the same direction so this is right?

\begin{array}{l}
F_{net}=F_N-F_g \\
F_{net}=F_N-mg \\
0 N=F_N-mg \cos \theta \\
F_N=mg\cos \theta \\
F_N=(10)(9.8)cos21.8 \\
F_N=90.99 \\
F_N=91 N
\end{array}

You keep confusing coefficient of friction with frictional force. @jbriggs444 already corrected you on this in post #18.
If the coefficient changes by 0.1 and the normal force is N, how much does the frictional force change by?

How do I find the change in frictional force? ##F_f=\mu F_N## ?
 
  • #27
Specter said:
So how about this? ##F_N## and ##mg\cos (\theta)## both act in the same direction so this is right?

\begin{array}{l}
F_{net}=F_N-F_g \\
F_{net}=F_N-mg \\
0 N=F_N-mg \cos \theta
\end{array}
Before you write down an equation, you need to decide what the terms in the equation mean. For instance, what is ##F_{net}##? Is it a vector? Is it a component of a vector? What forces does it sum over?

This becomes important when we find a ##\cos \theta## term springing up from no where.
 
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  • #28
jbriggs444 said:
Before you write down an equation, you need to decide what the terms in the equation mean. For instance, what is ##F_{net}##? Is it a vector? Is it a component of a vector? What forces does it sum over?

This becomes important when we find a ##\cos \theta## term springing up from no where.

##F_{net}## is a sum of ##F_N## and ##\displaystyle F_{g_y}##, but ##F_N=0## because there is no acceleration along the y-axis. Is that correct? I used ##mg\cos \theta## because ##F_{g_y}=mg\cos \theta## . I don't really understand this much but I'm trying my best to find problems in my text and base what I am writing down here off of those problems.
 
  • #29
Specter said:
So how about this? ##F_N## and ##mg\cos (\theta)## both act in the same direction so this is right?
FN and Fgy are in the same direction, and the magnitude of Fgy is ##mg\cos(\theta)##. But "##mg\cos(\theta)##" Is not a force, so does not have direction.
 
  • #30
Specter said:
How do I find the change in frictional force? ##F_f=\mu F_N## ?
There arevtwo different frictional forces, so we need to name them differently.
In the static case, the general formula is ##F_{s}\leq\mu_sN##. Since we know it is just about to slip that becomes ##F_{s}=\mu_sN##.
In the kinetic case, ##F_{k}=\mu_kN##.
So how much has frictional force changed by?
 
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  • #31
haruspex said:
FN and Fgy are in the same direction, and the magnitude of Fgy is ##mg\cos(\theta)##. But "##mg\cos(\theta)##" Is not a force, so does not have direction.
Ahhh I am so lost. So ##mg\cos \theta## does not have a direction because it is not a force, but instead just a component of ##mg##.

Would ##F_N=F_g## then? I don't get it.

As for the change in frictional force I need to find the normal force before doing that. Once I have found ##F_N## would I find both frictional forces and then subtract the two to find the difference or change in frictional force?
 
  • #32
Specter said:
Ahhh I am so lost. So ##mg\cos \theta## does not have a direction because it is not a force, but instead just a component of ##mg##.
That is not quite it.
The "component of the force of gravity in the normal direction" is a vector. It has a direction. Trivially, that direction is normal to the surface.

But ##mg \cos \theta## is not a vector. It is a scalar value. You could multiply it by the unit vector to give it a direction. If you used the unit normal vector, that direction would be normal to the surface.
Would ##F_N=F_g## then?
##F_g## is the force of gravity. Its direction is downward.
##F_N## is the normal component of the contact force. Its direction is normal to the surface.
Clearly the two are unequal because they do not even point in the same direction.

However, the normal component of the gravitational force plus the normal component of the contact force will total to give you the net force in the normal direction.
 
  • #33
jbriggs444 said:
That is not quite it.
The "component of the force of gravity in the normal direction" is a vector. It has a direction. Trivially, that direction is normal to the surface.

But ##mg \cos \theta## is not a vector. It is a scalar value. You could multiply it by the unit vector to give it a direction. If you used the unit normal vector, that direction would be normal to the surface.

##F_g## is the force of gravity. Its direction is downward.
##F_N## is the normal component of the contact force. Its direction is normal to the surface.
Clearly the two are unequal because they do not even point in the same direction.

However, the normal component of the gravitational force plus the normal component of the contact force will total to give you the net force in the normal direction.

Yeah at that point I was just guessing. All I need to do is add the components of mg?So ##F_g## is the force of gravity. ##mg\sin \theta## would be the normal component of the gravitational force.

##mg\cos \theta## is the normal component of the contact force.

So,

##F_N=mg\sin(\theta)+mg\cos(\theta)##
 
  • #34
Specter said:
Yeah at that point I was just guessing. All I need to do is add the components of mg?So ##F_g## is the force of gravity. ##mg\sin \theta## would be the normal component of the gravitational force.
What makes you think so?
##mg\cos \theta## is the normal component of the contact force.
What makes you think so?
##F_N=mg\sin(\theta)+mg\cos(\theta)##
##F_N## is the normal component of the total contact force. The tangential component is named "friction".
 
  • #35
jbriggs444 said:
What makes you think so?

What makes you think so?

##F_N## is the normal component of the total contact force. The tangential component is named "friction".

You said that the normal component of the gravitational force plus the normal component of the contact force will total to give you the net force in the normal direction.

I now see that the normal component of the contact force is ##F_N## because ##F_N## is perpendicular to ##F_s##. I should have read your reply above more carefully, sorry! I don't know what the normal component of gravitational force would be. I thought it would be ##mg\sin \theta## because it is perpendicular to ##mg\cos \theta## and ##F_N=mg\cos \theta##. If that's not the case, would the normal component of gravitational force be ##mg## then?
 
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  • #36
Specter said:
You said that the normal component of the gravitational force plus the normal component of the contact force will total to give you the net force in the normal direction.

I now see that the normal component of the contact force is ##F_N##. I don't know what the normal component of gravitational force would be. I thought it would be ##mg\sin \theta## because it is perpendicular to ##mg\cos \theta## and ##F_N=mg\cos \theta##. If that's not the case, would the normal component of gravitational force be ##mg## then?
I am having trouble following your logic.

The point of this little exercise is to determine ##F_N##. If you've already decided that ##F_N=mg\cos \theta## then you are done.

The appropriate reasoning would be as follows:

Write down a force balance for the object in the normal direction. There are two relevant forces in this direction. One is the normal component of the gravitational force. That component is ##F_{gy}##. It is directed negative (into the surface) and has a magnitude of ##mg \cos \theta##. The other is the normal force (aka the normal component of the total contact force). It is directed positive (out of the surface). Its magnitude is unknown. The whole point of this little sub-exercise is to determine ##F_N##.

We can write down the force balance equation. ΣF=ma: ##-mg \cos \theta + F_N = ma_N##

The object is not accelerating perpendicular to the surface. It is simply sitting there or sliding. So ##a_N = 0##.

Solve for ##F_N##. What do you get?

Nowhere in this line of reasoning does ##mg \sin \theta## come in. It is irrelevant to the normal direction. It might have some relevance if we were doing a force balance in the tangential direction. But we are not doing any such thing at this point. Put it right out of your head.
 
Last edited:
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  • #37
jbriggs444 said:
I am having trouble following your logic.

The point of this little exercise is to determine ##F_N##. If you've already decided that ##F_N=mg\cos \theta## then you are done.

The appropriate reasoning would be as follows:

Write down a force balance for the object in the normal direction. There are two relevant forces in this direction. One is the normal component of the gravitational force. That component is ##F_{gy}##. It is directed negative (into the surface) and has a magnitude of ##mg \cos \theta##. The other is the normal force (aka the normal component of the total contact force). It is directed positive (out of the surface). Its magnitude is unknown. The whole point of this little sub-exercise is to determine ##F_N##.

We can write down the force balance equation. ΣF=m: ##-mg \cos \theta + F_N = ma_N##

The object is not accelerating perpendicular to the surface. It is simply sitting there or sliding. So ##a_N = 0##.

Solve for ##F_N##. What do you get?

Nowhere in this line of reasoning does ##mg \sin \theta## come in. It is irrelevant to the normal direction. It might have some relevance if we were doing a force balance in the tangential direction. But we are not doing any such thing at this point. Put it right out of your head.
##F_N=(10 kg)(9.8m/s^2)\cos21.8##
##=91 N##

I've got this answer before, am I still solving incorrectly? I also have two masses, 10 kg and 20 kg, so do I find the normal force for both?
 
  • #38
Specter said:
##F_N=(10 kg)(9.8m/s^2)\cos21.8##
##=91 N##

I've got this answer before, am I still solving incorrectly? I also have two masses, 10 kg and 20 kg, so do I find the normal force for both?
You might want to review post #16. It is 22 posts later and we are finally circling back to it.

The 91 N number tells you the normal force for the 10 kg mass.

[Personally, I would prefer to leave it as ##mg \cos \theta##. It is too early to be putting numbers in. We are working algebra still. Leaving "m" in the result will turn out to be helpful a few posts from now -- it is going to cancel out]

Recall what you are trying to determine: "how fast will the boxes accelerate along the plank, once they start to slide"

Recall that we already know how much less the coefficient of kinetic friction (0.3) is than the coefficient of static friction (0.4).

How much less is the force of kinetic friction than the force of static friction?
 
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  • #39
jbriggs444 said:
You might want to review post #16. It is 22 posts later and we are finally circling back to it.

The 91 N number tells you the normal force for the 10 kg mass.

[Personally, I would prefer to leave it as ##mg \cos \theta##. It is too early to be putting numbers in. We are working algebra still. Leaving "m" in the result will turn out to be helpful a few posts from now -- it is going to cancel out]

Recall what you are trying to determine: "how fast will the boxes accelerate along the plank, once they start to slide"

Recall that we already know how much less the coefficient of kinetic friction (0.3) is than the coefficient of static friction (0.4).

How much less is the force of kinetic friction than the force of static friction?

Alright. So I have the normal force which is ##F_N=mg\cos \theta##.

To find the force of kinetic friction I would do ##F_{f_k}=\mu_kmg\cos \theta##
To find the force of static friction I would do ##F_{f_s}=\mu_smg\cos \theta##

Is this the correct way to figure that out?
 
  • #40
Specter said:
Alright. So I have the normal force which is ##F_N=mg\cos \theta##.

To find the force of kinetic friction I would do ##F_{f_k}=\mu_kmg\cos \theta##
To find the force of static friction I would do ##F_{f_s}=\mu_smg\cos \theta##

Is this the correct way to figure that out?
That is all correct. So what do you get when you subtract the two equations above from each other?
 
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  • #41
jbriggs444 said:
That is all correct. So what do you get when you subtract the two equations above from each other?
When solving for the force of friction in each of the equations above, does it matter which mass I use (10 kg or 20 kg)?

This is what I've done. I really hope this is correct.

##F_{f_s}=\mu_smg\cos (\theta)##
##F_{f_s}=(0.4)98\cos21.8##
##F_{f_s}=36.4 N##

##F_{f_k}=\mu_kmg\cos (\theta)##
##F_{f_k}=(0.3)08\cos21.8##
##F_{f_k}=27.3 N##

##F_{f_s}-F_{f_k}##
##36.4-27.3=9.1 N##
 
  • #42
Specter said:
When solving for the force of friction in each of the equations above, does it matter which mass I use (10 kg or 20 kg)?

This is what I've done. I really hope this is correct.

##F_{f_s}=\mu_smg\cos (\theta)##
##F_{f_s}=(0.4)98\cos21.8##
##F_{f_s}=36.4 N##

##F_{f_k}=\mu_kmg\cos (\theta)##
##F_{f_k}=(0.3)08\cos21.8##
##F_{f_k}=27.3 N##

##F_{f_s}-F_{f_k}##
##36.4-27.3=9.1 N##
It is correct, as far as it goes for the 10 kg mass. But I had hoped you would subtract the equations algebraicly. That way you would still have the "m" in there.
 
  • #43
jbriggs444 said:
It is correct, as far as it goes for the 10 kg mass. But I had hoped you would subtract the equations algebraicly. That way you would still have the "m" in there.
How would I subtract those algebraically?

##F_f=\mu_smg\cos (\theta)-\mu_kmg\cos (\theta)##

If I use the same mass for each equation wouldn't I get the same result as above?
 
  • #44
Specter said:
How would I subtract those algebraically?

##F_f=\mu_smg\cos (\theta)-\mu_kmg\cos (\theta)##

If I use the same mass for each equation wouldn't I get the same result as above?
You have a common factor of ##mg \cos \theta##. Factor that out.

Edit. Missed this on the first pass. If you are going subtract equations, you need to subtract equations.

You subtracted the right hand sides but simply made up a name and put it on the left hand side. The result is meaningless since you have not identified the thing on the left hand side.

Please try again, subtracting the left hand sides and putting the result on the left hand side and subtracting the right hand sides and putting the result in the right hand side. This is algebra. We have to play by the rules of algebra, making sure that the manipulations are valid.
 
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  • #45
jbriggs444 said:
You have a common factor of ##mg \cos \theta##. Factor that out.

Edit. Missed this on the first pass. If you are going subtract equations, you need to subtract equations.

You subtracted the right hand sides but simply made up a name and put it on the left hand side. The result is meaningless since you have not identified the thing on the left hand side.

Please try again, subtracting the left hand sides and putting the result on the left hand side and subtracting the right hand sides and putting the result in the right hand side. This is algebra. We have to play by the rules of algebra, making sure that the manipulations are valid.

So when subtracting the two equations:

##\mu_smg\cos (\theta)-\mu_kmg\cos (\theta)##

Wont I just be left with ##\mu_s-\mu_k##?
 
  • #46
Specter said:
So when subtracting the two equations:

##\mu_smg\cos (\theta)-\mu_kmg\cos (\theta)##

Wont I just be left with ##\mu_s-\mu_k##?
No. You will not. You should subtract the two left hand sides and leave the difference on the left hand side. You should subtract the two right hand sides and leave the difference on the right hand side. You should leave an equal sign in the middle. You will wind up with an equation with something on the left, something on the right and an equal sign in the middle.

Let us back up to #39 where you almost had it.

To find the force of kinetic friction I would do ##F_{f_k}=\mu_kmg\cos \theta##
To find the force of static friction I would do ##F_{f_s}=\mu_smg\cos \theta##
$$F_{f_k}=\mu_kmg\cos \theta$$ $$F_{f_s}=\mu_smg\cos \theta$$ $$F_{f_k} - F_{f_s}= \mu_k mg \cos \theta - \mu_s mg \cos \theta$$ That is how you subtract two equations.
 
  • #47
jbriggs444 said:
No. You will not. You should subtract the two left hand sides and leave the difference on the left hand side. You should subtract the two right hand sides and leave the difference on the right hand side. You should leave an equal sign in the middle. You will wind up with an equation with something on the left, something on the right and an equal sign in the middle.

Let us back up to #39 where you almost had it.$$F_{f_k}=\mu_kmg\cos \theta$$ $$F_{f_s}=\mu_smg\cos \theta$$ $$F_{f_k} - F_{f_s}= \mu_k mg \cos \theta - \mu_s mg \cos \theta$$ That is how you subtract two equations.

Sorry but I am still confused.

##F_{f_k} - F_{f_s}= \mu_k mg \cos \theta - \mu_s mg \cos \theta##
##F_{f_k} - F_{f_s}=mg\cos \theta(\mu_k-\mu_s)##
##F_{f_k} - F_{f_s}=\mu_k-\mu_s##

If I factor ##mg\cos \theta## out of the right hand side, I am left with just ##\mu_k-\mu_s## and honestly, I am not sure how I would subtract the left side. Wouldn't I need to know what ##\mu mg\cos \theta## equals in order to do that?
 
  • #48
Specter said:
##F_{f_k} - F_{f_s}=mg\cos \theta(\mu_k-\mu_s)##
##F_{f_k} - F_{f_s}=\mu_k-\mu_s##
That is an invalid algebraic manipulation. You are not allowed to simply remove terms from an expression. This is algebra. You have to play by the rules.

By "factor out", I mean to take the common factor and use the distributive law to put it to one side. You had already done that in the first equation above. Removing it entirely in the second equation was erroneous. The second equation does not follow from the first.

Now then, you also asked:

I am not sure how I would subtract the left side.
That has already been done. The result is above. It is ##F_{f_k} - F_{f_s}##. No need to go further.

Wouldn't I need to know what ##\mu mg\cos \theta## equals in order to do that?
As it turns out, you do know what ##\mu mg \cos \theta## equals. Every one of the terms in that formula are known. You know ##\mu_s## and ##\mu_f##. You know ##\theta## and, therefore, ##\cos \theta##. You know ##m##. You know ##g##.
 
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  • #49
Factoring out does not mean throwing away. It just means rewriting a sum of terms as a product of a simpler sum and a common factor shared by the two original terms
Specter said:
.
##F_{f_k} - F_{f_s}= \mu_k mg \cos \theta - \mu_s mg \cos \theta##
##F_{f_k} - F_{f_s}=mg\cos \theta(\mu_k-\mu_s)##
That's the factoring out done.
 
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