# Calculating acceleration from the coefficient of kinetic friction

• Specter
In summary, the conversation discusses a scenario where a crane is raising two boxes of nails on a plank to a roof. One box is already open and half full, while the other is new and both boxes weigh 10 kg and 20 kg respectively. The coefficient of static friction is given as 0.4 and the coefficient of kinetic friction is 0.3. The conversation then goes on to discuss the calculation of the angle at which each box begins to slide and the acceleration of the boxes along the plank once they start to slide. The summary also includes a detailed explanation of the forces involved, including the weight and normal components of the boxes and the use of trigonometry to calculate these components.
Specter

## 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:

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?

## 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?

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.

#### Attachments

• BLCZgpB.png
1.7 KB · Views: 1,131
• hr6DMXh.png
3.4 KB · Views: 1,141
Last edited:
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).

Specter
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?

#### Attachments

• hv7HlsU.png
5 KB · Views: 679
Last edited:
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##.

Specter
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?

Specter
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.

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.

Specter
haruspex said:

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?

##\Sigma## is a capital "sigma". It means "sum." ##ma = \Sigma F## means "##ma## is the sum of all the forces."

Specter
RPinPA said:
##\Sigma## is a capital "sigma". It means "sum." ##ma = \Sigma F## means "##ma## is the sum of all the forces."
Thank you.

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?

Specter
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?

Yes, That is what @haruspex was trying to get you to realize.

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.

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.

Specter
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?

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.

Specter
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}

Good. Now with that in hand, can you take the next steps?

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##

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?

Specter
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.

Last edited:
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?

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.

Specter and jbriggs444
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## ?

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.

Specter
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.

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.

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?

Specter
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?

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.

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)##

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".

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?

Last edited:

• Introductory Physics Homework Help
Replies
18
Views
2K
• Introductory Physics Homework Help
Replies
9
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
859
• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
16
Views
3K
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
10
Views
4K
• Introductory Physics Homework Help
Replies
1
Views
922
• Introductory Physics Homework Help
Replies
14
Views
2K