Calculating acceleration from the coefficient of kinetic friction

[jbriggs444]

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.

 

[Specter]

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?

 

[jbriggs444]

$F_N=(10 kg)(9.8m/s^2)\cos21.8$
$=91 N$

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?

 

[Specter]

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?

 

[jbriggs444]

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?

 

[Specter]

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.

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

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

$F_{f_s}-F_{f_k}$
$36.4-27.3=9.1 N$

 

[jbriggs444]

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.

 

[Specter]

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?

 

[jbriggs444]

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.

 

[Specter]

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 $\mu_s-\mu_k$?

 

[jbriggs444]

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.

 

[Specter]

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?

 

[jbriggs444]

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

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 $g$.

 

[haruspex]

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

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

That's the factoring out done.