What is the minimum force needed to move a block with friction present?

[Herman Trivilino]

Are you saying applying a force at an angle can decrease the friction without actually lifting it. That there is some $0<F_N<mg$

You can easily verify this for yourself. Place a bathroom scale on the floor next to a table. Stand on the scale and it reads $mg$. Place your hand on the table top and push down. The reading on the scale will now be less than $mg$.

 

[Herman Trivilino]

When μ>1, the first 3 choices are clearly greater than the weight mg. In that case, one can "make the body move" by pulling straight up with a force that is equal to the weight and less than any of the first 3 choices. Thus, none of these can be a minimum. This leaves the fourth choice as the answer.

Do you really think that a student who doesn't understand static friction and FBD's could carry out a line of reasoning like this? It's possible, but in my experience quite rare.

 

[Helios047]

F= m(a+g*Mus)/(cos(theta)-Mus*sin(theta))
You will have to find acceleration a from 1D motion with constant acceleration. Or if you're computer literate then simple numerical computation will do the trick. As to answering the question, the value of F has to be larger than the expression above.

 

[Helios047]

F= m(a+g*Mus)/(cos(theta)-Mus*sin(theta))
You will have to find acceleration a from 1D motion with constant acceleration. Or if you're computer literate then simple numerical computation will do the trick. As to answering the question, the value of F has to be larger than the expression above.

Edit: you can find a from a= ((v^2)-(x-xnot)/2)^1/2 expression.

 

[jbriggs444]

F= m(a+g*Mus)/(cos(theta)-Mus*sin(theta))

Let me type set that for you. It hurts my eyes otherwise.

$F=\frac{m(a+\mu_s g)}{\cos \theta - \mu_s \sin \theta}$

However, it still smells like gibberish to me.

My best guess at your formula for a is $a=\sqrt{v^2 - \frac{x - \hat{x}}{2}}$

But what is $a$ in a problem that involves no acceleration? And what is $x$ or $\hat{x}$?

You will have to find acceleration a from 1D motion with constant acceleration. Or if you're computer literate then simple numerical computation will do the trick. As to answering the question, the value of F has to be larger than the expression above.

 

[haruspex]

Sorry, I did a blunder. Second eqn should be $F\cos \theta-\mu F_N=0$. This should produce the correct eqn.

I assume you then got the equation $F=\frac{mg}{\sin(\theta)+\mu\cos(\theta)}$.
When you see an expression like $G=A\sin(\theta)+B\cos(\theta)$, theta being unknown, there is a trick that can be very useful. Let $\tan(\alpha)=B/A$. So $G/A=\sin(\theta)+\tan(\alpha)\cos(\theta)$
$(G/A)\cos(\alpha)=\sin(\theta)\cos(\alpha)+\sin(\alpha)\cos(\theta)=\sin(\theta+\alpha)$.
Notice that there is now only one occurrence of theta in the equation for G. It should be obvious what values of theta minimise or maximise G.
See if you can apply that here to avoid doing any calculus.

 

[Helios047]

Let me type set that for you. It hurts my eyes otherwise.

$F=\frac{m(a+\mu_s g)}{\cos \theta - \mu_s \sin \theta}$

However, it still smells like gibberish to me.

My best guess at your formula for a is $a=\sqrt{v^2 - \frac{x - \hat{x}}{2}}$

But what is $a$ in a problem that involves no acceleration? And what is $x$ or $\hat{x}$?

Sorry I was using the formula for velocity from my memory. The actual acceleration formula is (v^2-v0^2)/2*(x-x0)=a

X0 is initial distance and v0 is initial velocity. With proper assumptions one can work out the math. And we need acceleration to overcome static friction. If this was dynamic friction, then my good Sir, your assumption would be correct.

 

[jbriggs444]

Sorry I was using the formula for velocity from my memory. The actual acceleration formula is (v^2-v0^2)/2*(x-x0)=a

X0 is initial distance and v0 is initial velocity. With proper assumptions one can work out the math

Sorry I was using the formula for velocity from my memory. The actual acceleration formula is (v^2-v0^2)/2*(x-x0)=a

X0 is initial distance and v0 is initial velocity. With proper assumptions one can work out the math. And we need acceleration to overcome static friction. If this was dynamic friction, then my good Sir, your assumption would be correct.

This seems to seriously miss the point of the exercise. Any motion will do to break static friction. Any non-zero acceleration, no matter how small, will suffice. The limit is clearly when $a=0$. That is the limit that one must solve for. By Newton's second law, $\sum F = ma$. With $a=0$ this occurs when $\sum F = 0$. And we want static friction to be at its limit, so $F_s = \mu F_n$. Easy peazy.

 

[rudransh verma]

This seems to seriously miss the point of the exercise. Any motion will do to break static friction. Any non-zero acceleration, no matter how small, will suffice. The limit is clearly when $a=0$. That is the limit that one must solve for. By Newton's second law, $\sum F = ma$. With $a=0$ this occurs when $\sum F = 0$. And we want static friction to be at its limit, so $F_s = \mu F_n$. Easy peazy.

This seems to be a confusion in me too. If we take a=0 we are calculating net force that will not make the body move. And we want to calculate minimum force required to move the block. There would not be even a slightest non noticeable acceleration.
And I said the same thing in previous post as you and yet you were skeptical. Why?

 

[Delta2]

This seems to be a confusion in me too. If we take a=0 we are calculating net force that will not make the body move. And we want to calculate minimum force required to move the block. There would not be even a slightest non noticeable acceleration.
And I said the same thing in previous post as you and yet you were skeptical. Why?

This is like a problem in math where the minimum doesn't exist , but the infimum exists (the minimum acceleration or force doesn't exist for this problem but the infimum exists and it is zero(for acceleration)). Tell me if you are interested to hear more.

 

[rudransh verma]

Tell me if you are interested to hear more.

Proceed!

 

[Delta2]

I ll have to start with the difference between minimum and infimum (or maximum and supremum). Both are the smaller elements but with the difference that the minimum has to exist within the set of reference, while the infimum can be outside of the set. The infimum in any bounded subset of reals exists but the minimum might not exist
Example 1: The set {1,2,3,4,5,6} minimum and infimum are both 1 (and maximum and supremum 6).
Example 2:The set of all positive reals. This set has infimum 0 but the minimum isn't 0 because 0 isn't positive so it doesn't belong to set. Also for any positive real we can find a smaller, so that's why minimum doesn't exist for this set.

Coming back to this physics problem. It asks for the minimum force to move the block, while I *THINK* (not sure) it should ask for the infimum force. The minimum force to move the block (here our set is the set of all forces that can move the block) doesn't exist in my opinion because for any force $F>\frac{mg}{\sqrt{1+\mu^2}}=\epsilon+\frac{mg}{\sqrt{1+\mu^2}}$ we can find a smaller one we just have to take $\epsilon>0$ smaller. The infimum force is $\frac{mg}{\sqrt{1+\mu^2}}$ but this force doesn't belong in our set because it can not move the block.

 

[rudransh verma]

The minimum force to move the block (here our set is the set of all forces that can move the block) doesn't exist in my opinion because for any force F>mg1+μ2 there exists ϵ+mg1+μ2 we can find a smaller one we just have to take ϵ>0 smaller.

Ok. You mean 'There exists'. For real set of all the forces that can move the block there is no minimum force.

That is the limit that one must solve for.

So we must solve for the limit where static friction is at max and it is at $F=\frac{\mu mg}{\sqrt{1+\mu ^2}}$. Any force greater than this will do the job.

 

[Delta2]

Ok. You mean 'There exists'. For real set of all the forces that can move the block there is no minimum force.

Yes there is no minimum force because any force greater than the limit $\frac{\mu mg}{\sqrt{1+\mu^2}}$can move the block, but the limit itself cannot move the block.

 

[Herman Trivilino]

Yes there is no minimum force because any force greater than the limit $\frac{\mu mg}{\sqrt{1+\mu^2}}$can move the block, but the limit itself cannot move the block.

I disagree. If a force of $\frac{\mu mg}{\sqrt{1+\mu^2}}$ is applied the net force is zero. The object could be at rest or it could be moving at a steady speed.

 

[Delta2]

I disagree. If a force of $\frac{\mu mg}{\sqrt{1+\mu^2}}$ is applied the net force is zero. The object could be at rest or it could be moving at a steady speed.

case 1 if the object is at rest , then this force is not enough to move the block (as it will remain at rest)so case 1 dismissed.
case 2 if the object is moving at constant speed (after the net applied forces becomes zero) then it must be moving (with constant or not constant speed) before the net applied force becomes zero. So i feel a conflict with the statement of the problem, that says "force to move the block". I mean if it is already moving, what's the point to apply a force, to move it further with constant speed? I don't think so. I think the problem statement implies that the body has zero speed before we apply the force.

 

[haruspex]

case 1 if the object is at rest , then this force is not enough to move the block (as it will remain at rest)so case 1 dismissed.

In the real world, if a system is static then we can deduce the forces exactly balance, but we cannot argue it the other way.
When we write that the limit of static frictional force is $\mu_sF_N$ we mean that as a threshold: if the other forces tending to make the bodies slide against each other add up to less than this then they won’t slip, and if they add up to more they will. If we attempt to apply a force of exactly that we cannot tell what will happen.

 

[Helios047]

This seems to seriously miss the point of the exercise. Any motion will do to break static friction. Any non-zero acceleration, no matter how small, will suffice. The limit is clearly when $a=0$. That is the limit that one must solve for. By Newton's second law, $\sum F = ma$. With $a=0$ this occurs when $\sum F = 0$. And we want static friction to be at its limit, so $F_s = \mu F_n$. Easy peazy.

This is the same thing I typed and tried to explain. I included the formula for acceleration if you want to get more accurate results. How you interpret it, is up to the person of interest.

 

[haruspex]

I included the formula for acceleration if you want to get more accurate results.

It does not provide any more accurate result for the question posed in post #1. Any acceleration greater than zero is not the result of the minimum necessary force.
Your post was effectively off-topic.

 

[Tom.G]

Another way of answering the original problem would be:

(force to move the block) >[/size] (some expression you have obtained)

 

[Delta2]

In the real world, if a system is static then we can deduce the forces exactly balance, but we cannot argue it the other way.
When we write that the limit of static frictional force is $\mu_sF_N$ we mean that as a threshold: if the other forces tending to make the bodies slide against each other add up to less than this then they won’t slip, and if they add up to more they will. If we attempt to apply a force of exactly that we cannot tell what will happen.

I got no clue what's your point here. My point is that the calculated minimum force does not move the block if the block is initially at rest. If initially (before we apply the force) it is moving then what's the point of the problem, minimum force to move a block that is already moving?

 

[haruspex]

My point is that the calculated minimum force does not move the block if the block is initially at rest.

My point is that you cannot say that. You can only say that anything less will not move the block, and anything more will move the block. What would happen if it were exactly equal, a practical impossibility, is indeterminate.

 

[Delta2]

What would happen if it were exactly equal, a practical impossibility, is indeterminate.

I don't know if to be exactly equal is practical impossible, but the laws of mainstream physics say that the block will remain at rest in that case.

I would just rather say that for physicists minimum and infimum are practically the same thing, as we are used in the world of physics to do many mathematical innovations e hehe.

 

[rudransh verma]

I disagree. If a force of $\frac{\mu mg}{\sqrt{1+\mu^2}}$ is applied the net force is zero. The object could be at rest or it could be moving at a steady speed.

The question clearly states that $\mu$ is static friction coefficient. So the block should be initially at rest.

 

[rudransh verma]

Yes, that quickly becomes messy, which is why I suggested checking whether the denominator had a minimum or maximum since the numerator is constant. Those are tricks that one picks up with experience. Now the reason it works is the following, consider two (non-zero) functions $F(x)$ and $f(x) = 1/F(x)$. We then have ($x$ being some arbitrary variable the functions depend on)
$$
f' = \frac{df}{dx} = \frac{d(1/F)}{dx} = -\frac {F'}{F^2}
$$
and so $f'(x_0) = 0$ if $F'(x_0) = 0$. It then also follows that
$$
f'' = -\frac{d}{dx}\frac{F'}{F^2} = - \frac{F''}{F^2} + 2\frac{F'^2}{F^3}.
$$
If we evaluate this for the extreme point $x_0$, then $F'(x_0) = 0$ and therefore
$$
f''(x_0) = -\frac{F''(x_0)}{F(x_0)^2}.
$$
Hence, $f''(x_0)$ is positive if $F''(x_0)$ is negative and vice versa. So if $x_0$ is a max of $f$, it is a min of $F$, etc.

Finalizing,
$$F\cos \theta-\mu_s F_N=0$$
$$F\sin \theta+F_N-mg=0$$
After solving, $$F=\frac{\mu_s mg}{\cos \theta+\mu_s\sin \theta}$$
Taking $dF/d\theta=0$
$$\mu_s=\tan \theta$$
Now taking $$f=1/F=\frac{\cos \theta+\mu_s \sin \theta}{\mu_s mg}$$
$$d^2f/d\theta^2=-\frac1{\sin \theta mg}$$
So f is max at $\mu_s=\tan \theta$ and F is min.

 

[Orodruin]

Finalizing,
$$F\cos \theta-\mu_s F_N=0$$
$$F\sin \theta+F_N-mg=0$$
After solving, $$F=\frac{\mu_s mg}{\cos \theta+\mu_s\sin \theta}$$
Taking $dF/d\theta=0$
$$\mu_s=\tan \theta$$
Now taking $$f=1/F=\frac{\cos \theta+\mu_s \sin \theta}{\mu_s mg}$$
$$d^2f/d\theta^2=-\frac1{\sin \theta mg}$$
So f is max at $\mu_s=\tan \theta$ and F is min.

But what is the minimal value of F? This was the original question. You have shown which angle it occurs at and that it is a min, but not quoted the actual answer.

 

[Herman Trivilino]

If initially (before we apply the force) it is moving then what's the point of the problem, minimum force to move a block that is already moving?

To test to see whether or not the student understands that zero net force is required to keep an object moving.

But, if the object were already moving we'd be dealing with the coefficient of kinetic friction, not the coefficient of static friction.

 

[haruspex]

I don't know if to be exactly equal is practical impossible,

There are always vibrations. Go down to the quantum level if necessary. Exactitude of continuous phenomena is a mathematical idealisation.

the laws of mainstream physics say that the block will remain at rest in that case.

Do they? As I understand them, they say only that $\mu_sN$ is the limiting magnitude of the frictional force. Some sources might imply it will only slide if that limit is exceeded, but I can see how it would be easy to fall into implying that without proper consideration. I can also see how it is easy to gain that impression even if it is not implied in a text.

 

[jbriggs444]

Do they? As I understand them, they say only that $\mu_sN$ is the limiting magnitude of the frictional force.

Indeed, the engineering approximation that we know as "static friction" becomes less well defined as one looks more closely into the details.

A problem with static friction is that it may be conceptually ill-defined. First, Fs is not single-valued even if the materials in contact, the load, and a potentially present lubricant are well specified. Instead static friction is known to depend on the age of the contact (the increase is logarithmic in time over a broad range of contact ages) and the rate with which the shear stress is increased. Second, static friction may not even be static. Transient creep-like motion, difficult to detect at the macroscopic scale, can take place before the rapid slip event (1). To probe the fundamental laws of static friction, one therefore needs to study extremely small sliding velocities vs. Going down to vs slightly <1 μm/s for a paper-on-paper system, Baumberger and coworkers (2, 3) showed that creep occurs in those systems during the stick phase, although the lateral forces were well below Fs. In this issue of PNAS, Yang, Zhang, and Marder (4) push the envelope even more slowly and manage to resolve sliding velocities down to 10−5 μm/s. Their analysis of this experimental data in terms of a rate and state model for friction suggests that slip precedes static friction and furthermore confirms the expectation that creep takes place at shear forces much below the static friction.

 

[Orodruin]

Let’s ask ourselves if this discussion really helps the OP or if it is just an aside that becomes yet another smoke screen of details not relevant for the OP’s understanding of the problem ...

 

[haruspex]

Let’s ask ourselves if this discussion really helps the OP or if it is just an aside that becomes yet another smoke screen of details not relevant for the OP’s understanding of the problem ...

It does relate to posts #59 and #63 by the OP, but I agree it should be moved to a private discussion until we have consensus.

 

[Orodruin]

It does relate to posts #59 and #63 by the OP, but I agree it should be moved to a private discussion until we have consensus.

I believe that for the purposes of the OP’s level and current knowledge it is sufficient to state that it does not really matter because if it doesn’t move at that force, it will be the infimum of the moving forces. Regardless, applying an exact force is not really physically possible so the question is quite moot either way.

 

[Dadface]

Do you really think that a student who doesn't understand static friction and FBD's could carry out a line of reasoning like this? It's possible, but in my experience quite rare.

If a question of this type was set in a UK multiple choice exam the examiners would expect the students to come up with an answer in an average time of just over one minute. In other words they would expect the students to reach the answer without going through a time consuming analysis of the type being carried out in this thread. For this particular question the answer is quickly arrived at by a process of elimination as pointed out by Kuruman.

 

[Herman Trivilino]

If a question of this type was set in a UK multiple choice exam the examiners would expect the students to come up with an answer in an average time of just over one minute.

And the examiners would also expect the students to have an understanding of static friction and FBD's.

 

[Dadface]

And the examiners would also expect the students to have an understanding of static friction and FBD's.

True, but although it can be more interesting and educationally beneficial to carry out a proper analysis the majority of students who go along that route for problems of this type would probably run out of time and end up looking at the question again or taking a guess.

 

[Herman Trivilino]

True, but although it can be more interesting and educationally beneficial to carry out a proper analysis the majority of students who go along that route for problems of this type would probably run out of time and end up looking at the question again or taking a guess.

You seem to be missing my point. Here is the logic @kuruman demonstrated in Post #36 that solves the problem quickly and cleverly:

When μ>1, the first 3 choices are clearly greater than the weight mg. In that case, one can "make the body move" by pulling straight up with a force that is equal to the weight and less than any of the first 3 choices. Thus, none of these can be a minimum. This leaves the fourth choice as the answer.

My point is that in order for a student to carry out that analysis, the student would need to posses a working knowledge of static friction and FBD's.

 

[Dadface]

You seem to be missing my point. Here is the logic @kuruman demonstrated in Post #36 that solves the problem quickly and cleverly:

My point is that in order for a student to carry out that analysis, the student would need to posses a working knowledge of static friction and FBD's.

Those topics would be on the syllabus of any exam board that sets the question so, ideally, students should have a good working knowledge of them. But look at answers 1,2 and 3 given in post 15. If μ is bigger than one, each of the three answers would give a minimum force bigger than mg, in other words bigger than the minimum force needed to lift the block. So each of those answers can be discounted.

 

[Herman Trivilino]

But look at answers 1,2 and 3 given in post 15. If μ is bigger than one, each of the three answers would give a minimum force bigger than mg, in other words bigger than the minimum force needed to lift the block. So each of those answers can be discounted.

And to be able to come up with that line of reasoning a student would have to understand static friction and FBD's. In my opinion. Of course you can disagree with my point, but I don't see anything you've written so far that even attempts to do that.

 

[Dadface]

And to be able to come up with that line of reasoning a student would have to understand static friction and FBD's. In my opinion. Of course you can disagree with my point, but I don't see anything you've written so far that even attempts to do that.

Using the shortcut method requires a knowledge that the minimum force needed to just lift the mass is equal to mg. Students could illustrate that with a force diagram but I can't see any advantage in doing so.

The line of reasoning involves looking at the four equations given and seeing that when μ has a value greater than one the force calculated from each of the first three equations would be greater than mg. A knowledge of what μ is and stands for is not required.

Most students given this as an exam question would be familiar with free body diagrams and static friction and I would bet that on the day of the exam a majority would jump in using a diagram of the type shown in the question in order to try and work out the answer. It can be a fun and interesting thing to try but in a multiple choice exam can take up too much valuable time.

 

[Herman Trivilino]

Using the shortcut method requires a knowledge that the minimum force needed to just lift the mass is equal to mg. Students could illustrate that with a force diagram but I can't see any advantage in doing so.

Doesn't matter. Prior to taking the test the prepared students would be familiar with FBD's. This one is so simple they could picture it in their head without drawing it. And even if they did draw it, it would take only a few seconds.

The line of reasoning involves looking at the four equations given and seeing that when μ has a value greater than one the force calculated from each of the first three equations would be greater than mg. A knowledge of what μ is and stands for is not required.

But they would need to know something about $\mu$ to understand the significance of it being greater than one.

Most students given this as an exam question would be familiar with free body diagrams and static friction and I would bet that on the day of the exam a majority would jump in using a diagram of the type shown in the question in order to try and work out the answer.

Yes. and the clever ones would take the shortcut. But only if they had previously built up a knowledge of static friction and FBD's.

You have still not addressed the pedagogical issue of how a student could reason through that shortcut without first having built up a knowledge of static friction and FBD's.

 

[Dadface]

Doesn't matter. Prior to taking the test the prepared students would be familiar with FBD's. This one is so simple they could picture it in their head without drawing it. And even if they did draw it, it would take only a few seconds.

We seem agree on this point. If its understood that the minimum force cannot be bigger than mg no other FBD is needed.

But they would need to know something about $\mu$ to understand the significance of it being greater than one.

Even if there were students who didn't know about μ they could still come to the right response by looking at the four given equations. By doing so they should realize that μ is a dimensionless number and that the correct answer should apply whatever its value.

Yes. and the clever ones would take the shortcut. But only if they had previously built up a knowledge of static friction and FBD's.

You have still not addressed the pedagogical issue of how a student could reason through that shortcut without first having built up a knowledge of static friction and FBD's.

You don't need a knowledge of FBDs and static friction. You do need to know that the minimum force cannot be bigger than mg and that μ is an unspecified number.

it might seem to be a strange sort of question in that it's not really testing students knowledge of FBDs and friction but it is a multiple choice question and there would be a limited average available time for its solution.

 

[Dadface]

Mister T, I don't think we would be having this conversation if the wording of the question was more specific.

I suspect that a lot of peoples initial reaction on reading the question is that it requires one to find the minimum force needed to move the block along the surface. If that were the case students would have to apply their knowledge of FBDs and static friction.

But the question doesn't ask for the minimum force to move the block along the surface, it asks for the minimum force needed to move the block, in other words move it along, or move it up or tilt it or move it in some other way. Of course the minimum force needed to lift the block completely from the surface is mg.

 

[Dadface]

I would have liked to edit the above post by re writing the first sentence but was unable to do so. I guess there is a time limit for editing. The following is what I would have written.

Mr T, I think we wouldn't be having this conversation if the wording of the question was more specific.