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

[kuruman]

Now that OP has announced that he "gets it", I think it is time for me to note that this question is not very well designed. Presumably it tests the extent to which the solver has mastered static friction and FBDs. However, its presentation as a multiple choice question leaves something to be desired because the entire process of finding the minimum force through an FBD, an understanding of $f_s^{\text{max}}$ and optimization, can be bypassed by a "solver" who knows nothing about these subjects but can interpret mathematical equations.

When $\mu>1$, 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.

 

[haruspex]

You should be able to figure out the answer to this. What does force and torque balance on the block tell you? For simplicity, consider a homogeneous block with mass $m$ with a vertical string being pulled up attached to one of its sides. Will the block start tilting at the first minimal application of force? If not, how hard can you pull before it does?

Depending on the block dimensions, it might not be possible to move the block with minimum force so applied without its tilting.
Just as we are free to choose the angle of the force, we can choose where to a
apply it. By exerting it along a line passing through the point at floor level directly below the mass centre we can avert any risk of tilting.

Not that tilting is necessarily a disadvantage. If it turns out that it can allow a yet lower force magnitude then the problem is not well posed.

Specifying a rectangular block of given dimensions and a string attached to the base (say) at one end could be the basis of a more advanced question.

 

[jbriggs444]

Specifying a rectangular block of given dimensions and a string attached to the base (say) at one end could be the basis of a more advanced question.

Similar to a scenario that you'd put some work into just recently as I recall.

https://www.physicsforums.com/threa...entrically-loaded-block.1013337/#post-6611355

 

[rudransh verma]

That is fine. Just as long as you are aware that until you reach and have learned those chapters, you will not really be able to tell when things tip over etc.

If we think about it when we rather apply force at an angle we do decrease the force of friction but we now are applying some force against gravity. So are we really minimising any applied force?
Is friction playing bigger role than gravity?

 

[Orodruin]

If we think about it when we rather apply force at an angle we do decrease the force of friction but we now are applying some force against gravity. So are we really minimising any applied force?
Is friction playing bigger role than gravity?

The entire point of this problem is to figure this out and to figure out which angle of application gives you the minimal force requirement.

 

[rudransh verma]

The entire point of this problem is to figure this out and to figure out which angle of application gives you the minimal force requirement.

Right! So we make some eqns and find $dF/d\theta$ ie rate of change of force with respect to angle change and then take the derivative equal to zero to get the minima/maxima force.
So, $dF/d\theta=0$, $\mu =\tan \theta$
But how do you decide if its maxima or minima force at $\mu =\tan \theta$?

 

[jbriggs444]

Right! So we make some eqns and find $dF/d\theta$ ie rate of change of force with respect to angle change and then take the derivative equal to zero to get the minima/maxima force.
So, $dF/d\theta=0$, $\mu =\tan \theta$
But how do you decide if its maxima or minima force at $\mu =\tan \theta$?

Do not just show your answers. Show your work. That would make it easier to suggest the second derivative test.

 

[rudransh verma]

That would make it easier to suggest the second derivative test.

$$F\sin \theta +F_N-mg=0$$
$$F\cos \theta- \mu mg=0$$
So,$$F=\frac{\mu mg}{\cos \theta+\mu \sin \theta}$$
Finding max/min force, $$dF/d\theta=0$$
After solving,
$$\mu=\tan \theta$$

 

[jbriggs444]

$$F\sin \theta +F_N-mg=0$$
$$F\cos \theta- \mu mg=0$$
So,$$F=\frac{\mu mg}{\cos \theta+\mu \sin \theta}$$
Finding max/min force, $$dF/d\theta=0$$
After solving,
$$\mu=\tan \theta$$

You have not shown your work taking the derivative of F and solving for the zero(es) of the resulting function.

 

[rudransh verma]

You have not shown your work taking the derivative of F and solving for the zero(es) of the resulting function.

$$dF/d\theta=d/d\theta(\frac{\mu mg}{\cos \theta+\mu \sin \theta})$$
$$=-\frac{\mu mg}{(\cos \theta+\mu \sin \theta)^2}d/d\theta (\cos \theta+\mu \sin \theta)$$
$$=-\frac{\mu mg}{(\cos \theta+\mu \sin \theta)^2}(-\sin \theta+\mu \cos \theta)=0$$
$$=\mu mg\sin \theta-\mu ^2 mg\cos \theta=0$$
$$=\sin \theta-\mu \cos \theta=0$$
$$\mu=\tan \theta$$

 

[Orodruin]

$$F\sin \theta +F_N-mg=0$$
$$F\cos \theta- \mu mg=0$$
So,$$F=\frac{\mu mg}{\cos \theta+\mu \sin \theta}$$
Finding max/min force, $$dF/d\theta=0$$
After solving,
$$\mu=\tan \theta$$

Your second equation seems to have assumed that the frictional force is $\mu mg$, but this is no longer the case if the normal force is not $mg$. It would imply $F = \mu mg/\sin\theta$, but this is not what you obtain. Please show your intermediate steps.

 

[rudransh verma]

Your second equation seems to have assumed that the frictional force is μmg, but this is no longer the case if the normal force is not mg. It would imply F=μmg/sin⁡θ, but this is not what you obtain. Please show your intermediate steps.

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

 

[Orodruin]

In order to find whether it is a max or a min you need to take the second derivative of $F$ and find if it is positive or negative for the value of $\theta$ that you found. However, since the dependence on $\theta$ is in the denominator, it is probably easier to consider $f = 1/F$ instead. Since $1/x$ is a monotonously decreasing function (for $x > 0$), $f$ and $F$ will share their extreme points. Where $F$ has a max, $f$ has a min and vice versa.

 

[rudransh verma]

In order to find whether it is a max or a min you need to take the second derivative of $F$ and find if it is positive or negative for the value of $\theta$ that you found. However, since the dependence on $\theta$ is in the denominator, it is probably easier to consider $f = 1/F$ instead. Since $1/x$ is a monotonously decreasing function (for $x > 0$), $f$ and $F$ will share their extreme points. Where $F$ has a max, $f$ has a min and vice versa.

First I started taking the usual case $d^2F/d\theta^2=-mg\sin \theta(\cos^2 \theta(3\cos ^2 \theta-2)-\sin^2\theta(1+cos^2 \theta))$.
But its difficult to tell its -ve or +ve.
So as you said $df/d\theta=(1/\mu mg)(-\sin \theta+\mu \cos \theta)$
and second derivative is $-1/mg\sin \theta$ after taking $\mu =\tan \theta$
So its -ve. That means what?
Can you explain the maths in more detail here?

 

[Orodruin]

First I started taking the usual case $d^2F/d\theta^2=-mg\sin \theta(\cos^2 \theta(3\cos ^2 \theta-2)-\sin^2\theta(1+cos^2 \theta))$

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.

 

[Mister T]

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 $mg$.

 

[Mister T]

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.

 

[Mister T]

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) > (some expression you have obtained)