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

[rudransh verma]

Homework Statement

A body of mass m results on horizontal floor with which it has coefficient of static friction $\mu$. It is desired to move the block with minimum possible force. Find the applied force F.

Relevant Equations

$F_n=ma$
$F_s=\mu F_N$

We know the minimum force to move the body in presence of friction will be $F>\mu F_N$
But it’s not right.

 

[Doc Al]

We know the minimum force to move the body in presence of friction will be $F>\mu F_N$
But it’s not right.

Why not? (What does $F_N$ equal?)

 

[Vanadium 50]

Where is the effort here?

 

[Delta2]

I am intrigued, what is the answer key @rudransh verma . Why you say its not correct? According to my opinion the minimum force is $\mu mg$.

 

[rudransh verma]

Why not? (What does $F_N$ equal?)

I am intrigued, what is the answer key @rudransh verma . Why you say its not correct? According to my opinion the minimum force is $\mu mg$.

Yes $\mu mg$. But the answer in my book says $F=\frac{\mu mg}{\sqrt{1+\mu^2}}$

 

[Delta2]

Yes $\mu mg$. But the answer in my book says $F=\frac{\mu mg}{\sqrt{1+\mu^2}}$

Something is fishy here, can you post some sort of figure that perhaps exists in the book for this problem?

 

[Doc Al]

But the answer in my book says $F=\frac{\mu mg}{\sqrt{1+\mu^2}}$

Did you provide the complete problem statement?

 

[rudransh verma]

Something is fishy here, can you post some sort of figure that perhaps exists in the book for this problem?

 

[Doc Al]

Seriously? You only now thought it relevant to tell us that the applied force F is at an angle?

 

[rudransh verma]

Seriously? You only now thought it relevant to tell us that the applied force F is at an angle?

No. This is the complete question. There is no mentioning of any angle as you can see.I don’t know why they are applying force at an angle.

 

[Doc Al]

No. This is the complete question. There is no mentioning of any angle as you can see.I don’t know why they are applying force at an angle.

So you are saying that the diagram you just posted has nothing to do with this problem?

 

[rudransh verma]

So you are saying that the diagram you just posted has nothing to do with this problem?

I am saying they have solved the problem assuming force at an angle. That diagram is not the part of question but of the answer.

 

[Doc Al]

I am saying they have solved the problem assuming force at an angle. That diagram is not the part of question but of the answer.

That's a bit hard to believe.

(1) Post the complete problem statement
(2) Post the complete solution provided.

 

[Doc Al]

Ah, I see the issue now. You were not told anything about the direction of the applied force, so you are free to apply it in any direction to solve for the minimum force needed. Assuming that the applied force must be horizontal was a mistake.

 

[rudransh verma]

1) Post the complete problem statement

Post the complete solution provided.

 

[Doc Al]

See my last post. Do you see how being allowed to apply the force at an angle can affect the answer?

 

[Orodruin]

I don’t know why they are applying force at an angle.

Because that lowers the normal force and therefore the horizontal force required to move the block. Since the problem did not say anything about the direction, in order to find the minimal force, you need to find an expression for the required force as a function of the angle and then minimise with respect to the angle.

 

[SammyS]

Is this more readable?

 

[Delta2]

OK , well I am not sure if that "the force applied must be at an angle " should be part of the problem statement or the student would have to think it. I was not smart enough to think it, and I am deeply sorry for myself .

 

[Orodruin]

OK , well I am not sure if that "the force applied must be at an angle " should be part of the problem statement or the student would have to think it. I was not smart enough to think it, and I am deeply sorry for myself .

Everybody makes bloopers or misread problems sometimes. Obviously, stating explicitly that the force could be at an angle would make it more obvious, but technically it is also fair game with the problem just stating to find the minimal force.

 

[Nugatory]

Obviously, stating explicitly that the force could be at an angle would make it more obvious, but technically it is also fair game with the problem just stating to find the minimal force.

The problem asks specifically for the magnitude of the force. That's a very broad hint here.

 

[haruspex]

OK , well I am not sure if that "the force applied must be at an angle " should be part of the problem statement or the student would have to think it. I was not smart enough to think it, and I am deeply sorry for myself .

It can help to put yourself into the problem. If you had to tug a heavy mass across a floor, how would you do it?

 

[Delta2]

It can help to put yourself into the problem. If you had to tug a heavy mass across a floor, how would you do it?

I don't know how you think but according to my thinking and intuition I would push it with horizontal force.

 

[Orodruin]

I don't know how you think but according to my thinking and intuition I would push it with horizontal force.

So you just learned something practically useful!

In addition, pulling at an angle will also mean helping to overcome issues of bad grip on the ground on your part as you also increase your own normal force.

 

[haruspex]

I don't know how you think but according to my thinking and intuition I would push it with horizontal force.

Maybe you need to do more manual labour!

Two physicists were watching workmen on a building site. "See how he pulls the wheelbarrow behind himself instead of pushing it in front. He has obviously worked out that because he is pulling up at a slight angle it reduces the rolling resistance", observed one.
The other physicist asked the workman why he did it. "I've pushed wheelbarrows for thirty years," he replied, "and I'm fed up with the sight of 'em."

 

[rudransh verma]

Because that lowers the normal force and therefore the horizontal force required to move the block.

To apply minimum force the force of friction has to be less and will become less when we decrease the normal force and that would mean a tilted block.
But the eqns are $F_N+F\sin \theta =mg$
$F\cos \theta=\mu F_N$ which I don’t think is consistent with the tilted block.

 

[Orodruin]

and that would mean a tilted block.

No, it would not necessarily tilt the block. Why do you think so? There is a separate set of equations (based on torque balance) that would tell you whether the block tilts or not.

which I don’t think is consistent with the tilted block.

Why not? They are valid as long as the coefficient of friction remains the same.

 

[rudransh verma]

No, it would not necessarily tilt the block. Why do you think so?

If it will not leave the ground at least from one side how the friction will decrease ?

 

[Orodruin]

If it will not leave the ground at least from one side how the friction will decrease ?

By decreasing the normal force. Maximal friction is proportional to normal force.

 

[rudransh verma]

By decreasing the normal force. Maximal friction is proportional to normal force.

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

 

[Orodruin]

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 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?

 

[rudransh verma]

Will the block start tilting at the first minimal application of force? If not, how hard can you pull before it does?

No. I think if we apply some upward force block will not move because $F_N+ F_a=mg$. We need to increase the $F_a=mg$. Then block will move. So we can apply a force to decrease the normal without actually lifting it up.
$F_a$ is applied force.

 

[Orodruin]

No. I think if we apply some upward force block will not move because $F_N+ F_a=mg$. We need to increase the $F_a=mg$. Then block will move. So we can apply a force to decrease the normal without actually lifting it up.
$F_a$ is applied force.

Not be lifted up, when does it start tilting? Did you draw a free body diagram and do the torque analysis?

 

[rudransh verma]

Not be lifted up, when does it start tilting? Did you draw a free body diagram and do the torque analysis?

I have not studied about torque. Sorry. It’s in future chapters.
But I get it. Force at an angle will have its component in the direction of normal and that will decrease the normal force and thus friction. It will be enough to move the block forward.

 

[Orodruin]

I have not studied about torque. Sorry. It’s in future chapters.

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.

 

[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.