Can You Solve This Motion Problem Using a Simple Trick?

[bob012345]

I am saying when you place your hand on the surface which force do you apply. Weight! but when you throw your hand on the surface there is another force that is generated due to acceleration.

You do not apply weight. That is a very confusing way to think of it. Define another force that is generated due to acceleration. Acceleration of what? Having your own definitions and terms will get you into trouble with physics questions and it makes it difficult to help you.

 

[rudransh verma]

You do not apply weight. That is a very confusing way to think of it. Define another force that is generated due to acceleration. Acceleration of what? Having your own definitions and terms will get you into trouble with physics questions.

I mean its obvious. There is a force on the surface and that is W. IF you place your hand on surface does it break. Might not. But if you punch, then?

 

[jbriggs444]

I mean its obvious. There is a force on the surface and that is W.

As I use the term, "weight" (or "apparent weight") is the net inertial force resulting from gravity together with the choice of a particular rest frame.

My weight is a downward force on me. The force of my hand on a table is a separate force entirely. It is a contact force between my hand and the table.

When we "weigh" ourselves with a scale, we make use of Newton's second law. Since we are not accelerating, we can deduce that the supporting force from the scale (which the scale directly measures) matches the downward force from gravity (which the scale, thus, measures indirectly).

 

[bob012345]

I mean its obvious. There is a force on the surface and that is W. IF you place your hand on surface does it break. Might not. But if you punch, then?

It's obvious that if you hit something it might break but that is imprecise language to describe this physics problem.

 

[rudransh verma]

Better to simply say gravity acts on th eknife at all times while friction resists the motion of the knife as it falls through the cardboard.

Yes! Frictional force ${F_{fric}} = -(mg+force due to falling)= 7 N$.

 

[jbriggs444]

Yes! Frictional force ${F_{fric}} = -(mg+force due to falling)= 7 N$.

By "force due to falling", do you mean "total force required to explain the observed deceleration of the knife"?

 

[vela]

Forces are how pairs of objects interact. The weight is due to the interaction of the knife with the Earth. Friction is due to the interaction between the knife and cardboard. These two forces acting on the knife sum to produce a net force which results in the knife's acceleration.

In a sense, the discussion is about the language you're using, but your choice of words does seem to reveal a misunderstanding about forces. In particular, there is no force due to acceleration. Just because the knife is accelerating doesn't mean a third force is acting on the knife in addition to gravity and friction.

 

[bob012345]

Yes! Frictional force ${F_{fric}} = -(mg+force due to falling)= 7 N$.

The friction has nothing to do with falling. Calling it the force due to falling only adds confusion. It has to do with the kinetic energy the knife possesses as it hits the cardboard. That energy goes into ripping bonds as well as making heat and noise. In this case the kinetic energy comes about because the knife falls under the influence of gravity but the friction is not a force due to falling.

 

[Steve4Physics]

Yes! Frictional force ${F_{fric}} = -(mg+force due to falling)= 7 N$.

In what direction does this mysterious 'force due to falling' act?

If it is in the opposite direction to mg, your equation gives:
$F_{fric} = -(0.20\times10-5) = 3N$, not $7N$.

If it is in the same direction as mg, your equation gives:
$F_{fric} = -(0.20\times10+5) = -7N$, not $7N$.

Anyway, the expression 'force due to falling' is meaningless and highly confusing in this situation. You need to avoid it if you hope to be able to solve other problems of this sort.

There are only 2 forces acting on th knife as it cuts through the cardboard:
knife's weight (W)
resistive force of cardboard (R)
And their vector sum is $F_{net}$,

You need to work out how to answer the question using W, R and $F_{net}$.

 

[haruspex]

That is the important part, yes. But you have to divide by total distance, of course.

I think that's the $\Delta x$.

 

[haruspex]

I think that the two methods of averaging should be placed on an equal footing

As I posted, it does seem reasonable to require that (unqualified) average force = mass x (unqualified) average acceleration. Would you dispute that average acceleration means $\frac{\Delta v}{\Delta t}$?

Other problems arise when you use vectors. $\vec F_{avg}=\frac{\Delta\vec P}{\Delta t}$ works. Try doing that with energy and displacement.

 

[bob012345]

As I posted, it does seem reasonable to require that (unqualified) average force = mass x (unqualified) average acceleration. Would you dispute that average acceleration means $\frac{\Delta v}{\Delta t}$?

Other problems arise when you use vectors. $\vec F_{avg}=\frac{\Delta\vec P}{\Delta t}$ works. Try doing that with energy and displacement.

Assuming the acceleration and force is not constant but is some well behaved function over an interval, can we derive a general relationship between the two methods of averaging?

 

[haruspex]

Assuming the acceleration and force is not constant but is some well behaved function over an interval, can we derive a general relationship between the two methods of averaging?

As I mentioned in post #14, it is an interesting exercise to compare the two averages for a quarter cycle of SHM. As I recall, the ratio is $4:\pi$. I have seen this error committed in cases where SHM is a more reasonable model for the impact than is constant force.

If a body is at rest for time $t_0$ then accelerates at a for time $t_1$, the distance based average is $a$ instead of $a\frac{t_1}{t_0+t_1}$.
Swapping around to accelerating for time $t_0$ etc., it gives $a\frac{t_0}{t_0+2t_1}$ instead of $a\frac{t_0}{t_0+t_1}$.
So it can give a larger or smaller result.

 

[kuruman]

Would you dispute that average acceleration means $\frac{\Delta v}{\Delta t}$?

I would say that $\frac{\Delta v}{\Delta t}$ is indisputably time-averaged acceleration. I do not disagree with you. I have said all I have to say but, if you wish to continue this conversation, I think it should be done aside either in another thread or by PM to keep the main thread and the OP focused.

 

[rudransh verma]

In what direction does this mysterious 'force due to falling' act?

So is it wrong to say “force is also created through acceleration”?

If it is in the same direction as mg, your equation gives:
Ffric=−(0.20×10+5)=−7N, not 7N.

No! How? mg is -ve: -mg and force due to falling or acceleration is -5 not 5. So -(-2-5)=7 N (friction force).

 

[rudransh verma]

Just because the knife is accelerating doesn't mean a third force is acting on the knife in addition to gravity and friction.

I want to correct you. Force is applied by the knife on the cardboard due to its acceleration caused by its weight. This force is force due to acceleration.
All that acceleration in air creates that force.

 

[rudransh verma]

By "force due to falling", do you mean "total force required to explain the observed deceleration of the knife"?

No! Plus weight!

 

[haruspex]

This force is force due to acceleration.

That acceleration was caused by gravitational force. It now has momentum, which is different from force.
As the knife goes through the cardboard, gravitational force still acts, as well as friction from the cardboard. There is no third force caused by the acceleration it previously underwent.

 

[rudransh verma]

That acceleration was caused by gravitational force. It now has momentum, which is different from force.
As the knife goes through the cardboard, gravitational force still acts, as well as friction from the cardboard. There is no third force caused by the acceleration it previously underwent.

Are you saying while in air there is no force but momentum. As it collides with cardboard momentum is transferred not force. There are just two forces gravity and friction. Gravity in air and cardboard. Friction in cardboard.

 

[haruspex]

Are you saying while in air there is no force but momentum.

No, while falling through the air there is one force, gravity. This gives the knife momentum: momentum=force x time.
The momentum gained allows it to continue descending despite the added force of friction. But it will lose momentum because the frictional force exceeds the gravitational force, leading to a net upward force.

 

[rudransh verma]

No, while falling through the air there is one force, gravity. This gives the knife momentum: momentum=force x time.

My mistake! Yeah gravity too. Ok! So as time passes in air more and more momentum is acquired by body.( I called it force!)

The momentum gained allows it to continue descending despite the added force of friction.

So are the two forces equal at all times and because of momentum it’s going through the cardboard?

But it will lose momentum because the frictional force exceeds the gravitational force, leading to a net upward force.

I don’t get this line.
By the way momentum is next chapter after laws of motion. Why have they given this question here or can it be understood without momentum?
So moving on friction F= -(-mg-mv/t)=-(-2+-2/t)= ?

 

[haruspex]

So are the two forces equal at all times

No. While gravity was unopposed, the knife accelerated, gaining downward momentum. When it hit the cardboard, it encountered a frictional force greater than the force from gravity. The net force now being upwards, that added upward momentum, i.e. it reduced downward momentum. Eventually all the momentum is lost and the knife comes to rest.

can it be understood without momentum?

Yes, but I mentioned it because you wrote that the downward acceleration gave the knife "force". What you were intuitively grappling with was the concept of momentum, which is different from force.

friction F= -(-mg-mv/t)

You will find it less prone to error if you always write such equations in the Newtonian form, (sum of forces) = mass * acceleration, and stick to a sign convention. In this case, the motion being vertical, you should choose to make either upwards positive for all accelerations, forces, velocities.., or downwards positive for all.

 

[rudransh verma]

Yes, but I mentioned it because you wrote that the downward acceleration gave the knife "force". What you were intuitively grappling with was the concept of momentum, which is different from force.

Ok.

While gravity was unopposed, the knife accelerated, gaining downward momentum. When it hit the cardboard, it encountered a frictional force greater than the force from gravity. The net force now being upwards, that added upward momentum, i.e. it reduced downward momentum. Eventually all the momentum is lost and the knife comes to rest.

Ok. Hence the retardation. So downward momentum starts from initial point and is there till all is lost by upward momentum. Net momentum = net force=0 . Body comes to rest.

You will find it less prone to error if you always write such equations in the Newtonian form, (sum of forces) = mass * acceleration, and stick to a sign convention.

true!

 

[jbriggs444]

Ok. Hence the retardation. So downward momentum starts from initial point and is there till all is lost by upward momentum. Net momentum = net force=0 . Body comes to rest.

Be careful. Momentum and force do not have the same units. Saying that momentum is equal to force will never be correct. They are both zero. But they are incommensurable.

 

[haruspex]

Net momentum = net force=0

Since a force cannot equal a momentum, being dimensionally different, it is clearer to write that as momentum =0 (so no velocity), net force=0 (so no acceleration).
Note the parallel : just as acceleration is the rate of change of velocity, force is the rate of change of momentum.

 

[rudransh verma]

Note the parallel : just as acceleration is the rate of change of velocity, force is the rate of change of momentum.

Ok but please solve the problem with or without p.

 

[Steve4Physics]

You may have resolved these questions through what @haruspex has said. But since you asked me directly, here's my reply.

So is it wrong to say “force is also created through acceleration”?

It is wrong. Net force (if non-zero) produces acceleration. Not the other way round.

And do not confuse 'acceleration' (units m/s²) with 'momentum' (units kg.m/s).

No! How? mg is -ve: -mg and force due to falling or acceleration is -5 not 5. So -(-2-5)=7 N (friction force).

The 5N force is the net force as the knife passes through the cardboard. It acts upwards (which is why it slows down the knife) so has a positive value, +5N not -5N.

 

[haruspex]

Ok but please solve the problem with or without p.

You do not need to invoke momentum. Just correct the equations below by including gravity.

So $F=(200*25)/1000$
$F=5 N$

 

[rudransh verma]

You do not need to invoke momentum. Just correct the equations below by including gravity.

the third eqn of motion give acceleration based on the values v, u, s. The question is imagery. It’s mere algebra. But what if it were true in real life. Body falling through cardboard, the distance covered and v and u all combined should have given correct acceleration including g. It looks like while the body is traveling through cardboard, it travels a distance s and the acceleration it undergoes is in the absence of gravity.
Math wise it’s correct but physics wise it’s not, I guess!

 

[jbriggs444]

Math wise it’s correct but physics wise it’s not, I guess!

The math, the physics and the reality all match. It is your understanding of them that is astray.

the third eqn of motion give acceleration based on the values v, u, s. The question is imagery. It’s mere algebra. But what if it were true in real life. Body falling through cardboard, the distance covered and v and u all combined should have given correct acceleration including g.

If what you know is the distance covered $s$, the starting velocity $u$ and the ending velocity $v$ and if you have assumed that the acceleration is constant then you can indeed compute the required acceleration $a$.

One normally accounts for this acceleration by summing the forces and computing a total acceleration: $$F_1 + F_2 = ma$$ $$a = \frac{F_1 + F_2}{m}$$You seem to want to take the forces individually, compute accelerations individually and add the accelerations. $$a = a_1 + a_2 = \frac{F_1}{m} + \frac{F_2}{m}$$So your vision is that the knife is subject to two accelerations which combine to produce the true acceleration that is actually seen?

It looks like while the body is traveling through cardboard, it travels a distance s and the acceleration it undergoes is in the absence of gravity.

It undergoes an acceleration. That acceleration is less than it would have been in the absence of gravity.

 

[rudransh verma]

The math, the physics and the reality all match. It is your understanding of them that is astray.

It matches but after separately adding the of g to a. The eqn of motions should have given correct data. Not like adding the value of g after finding a to get total acceleration.
What is the meaning of third eqn? Third eqn says final velocity of the body is v whose initial velocity is u and it undergoes a acceleration and a displacement s. Relation is $v^2=u^2+2as$.
So it should have given us the final value of acceleration. There should have no need to add g afterwards.
In book they have done this solution which I think is not entirely correct.

 

[jbriggs444]

It matches but after separately adding the value of g to a. The eqn of motions should have given correct data. Not like adding the value of g after finding a to get total acceleration.
What is the meaning of third eqn? Third eqn says final velocity of the body is v whose initial velocity is u and it undergoes a acceleration and a displacement s. Relation is $v^2=u^2+2as$.
So it should have given us the final value of acceleration. There should have no need to add g afterwards.
In book they have done this solution which I think is not entirely correct.

You are not asked for the final value of acceleration. You are asked for the value of friction which would yield this acceleration while the knife is also subject to gravity.

The knife is subject to the force of gravity, so there is a need to model the effect of gravity.

The standard approach to the problem involves adding or subtracting forces. Not adding or subtracting accelerations. [Even though the math works either way]

 

[Steve4Physics]

it travels a distance s and the acceleration it undergoes is in the absence of gravity.

I hope @haruspex will excuse me for adding this...

Gravity always acts on the knife. The knife's weight is always -2N.

The resistance force in the cardboard is +7N.

The net force on the knife in the cardboard is therefore
$F_{net} = -2N +7N = 5N$

The knife in the cardboard behaves the same as if a single +5N force acted on it.

 

[rudransh verma]

[USER=422467]@jbriggs444[/USER] and @Steve4Physics I hope you get it in my post#81. What you are saying guys is also right.

The knife is subject to the force of gravity, so there is a need to model the effect of gravity.

Yeah! But why separately?
This was the original confusion in my OP. Now I guess it’s settled.

 

[jbriggs444]

Yeah! But why separately?

You are asked for friction. You observe the effect of friction and gravity combined. To calculate the effect of friction alone you have to somehow separate the effect of gravity from the effect of friction.

 

[haruspex]

the third eqn of motion give acceleration based on the values v, u, s. The question is imagery. It’s mere algebra. But what if it were true in real life. Body falling through cardboard, the distance covered and v and u all combined should have given correct acceleration including g. It looks like while the body is traveling through cardboard, it travels a distance s and the acceleration it undergoes is in the absence of gravity.
Math wise it’s correct but physics wise it’s not, I guess!

That has nothing to do with what I meant by:

Just correct the equations below by including gravity.

Those equations being:

So F=(200∗25)/1000
F=5N
Wrong answer!

The correction needed is to be clear and consistent about what F represents. Does it represent the net force on the knife or only the force of friction?

 

[vela]

Body falling through cardboard, the distance covered and v and u all combined should have given correct acceleration including g. It looks like while the body is traveling through cardboard, it travels a distance s and the acceleration it undergoes is in the absence of gravity.

It is kind of interesting that multiple people have told you gravity acts on the knife the entire time, but you seem to insist that gravity doesn't act on the knife once it's in the cardboard. Could you elaborate on what your thinking is here?

 

[bob012345]

It is kind of interesting that multiple people have told you gravity acts on the knife the entire time, but you seem to insist that gravity doesn't act on the knife once it's in the cardboard. Could you elaborate on what your thinking is here?

I wonder if it is how a student might interpret a problem if the book refers to an equation without $g$ explicitly in it as meaning gravity is not involved. But the OP has since seemed to understand that gravity is involved through the whole problem and this discussion is going in circles.

 

[rudransh verma]

Everyone, I am silly! My mistake . The eqn of motion has given the correct value of acceleration. That is net acceleration(25). And in the back of my head I was thinking it’s not. I cannot track everything where I went wrong but you guys understand. By the way I was not switching off the gravity. Might have by Mistake somewhere!
$$Net force= friction-mg$$
$$ma= ma’-mg$$
$$a’=a+g$$
$$a’=25+10$$
$$a’=35 m/s^2$$
So,$$friction= .2*35=7 N$$

 

[jbriggs444]

Everyone, I am silly! My mistake . The eqn of motion has given the correct value of acceleration. That is net acceleration(25). And in the back of my head I was thinking it’s not. I cannot track everything where I went wrong but you guys understand. By the way I was not switching off the gravity. Might have by Mistake somewhere!
$$Net force= friction-mg$$
$$ma= ma’-mg$$
$$a’=a+g$$
$$a’=25+10$$
$$a’=35 m/s^2$$
So,$$friction= .2*35=7 N$$

Why are you dividing by $m$ only to end up multiplying by $m$?

 

[rudransh verma]

Why are you dividing by m only to end up multiplying by m?

 

[Steve4Physics]

Knowing what you now know, what would your working be if this were a piece of homework or an examination question?

 

[rudransh verma]

Knowing what you now know, what would your working be if this were a piece of homework or an examination question?

Simple , upto a= 25 m/s^2 is right. Then just add the above part. Question done! 200 marks out of 100.

 

[kuruman]

Simple , upto a= 25 m/s^2 is right. Then just add the above part. Question done! 200 marks out of 100.

It's not that simple for me because I am having trouble sorting out the logical sequence "up to a = 25 m/s^2" in all these posts (93 and counting). Would it be possible for you to post the entire solution, from beginning to end, in one complete post? Thank you.

 

[Steve4Physics]

Simple , upto a= 25 m/s^2 is right. Then just add the above part. Question done! 200 marks out of 100.

Your working up to finding a=25m/s² did not correctly apply a stated sign-convention. Purely by chance, you ended up with the correct sign (+25m/s², assuming we are using upwards-is-positive).

What you call the 'above part' (assumed to mean your Post #89 working) shows a confused approach and will not earn all the 'method' marks.

As a (retired) physics teacher assessing a piece of homework, I award 6/10.

 

[Steve4Physics]

@rudransh verma, sorry to see the sad face. Maybe I was a bit harsh.

Do you understand why you didn't get 10/10?

To paraphrase George Santayana somewhat: if we don't learn from our mistakes, we are destined to repeat them.

 

[rudransh verma]

$$v^2=u^2+2as$$
$$v=+-10m/s$$
Taking -ve sign!
Again $$v^2=u^2+2as$$
$$a=+25m/s^2$$

So, $$Friction= F net+mg$$
$$(.2*25 + (.2)10)$$
$$Friction= +7 N$$

 

[Steve4Physics]

$$v^2=u^2+2as$$
$$v=+-10m/s$$
Taking -ve sign!
Again $$v^2=u^2+2as$$
$$a=+25m/s^2$$

So, $$Friction= F net+mg$$
$$(.2*25 + (.2)10)$$
$$Friction= +7 N$$

That’s much better. 9/10 (or 10/10 on a good day)!

But the easy/quick method (hinted at in earlier posts) is to forget about velocities, accelerations and forces. Using gravitational potential energy and work, the problem can easily be solved in about 3 lines. Can you do it that way?

 

[kuruman]

$$v^2=u^2+2as$$ $$v=+-10m/s$$ Taking -ve sign!
Again $$v^2=u^2+2as$$ $$a=+25m/s^2$$ So, $$Friction= F net+mg$$ $$(.2*25 + (.2)10)$$ $$Friction= +7 N$$

Thank you for the detailed solution. I think that you are unclear about the meaning of the symbols in your first equation. I understand that you have taken "up" as positive and "down" as negative. I also understand that in the expression $$v^2=u^2+2as$$ you have taken $v =-10$ m/s to be the velocity of the knife just before it hits the cardboard and $u=0$ is the final velocity of the knife. That is all correct and consistent. My question is, what about $s$? It should represent the displacement of the knife. Since the knife ends up below its initial position, the displacement must be negative, i.e. $s=-2~$m. If I substitute, I get $$(-10~\text{m/s})^2=0^2+2a(-2\text{m})$$ $$100~(\text{m/s})^2=-4~a~(\text{m})$$ $$a=-25~\text{m/}{\text{s}}^2.$$Something isn't right because we know that the acceleration is positive.

I am not trying to confuse you but to show you that there is still some confusion in your head about the equation $v^2=u^2+2as$, namely how you use it and what the symbols mean. Thinking of $v$ as the initial velocity and $u$ as the final velocity can lead you into trouble as it has here. A better way to write the same equation is to use subscripts to match points in space with velocities at these points: $$2a(s_2-s_1)=v_2^2-v_1^2.$$Then, if we assume that the edge of the cardboard where the knife hits is at zero height, we have $s_1=0$, $v_1=-10~\text{m/s}$, $s_2=-2~\text{m}$, $v_2=0$ so that $$2a(-2~\text{m}-0)=0^2-(-10~\text{m/s})^2$$ $$-4~a~(\text{m})=-100~(\text{m/s})^2$$ $$a=+25~\text{m/}{\text{s}}^2.$$Whether you continue using your equation or adopt the one I suggested is up to you. To quote Confucius, "If you make a mistake and do not correct it, this is called a mistake."

 

[Steve4Physics]

the displacement must be negative, i.e. $s=-2~$m. If I substitute, I get $$(-10~\text{m/s})^2=0^2+2a(-2\text{m})$$ $$100~(\text{m/s})^2=-4~a~(\text{m})$$ $$a=-25~\text{m/}{\text{s}}^2.$$Something isn't right because we know that the acceleration is positive.

I think @rudransh verma's working (in full) is
$v_f^2 = v_i^2+ 2as$ we get
$0^2 = (-10)^2 + 2a(-2)$
$4a = 100$
$a = 25m/s$ (positive as required)