Can You Solve This Motion Problem Using a Simple Trick?

[rudransh verma]

Homework Statement

An open knife edge of mass 200g is dropped from height 5 m on a cardboard. If the knife penetrates distance 2m into the cardboard, the average resistance offered by the cardboard to the knife edge is—

Relevant Equations

$v^2=u^2+2as$
$F=ma$

$v^2=u^2+2as$
$v=10m/s$
Again $v^2=u^2+2as$
$a=25 m/s^2$

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

 

[Doc Al]

As the knife penetrates the cardboard, what forces act on it?

 

[bob012345]

Homework Statement:: An open knife edge of mass 200g is dropped from height 5 m on a cardboard. If the knife penetrates distance 2m into the cardboard, the average resistance offered by the cardboard to the knife edge is—
Relevant Equations:: $v^2=u^2+2as$
$F=ma$

$v^2=u^2+2as$
$v=10m/s$
Again $v^2=u^2+2as$
$a=25 m/s^2$

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

A knife penetrating $2m$ into the cardboard? Who writes these problems? That seems completely unphysical unless it were moving unreasonable fast. I don't think it would go that far even into butter.

 

[Doc Al]

A knife penetrating $2m$ into the cardboard? Who writes these problems? That seems completely unphysical unless it were moving unreasonable fast.

I know! Some of these problems were just slapped together without much thought behind them.

 

[rudransh verma]

As the knife penetrates the cardboard, what forces act on it?

Gravity

 

[Doc Al]

Gravity

That's one of the forces acting on the knife. And it's the one you neglected in your solution.

 

[Orodruin]

This 2-meter long "knife" is twice as long as the medieval longsword which was a two-handed weapon with typical lengths of about 1 meter. Also, a blade of that length and mass made of iron would have a thickness of about 0.6 mm, hardly thick enough to penetrate cardboard without bending considering where its CM is at first contact. I, too, object to unphysical problems because they send the wrong message about reasonable sizes of numbers.

Nothing says that the knife is 2 meters long. I read the problem statement assuming it is of negligible length and plunges into the cardboard. Still quite unphysical though.

Homework Statement:: An open knife edge of mass 200g is dropped from height 5 m on a cardboard. If the knife penetrates distance 2m into the cardboard, the average resistance offered by the cardboard to the knife edge is—
Relevant Equations:: $v^2=u^2+2as$
$F=ma$

$v^2=u^2+2as$
$v=10m/s$
Again $v^2=u^2+2as$
$a=25 m/s^2$

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

Apart from what has already been said, there is absolutely no need to use the kinematic equations here. In fact, it only serves to confuse yourself.

 

[Doc Al]

Apart from what has already been said, there is absolutely no need to use the kinematic equations here. In fact, it only serves to confuse yourself.

Agreed. But since the OP started down that path, he may as well do it correctly.

Then we can suggest a much simpler approach to the problem. (For example, energy methods.)

 

[jbriggs444]

I read the problem (somewhat creatively) as speaking of a partially open folding knife. The blade has only folded out about 45 degrees or so instead of the full 180 degrees it would take to fully deploy. We imagine the blade locked in this position. Or having rusted so stiff that it does not open further.

A workman holding this knife on the second floor of a building under construction drops it while trying to open it. On the ground below is a piece of cardboard of unspecified thickness. It is positioned upright and held stationary with its top edge 5 meters below the workman's waist where he was fiddling with the knife. For sake of argument, the cardboard was protecting a 3 meter by 5 meter un-framed print of the Mona Lisa which is intended to hang on the wall of the CEO's office. The CEO is both pretentious and cheap.

The knife falls and arrives at the cardboard with the body of the knife on one side and the blade on the other and the top edge of the cardboard positioned nicely at the apex of the 45 degree angle that the two make with each other.

The knife slices cleanly through the cardboard, slowing as it proceeds further downward. It comes to rest 2 meters below the top of the cardboard.

 

[kuruman]

Nothing says that the knife is 2 meters long. I read the problem statement assuming it is of negligible length and plunges into the cardboard. Still quite unphysical though.

You are quite right. Post deleted.

 

[haruspex]

The other fault in the question is more serious in my book. It asks for the average force, i.e. $\frac{\Delta p}{\Delta t}$, but expects this to be calculated on the basis of energy or, equivalently, $v^2-u^2=2as$. That only gives the right answer if the force is constant. So it should instead say ”find the force, assuming it is constant ".

 

[bob012345]

The other fault in the question is more serious in my book. It asks for the average force, i.e. $\frac{\Delta p}{\Delta t}$, but expects this to be calculated on the basis of energy or, equivalently, $v^2-u^2=2as$. That only gives the right answer if the force is constant. So it should instead say ”find the force, assuming it is constant ".

I don't see that. The actual acceleration is probably not constant but one can calculate the average acceleration and use it to calculate an average force. The Work -Energy Theorem does not require a constant force over distance.

 

[jbriggs444]

I don't see that. The actual acceleration is probably not constant but one can calculate the average acceleration and use it to calculate an average force. The Work -Energy Theorem does not require a constant force over distance.

@haruspex is correct on this. It is one of his pet concerns.

Suppose that we want the average force weighted by time. We certainly can calculate the velocity with which the falling knife arrives at the cardboard. Which means that we have momentum. We could divide by elapsed time until the knife stops to arrive at the average (over time) force.

But we do not know how much time it will take. The knife could slow suddenly and then sag slowly through the cardboard with an average force very nearly equal to the weight of the knife. Or the knife could zip through the cardboard and stop suddenly at the bottom.

Yes, you are correct that if by "average force" we mean a distance-weighted average then all is well. But the default meaning of average force is usually taken to be a time-weighted average.

 

[haruspex]

I don't see that. The actual acceleration is probably not constant but one can calculate the average acceleration and use it to calculate an average force. The Work -Energy Theorem does not require a constant force over distance.

Further to @jbriggs444 explanation, it seems reasonable to require that average force = mass x average acceleration. This makes (change in momentum) / (elapsed time) the correct formula.
It is an interesting exercise to calculate the average force during a quarter cycle of SHM by the two methods. Indeed, in many instances of questions committing this solecism SHM is more likely than constant force.

 

[bob012345]

Further to @jbriggs444 explanation, it seems reasonable to require that average force = mass x average acceleration. This makes (change in momentum) / (elapsed time) the correct formula.
It is an interesting exercise to calculate the average force during a quarter cycle of SHM by the two methods. Indeed, in many instances of questions committing this solecism SHM is more likely than constant force.

I used the two methods and arrived at the same number but when you say it seems reasonable do you mean it's really not? I don't want to be more specific since the OP hasn't had a chance to finish the problem.

 

[bob012345]

Yes, you are correct that if by "average force" we mean a distance-weighted average then all is well. But the default meaning of average force is usually taken to be a time-weighted average.

Is this what you mean by a distance-weighted average force $\int_{x_i}^{x_f} F \,dx =\bar{F} Δx$?

 

[rudransh verma]

That's one of the forces acting on the knife. And it's the one you neglected in your solution.

But I have used the -g in first eqn where I got the value of v.

 

[haruspex]

But I have used the -g in first eqn where I got the value of v.

Did gravity switch off when the knife reached the cardboard?

 

[haruspex]

Is this what you mean by a distance-weighted average force $\int_{x_i}^{x_f} F \,dx =\bar{F} Δx$?

Yes.

 

[rudransh verma]

Did gravity switch off when the knife reached the cardboard?

But shouldn’t the eqn give total acceleration? Shouldn’t the eqn give correct acceleration?

 

[haruspex]

But shouldn’t the eqn give total acceleration? Shouldn’t the eqn give correct acceleration?

Yes, but only if you include all the forces acting at the time.

 

[Doc Al]

But I have used the -g in first eqn where I got the value of v.

Sure, but (as @haruspex points out) you fail to consider gravity once the knife enters the cardboard.

But shouldn’t the eqn give total acceleration? Shouldn’t the eqn give correct acceleration?

There's nothing wrong with your acceleration calculation. But when solving for the force that produces it, you must consider all forces.

 

[Steve4Physics]

@rudransh verma, what two forces act on the knife as it moves through the cardboard?

How does the value ‘$F=5N$’ (which you worked out in Post #1) relate to these two forces?

 

[jbriggs444]

Is this what you mean by a distance-weighted average force $\int_{x_i}^{x_f} F \,dx =\bar{F} Δx$?

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

 

[rudransh verma]

Yes, but only if you include all the forces acting at the time.

I am talking about third eqn of motion. There is no force in it. Are you talking about second law of Newton?

 

[rudransh verma]

what two forces act on the knife as it moves through the cardboard?

Gravity at all times and friction through the cardboard.

 

[Steve4Physics]

Gravity at all times and friction through the cardboard.

Yes. But it is not just simple friction. It is the resistive force that is referred to as "average resistance" in the original question. Gravity (weight) and the resistive force are the two forces.

So I repeat my earlier question:
How does the value ‘’$F=5N$" (which you worked out in Post #1) relate to these two forces?

 

[rudransh verma]

I got it @Steve4Physics If you ignore friction Then there is only one force. Due to this gravity it accelerates. And due to this acceleration there is a generation of another force. So there are actually two forces. Weight and the force due to acceleration.
So F Net= 5N + W =5+2= 7 N.

 

[jbriggs444]

I got it @Steve4Physics If you ignore friction Then there is only one force. Due to this gravity it accelerates. And due to this acceleration there is a generation of another force. So there are actually two forces. Weight and the force due to acceleration.
So F Net= 5N + W =5+2= 7 N.

You seem to be confusing yourself thoroughly. Certainly you are confusing me.

The 5 N you calculated in the first post is the net force. So the equation you write is $$F_\text{net}= 5\text{N}$$Now that $F_\text{net}$ is the sum of two individual forces: Gravity and Friction. So an equation we can write is:$$F_\text{net} = F_g + F_f$$The net force is 5 N, not 7 N.

 

[Steve4Physics]

Gravity at all times and friction through the cardboard.

Yes. And it is the friction force that you are being asked to find. So I repeat my earlier question:
"How does the value ‘’$F=5N$" (which you worked out in Post #1) relate to these two forces?

I got it @Steve4Physics If you ignore friction Then there is only one force. Due to this gravity it accelerates. And due to this acceleration there is a generation of another force. So there are actually two forces. Weight and the force due to acceleration.
So F Net= 5N + W =5+2= 7 N.

$F_{net}$ is not 7N.

 

[bob012345]

I got it @Steve4Physics If you ignore friction Then there is only one force. Due to this gravity it accelerates. And due to this acceleration there is a generation of another force. So there are actually two forces. Weight and the force due to acceleration.
So F Net= 5N + W =5+2= 7 N.

It would be helpful for you to sketch a force diagram to see what is happening in the cardboard. This thread is keeping me on a knife-edge.

 

[rudransh verma]

You seem to be confusing yourself thoroughly. Certainly you are confusing me.

The 5 N you calculated in the first post is the net force. So the equation you write is

Yeah I calculated that. But to correct myself that is not Fnet. That is the force generated through acceleration. So net force will be this force + weight.

 

[bob012345]

Yeah I calculated that. But to correct myself that is not Fnet. That is the force generated through acceleration. So net force will be this force + weight.

Again, please draw the force diagram.

 

[rudransh verma]

Again, please draw the force diagram.

 

[bob012345]

There are two parts to this problem, each requiring a different force diagram. The first part is the free fall for 5M. The second part is when the knife is cutting through the cardboard. Force diagrams show ALL the forces, not just the net force. It will always help you in solving problems to get a clear understanding of what is happening before trying to calculate things.