Newtonian physics word problem static friction

[late347]

Homework Statement

truck moves at 22,5 m/s

On board the truck, behind, is a box.

the static friction coefficient for the box, is 0,250 with regard to the friction between truck floor and the box floor.

calculate the shortest possible stopping distance, such that the box does not slide at all, on the floor of the truck.

Homework Equations

The Attempt at a Solution

The problem looks very difficult to do with kinematics equation because we lack the necessary known values.

My reasoning process was a little bit clouded because of this. It seemed like a tough problem to crack without many known values beforehand.

When car decelerates in the first place, then all the objects tend to keep going forward. Like... if you are driving around in the night. Then a moose comes to the road and you need to avoid collision. You have a dangerous situation on the road, and you emergency brake with full power at the brakes. Then, your body goes forward, and the seat belt activates, and stops your body from hitting the windshield and/or the steering wheel.The same thing would happen to the box possibly, indeed...
v0= 22,5m/s
t0= 0 seconds
v1=0 m/s
t1= unknown
s= unknown

In the beginning scenario 0

the box is at constant speed, as well as the car is at the constant speed. The sum of forces for the box seems to be 0. The box is in balanced state, and box's relative speed with regard to the car floor is 0.
Box feels gravity and normal force of the floor. Also it could be reasoned that because the box is staying still, in a moving car. Then the friction is enough to keep it there.

We know from experience that in the beginning scenario 0
Friction force= frictioncoefficient * normal force
Fn= G
G=mg

Ffriction = m_boxg *0,25

for our purposes, g= 10m/s^2

Ffriction = m_box* 2,5 m/s^2In the deceleration scenario 1

negative acceleration must be applied to the car. Force of breaking, causes negative acceleration for the car.

I suppose now the box would possibly fly toward the front of the car (mass keeps going forward.)
I suppose the Ffriction force, must resist this box's motion forwards? This seems to be the explanation based upon experience in real life and Newton's ideas...

(I'm little bit confused about static friction here, clarification could be useful)

I suppose with second thought. I had in my mind that the speed of the box, with regard to earth, and the speed of the car with regard to earth, must be equal speeds.If the box moves at greater ground speed, than the car does. Then the box slides forward on the floor. Therefore the relative speed of box with regard to the car should be zero.

I suppose that the overall acceleration (or deceleration) ought to be equal for the car and the box. It should be same value.

But then again forces themselves cause accelerations of any objects... Force causes acceleration.
Probably the box would require less force, to displace it by any amount. Then again boxes also tend to have hefty friction between the floor in many cases...

Probably the car, being the heavier object, usually, would require greater force, to decelerate.

I could use some clarification for my reasoning and problem solving process... I guess.

Yet I cannot escape the idea, that of course. The mass of the box, ought to keep moving forward, at the event of the deceleration (similar to the seat belt example). So, the friction must be enough to resist the forward velocity.

If there had been no friction in the beginning 0 scenario. Then to my mind it would be as though the box stands still with respect to the ground. And the car moves forward at 22,5m/s, from underneath the box... So the box would have slipped off the car. And the car would have gone forward without the box. But these are only my silly reflections about the subject.

In a deceleration scenario without enough friction. The box would have slipped forward toward the driver's cabin of the truck. Which would be dangerous indeed.

I tried to draw a freebody diagram but I don't know whether to focus on the forces of the box, or the forces of the car+ box system. Or something like that. Drawing a picture simply makes it more confusing to me, it seems.

If somebody could clarify the situation I would be happy.deceleration a= delta v/ delta t

2.5m/s^2 = (0- 22,5 m/s) / (t₁- 0 seconds)

t₁= 9 seconds duration for the deceleration to 0 m/s

distance = average speed * time

s= [(22,5-0) / 2] * 9
s=101,25m

 

[BvU]

The problem looks very difficult to do with kinematics equation because we lack the necessary known values

So you work with symbols, manipulate the relevant equations (where are they in your extremely lengthy story ?) and see if you can end up with something that only has knowns on the righthand side.

I could use some clarification for my reasoning and problem solving process... I guess

You definitely want to learn to summarize your line of thought in the same way you managed to summarize the problem statement .

I tried to draw a freebody diagram but I don't know whether to focus on the forces of the box, or the forces of the car+ box system. Or something like that. Drawing a picture simply makes it more confusing to me, it seems.

Still that would be a concise and effective way through this exercise. Since the question is about the box, I would start with an fbd of the box.

 

[Simon Bridge]

Try this for reasoning:
To stop something, you need to apply a force - which is $F=ma$.
You stop something in the shortest distance by making F as big as it can be. The acceleration is, therefore: $a=F/m$

In this case, the force comes from friction ... so what is the biggest force the friction can be?
Now you know the biggest the acceleration can be.

You are given the initial speed ... you know the final speed and the acceleration... so what is the distance taken to stop?

If you are unsure of which free body diagram to draw - draw all of them and see if one stands out.
After a bit of this practise you will get good at figuring out which ones to focus on.

 

[late347]

Try this for reasoning:
To stop something, you need to apply a force - which is $F=ma$.
You stop something in the shortest distance by making F as big as it can be. The acceleration is, therefore: $a=F/m$

In this case, the force comes from friction ... so what is the biggest force the friction can be?
Now you know the biggest the acceleration can be.

You are given the initial speed ... you know the final speed and the acceleration... so what is the distance taken to stop?

If you are unsure of which free body diagram to draw - draw all of them and see if one stands out.
After a bit of this practise you will get good at figuring out which ones to focus on.

It feels weird to think that static friction force has opposing direction towards something...

how could static friction cause force? What is the role of static friction when the box is upon the truck and the truck is moving at constant velocity? I suppose there must be friction, which causes the box to stay upon the truck in the first place when the truck moves forward.

What direction is static friction of the box, if the movement of the truck goes from left --> right.
I guess static friction resists the motion.

force of friction goes <---
this would be friction of truck floor at the box's bottom surface. (a force that affects the box)
I still think that the accelerations of the box, and acceleration of the car, must be equal value. If this is not so, then I think the box would move, with respect to the truck's floor. The goal of the problem was that the box should stay still with respect to the truck's floor.

I have difficulty with imagining the size of force for static friction though.

force = ma

But in reality the truck would have much greater mass than the mass of the box.
I suspect the braking forces (braking system of the car) must be quite high in Newtons.

If the static friction force is greater than a pulling force. The object does not accelerate(?), or does it?
For static friction to be overcome. The pulling force would have to momentarily overcome the static friction as far as I know

 

[BvU]

What is the role of static friction when the box is upon the truck and the truck is moving at constant velocity? I suppose there must be friction, which causes the box to stay upon the truck in the first place when the truck moves forward.

None. No role, there is no force needed to maintain a constant speed linear motion (Newton 1 !) Vertically, gravity keeps the box in place, offset by normal force.

What direction is static friction of the box, if the movement of the truck goes from left --> right.
I guess static friction resists the motion.

Uniform motion $\Rightarrow$ no net force. If there are no other horizontal forces, the friction force is zero. And yes, friction resists motion. That's how the box stays in place wrt the truck if the driver brakes gently. Your task is to find out how gently.

I still think that the accelerations of the box, and acceleration of the car, must be equal value. If this is not so, then I think the box would move, with respect to the truck's floor.

Correct.

I have difficulty with imagining the size of force for static friction though

You only have to calculate it .

the static friction coefficient for the box, is 0,250 with regard to the friction between truck floor and the box floor

So you need a relevant equation that has to do with this given. Let me help you:$$F_{\rm friction,\ max} = f\;N$$ where N is the normal force. (notice a difference with your sentence ?)

Draw the ... FBD !

Since the exercise asks for the shortest distance, you may assume this friction force is indeed at maximum value for the calculation you are going to do.

Post #4 is still very lengthy, but already a lot better than post #1 !

But in reality the truck would have much greater mass than the mass of the box.
I suspect the braking forces (braking system of the car) must be quite high in Newtons

True. And irrelevant. Why ?

If the static friction force is greater than a pulling force. The object does not accelerate(?), or does it?
For static friction to be overcome. The pulling force would have to momentarily overcome the static friction as far as I know

There is no pulling force. At first the box moves through the countryside at 22.5 m/s. At the end of the story it is at rest. So it decelerates. What force makes that possible ? You have one guess only.

 

[Simon Bridge]

It feels weird to think that static friction force has opposing direction towards something...

how could static friction cause force?

You have experienced how static friction can be a force ... you experience it every day as you walk: what is the force between your feet and the ground but static friction? If you remove friction, maybe you attempt to walk on ice, what happens?
Notice that you don't need to know how the force of friction comes about in order to see that it is there and use it in calculations.

What is the role of static friction when the box is upon the truck and the truck is moving at constant velocity?

In uniform motion there is zero net force. This is a consequence of Newton'd 1st Law.
If this is a real truck, though, there are many forces around - like bumps in the road and air resistance. Without friction, the box would likely slip off or around the truck during normal transport. If you have ever ridden in the back of a truck you know you don't need to hang on all the time.

I suppose there must be friction, which causes the box to stay upon the truck in the first place when the truck moves forward.

When the truck first starts moving forward, this is the case. At that stage the truck has to accelerate to speed ... if it accelerates too fast, then the box slides to the back of the truck: you have probably seen this happen. A moderate acceleration, though, and even an unsecured load will remaining place: you have probably also seen this.

Newtonian physics is mainly about stuff that you can see for yourself without much special equipment.

What direction is static friction of the box, if the movement of the truck goes from left --> right.
I guess static friction resists the motion.

force of friction goes <---
this would be friction of truck floor at the box's bottom surface. (a force that affects the box)

Here you should be careful: just look at the forces on the box.
There is only one: and it has to act in the direction of the acceleration.

I still think that the accelerations of the box, and acceleration of the car, must be equal value. If this is not so, then I think the box would move, with respect to the truck's floor. The goal of the problem was that the box should stay still with respect to the truck's floor.

This is correct - friction opposes the relative velocity between the contact surfaces.
The thing is, you don't need to know this to do the problem: you only need to know the relationship between force and acceleration.

I have difficulty with imagining the size of force for static friction though.

You have an equation which tells you how to work out the maximum size of the static friction force. It uses the coefficient of friction you are provided with. Check your notes - or look online.

force = ma

But in reality the truck would have much greater mass than the mass of the box.
I suspect the braking forces (braking system of the car) must be quite high in Newtons.

This is correct: fortunately you are not asked for the breaking force on the car - but on the maximum acceleration for the car, which you have already figured must be the same as the maximum acceleration of the box.
If F is the friction on the box, and a is the acceleration of the box, then m is the mass of ... (complete the sentence).

If the static friction force is greater than a pulling force. The object does not accelerate(?), or does it?

You need to check your notes about how static friction works ... it is a "reaction force".
It is always equal and opposite to the force trying to move the thing - until some maximum value is reached.
When you sit on a chair, the chair provides a force that exactly opposes your weight... unless, you are too heavy.
Similarly, static friction is equal and opposite to any force until the force is too big, then the object suddenly starts moving.
Again, this is something you have likely experienced.

For static friction to be overcome. The pulling force would have to momentarily overcome the static friction as far as I know

That's the idea - with the tweak that it is the maximum possible static friction that is to be overcome.
Fortunately, this is not very important to the problem - sure the truck bed exerts a force (via friction) on the box, but you do not need this to solve the problem.

Have a look at:
http://hyperphysics.phy-astr.gsu.edu/hbase/frict.html

Have you drawn a free body diagram for the box yet?

 

[late347]

I was discussing the problem witha good friend of mine today.

Im not sure that I completely understood the relevance conceptually regarding static friction force

I guess that the box must be decelerated at some force. It's impossible to stop the box, without applying a net force in the opposite direction of the initial velocity. If the net force of the box were zero, then the box would keep on moving forward.

Box and the truck were traveling at the same constant velocity originally. Box must stop at v1= 0 If the accelerations had been differerent between each other... then the box would have slided across the floor.

I'm not sure that I completely got the FBD correct but here's a picture at least... I am not sure what I'm supposed to see.
I suppose the forces that act on that particular body, are those forces that are summed up, to form the net force for that body.

I was looking at the book answer,

Book answer was such that.

(net force box, at horizontal axis ) = ma
F= ma

we know that
(Ffriction _box) = 0,25 * normalforce_box

normalforce _box= - G_box
normalforce_box = -mg
g= 10m/s^2

finally we achieve equation... Where does the left side of the equation come from algebraically speaking? maybe I'm dumb or something

ma = 0,25 * m * (-10m/s^2) ]] divide both by m

a= -2,5 m/s^2

If the acceleration of the truck and acceleration of box are same, then distance could be calculated now. Also the time taken, to decelerate, could be calculated

a= delta V/ delta t
delta t * a= delta V ]] divide both by a
delta t = deltaV/ a

=
t₁ -0 = (0- 22,5) / (-2,5)

t₁= 9 seconds

s= V₀t + 0,5* (-2,5)*(9)^2
s= 22,5*9 + (-101,25)

s=101,25meters

 

[Chestermiller]

Static friction works like this:

If you have a box on a horizontal flat surface and the box is not sliding horizontally relative to the surface, the friction force F is not equal to the coefficient of static friction times the normal force. It is less than the coefficient of static friction times the normal force. Only if the box is on the verge of sliding is the friction force equal to the normal force times the coefficient of static friction. So $$|F|\leq\mu_sN$$ with the equal sign applying when the box is on the verge of sliding.

In your problem, the only horizontal force acting on the box is the frictional force F (directed in the negative x direction). So the horizontal force balance is $$ma=-|F|$$where a is the acceleration in the positive x direction. Now, combining this with the inequality above, what do you get if the box is not sliding?

 

[Simon Bridge]

I'm not sure that I completely got the FBD correct but here's a picture at least... I am not sure what I'm supposed to see.

Your second diagram should be just the truck by itself: a fbd has only one mass in it.
The next step is to write $\sum F = ma$ for each diagram... also see post #8

Im not sure that I completely understood the relevance conceptually regarding static friction force

The friction force appears in your free body diagrams as one of the forces acting on the box (and also the truck). That is all there is to it.

Where does the left side of the equation come from algebraically speaking?

All the equations come from your first free body diagram and Newton's second law of motion.

 

[late347]

Static friction works like this:

If you have a box on a horizontal flat surface and the box is not sliding horizontally relative to the surface, the friction force F is not equal to the coefficient of static friction times the normal force. It is less than the coefficient of static friction times the normal force. Only if the box is on the verge of sliding is the friction force equal to the normal force times the coefficient of static friction. So $$|F|\leq\mu_sN$$ with the equal sign applying when the box is on the verge of sliding.

In your problem, the only horizontal force acting on the box is the frictional force F (directed in the negative x direction). So the horizontal force balance is $$ma=-|F|$$where a is the acceleration in the positive x direction. Now, combining this with the inequality above, what do you get if the box is not sliding?

ma = - F (times both sides by (-1)
-ma = F

-ma ≤ μ mg (assuming mass is nonnegative, and non-zero indeed, we divide both sides by m, the sign does not change)

-a ≤ μg (times both sides by (-1)
a ≥ -μg

a≥ -(0,25x10m/s^2

a≥ -2,5 m/s^2

I think that the most confusing things in this problem for me were three things

1. what does the truck do and what is the truck's net force, then? The problem stated that both the truck and the box essentually must come to a full stop.

2. does it even matter what the truck's force is, other than the simple assumption from Newton's laws that. The truck must be stopped also with respect to unmoving observer. The box must also be stopped with respect to unmoving observer. Therefore the truck also should have some net force, in the opposite direction of the the initial velocity vector.
The truck cannot stop, unless it has that net force, which would stop its initial velocity vo= 22,5 m/s

3.Is there enough information to ascertain what the truck's net force would be? Sucks that the masses of the truck and box were not provided...

 

[Chestermiller]

ma = - F (times both sides by (-1)
-ma = F

-ma ≤ μ mg (assuming mass is nonnegative, and non-zero indeed, we divide both sides by m, the sign does not change)

-a ≤ μg (times both sides by (-1)
a ≥ -μg

a≥ -(0,25x10m/s^2

a≥ -2,5 m/s^2

This result is correct. And, if the magnitude of the acceleration is high enough for the box to just be on the verge of slipping, then the equal sign applies.

I think that the most confusing things in this problem for me were three things

1. what does the truck do and what is the truck's net force, then? The problem stated that both the truck and the box essentually must come to a full stop.

To do this problem, you don't need to know anything about the truck other than that the truck and box move as a rigid body (without the box slipping), and the acceleration of the box is the same as the acceleration of the truck.

2. does it even matter what the truck's force is,

No.

other than the simple assumption from Newton's laws that. The truck must be stopped also with respect to unmoving observer. The box must also be stopped with respect to unmoving observer. Therefore the truck also should have some net force, in the opposite direction of the the initial velocity vector.
The truck cannot stop, unless it has that net force, which would stop its initial velocity vo= 22,5 m/s

This is all correct, but is not necessary to solve the problem as stated.

3.Is there enough information to ascertain what the truck's net force would be? Sucks that the masses of the truck and box were not provided...

If you knew the mass of the truck, you could determine the braking force from the ground and tires necessary to bring about the same acceleration you determined for the box. If the mass of the truck were M, then the force would be -2.5(M+m). Hopefully, this would not be large enough to exceed the coefficient of static friction between the ground and the tires, so that the truck would skid.

 

[BvU]

Dear Late,

Clear to me you now master this exercise. Do me (and self) a favour:
Write a concise (complete and minimal) path through the solution, as compact as, for example, the problem statement

A truck moves at 22.5 m/s. On the floor of the truck is a box. The static friction coefficient for the box is 0.25 with regard to the truck floor. Calculate the shortest possible stopping distance, such that the box does not slide at all.

Include the fbd for the box.

 

[late347]

Problem
Box is on board a truck, and the initial velocity of the truck v0= 22,5 m/s
truck v1= 0
box v1 = 0
box v0 = 22,5m/s relative to unmoving observer

calculate the shortest possible stopping distance such that a box onboard a truck, does not move with respect to the truck's floor during the braking event.
And the box has friction coefficient "mu" (μ)= 0,25

It is known that the truck and the box must decelerate as a rigid body such that both the box and truck decelerate at the same value.
It is known that both the truck and box must come to a full stop relative to an unmoving observer, in such a rigid body- fashion as above.

If it were not so... If it was such that the box had different acceleration value... then the box would have slided with respect to the truck's floor something which was forbidden.

If the truck would have decelerated too rapidly compard to the box... Then the result would have been that the truck accelerates away from underneath the box. Something which was also forbidden. This was the only reason I could come up with in Newtonian terms.It took me some time to realize this, even though it seems obvious to many people perhaps. For example, in a car crash, the car decelerates at such great force... the car would accelerate away from underneath the box, the latter of which would keep its forward initial velocity during car crashing event. The box would be slowed down by sliding friction force. But this scenario was forbidden, like already it was said. Forbidden because the box cannot slide, and must be fixed into place by the static friction.
Furthermore it should be said that the also the truck must become affected by a net force, in the same direction as the box's friction force. The truck's acceleration would become the same acceleraation as the box. However it seems there was insufficient information given for the truck and this force needs not be calculated even so... But it's good to note that the truck must have a net force in same direction as the box's static friction. The truck cannot stop without a force, which causes deceleration to the truck.

The net force for the forces affecting the box are in the vertical axis = 0
Those forces in Y-axis are normal force and gravity.

It seems indeed that initially before the braking event The balance of forces in the horizontal X-axis = 0. This is reasonable I suppose. If the balance of forces upon the box is zero, then the box is either at rest or moving at constant velocity. It seems that because the box is indeed traveling at constant velocity, therefore the balance of forces must be zero in the initial scenario before braking event.
In any event the net force affecting the box during braking event is the static friction force. For any object to come to a rest... That object must have been at rest in the first place or some force must be applied to that object, such that the force causes deceleration (negative acceleration) which slows the object to velocity of zero.

net force = mass x acceleration

the net force's magnitude for the box must be the value of the static friction force.
F_μ = μ * Fnormal
Where Fnormal is the normal force of that object.

We evidently create an equation, from which we solve the acceleration of the box itself. This acceleration ought to be the maximum acceleration value for the box ( this acceleration must also be used for the truck later).
This acceleration value ought to be the maximum allowed threshold for deceleration value. Car could be decelerate also at slower rate theoretically, but then the stopping distance would be greater as well... Distance was supposed to be the shortest possible stopping distance from speed, to full-stop at 0 m/s.
My only confusion at this late stage is from a conceptual point of view. How can it be argued that static friction forces the box's initial velocity to zero?
The whole bone of contention in this problem is that in actual fact in reality... what is happening to the box's velocity in the initial state of motion. The initial state of motion of the box is that the box has v0= 22,5 m/s with respect to earth. That is so because the box and truck move as a rigid body at the same constant velocity initially.

I guess I had difficulty with the idea that the so-called "static friction", is happening aboard moving platform (the truck itself moves relative to earth) and evidently this static friction would have slowed the box's ground speed to zero. It is sensible to think that sliding friction causes the truck itself to decelerate with respect to ground. The truck is connected directly to the ground after all, by the wheels and asphalt. I suppose because the box is not connected to the ground directly... the only realistic way of slowing down the box is to use the static friction to do this purpose. In some other scenarios sliding friction could also be possibly used, but it was not allowd in this problem.

To do this, we use the known definitions of Fnormal for the box. Normal force is equal and opposite in direction to the box's gravity. Box's gravity is known as
G=mg

F=ma
ma= FF_μ = μ * Fnormal

F_μ = μ * (-mg)

F=F
μ * (-gm) = ma

-gμ = a
mu and g, can be known, and plugged in at this point
mu equals 0,25
g equal -10 m/s^2

a= (0,25) (-10 m/s^2)

a= -2,5 m/s^2Stopping distance of the truck can be calculated as well, at this point using kinematics.
v1= v0 + at
v1-vo= at
(v1-v0)/ a= t

(0-22,5) / (-2,5) = 9 seconds

or alternatively
a= (ΔV) /(Δt)
a= (v1-v0) /( t1-t0)
a* (t1-t0) = v1-v0)
t1-t0= (v1-v0)/ a

t1- 0s = (0m/s - 22,5m/s) / (-2,5m/s^2)
t1= 9ss= Vaverage * t

Vavereage = (0+ 22,5) / 2= 11,25 m/s

s= 11,25 m/s * 9seconds
s= 101, 25 meters

round off to the three meaningful digits

ca. 101 meters.

Sometimes the confusing bit in these problems is to decide, which force is negative? Whether or not the acceleration equals negative acceleration, or positive deceleration. I think it's more common sense to talk about positivevalue of deceleration probably.
It is difficult to decide how to draw the force vectors into the force diagram for the box. Which direction is the Force plus, or minus? I think the requirement is simply that directly opposing forces are such that they must be negative number of the other force. For example Gravity and Fnormal are opposing in direction, and they should be equal magnitude in terms of their absolute value. (in those cases where the supportgin platform providing the normal force, is able to withstand the strain of supporting the object's weight)

Box's deceleration = 2,5m/s^2
stopping distance for car = 101m
time taken until stop (braking time) = 9s

 

[BvU]

Longest post so far , even without the fbd ...
Never mind. One last advice on signs and directions: Direction of vectors is established when you choose a coordinate system, so once you draw the 22.5 m/s. If that (usually x⁺) is to the right, then the only force in that direction on the box, the friction force, must be to the left in order to obtain a negative acceleration.

 

[late347]

Longest post so far , even without the fbd ...
Never mind. One last advice on signs and directions: Direction of vectors is established when you choose a coordinate system, so once you draw the 22.5 m/s. If that (usually x⁺) is to the right, then the only force in that direction on the box, the friction force, must be to the left in order to obtain a negative acceleration.

The post attempted to describe how to to go about the path to solving the problem.

So you cannot fault me for being thorough, at least.

I think typing it out that way certainly was useful to myself. Because it helped me to see how the problem-solving steps connected themselves to the specific word problem requirements.

Perhaps there were some extra bits there, but all-in-all I think it was a well-reasoned complete answer. A thorough path why the result should be obtained that way and how it was to be done.

This was one of the tougher practice problems in my book, at least for me.
My textbook didnt cover friction forces much. (And sadly I had forgotten good bit of that basic physics)

I was studying physics for the purpose of a collage entrance exam. Later I will know whether or not my physics study will have borne fruit in that sense. The exam is now over. But I definitely think that these types of heavily conceptual problems will increase my abilities as problemsolver. At least I hope that this deficiency of mine can be improved upon. (complicated word problems are tricky).

I was mostly attending lectures and self-studying math and physics problems and doing homework problems. I feel like I did quite well with my homework in terms of succcess rate. Yet I probably should have prepared more with exacting exam conditions (under time pressure etc)

The exam itself had multiple choice problems (often word problems) and otherwise tricky word problems.

 

[late347]

This is what I had come up with But I don't know if that's the proper way to do it.