Newton's second law -- Crate sliding in the back of an accelerating truck

[Franklie001]

Homework Statement

A truck of mass M1= 8000 kg is transporting a crate of mass M2 = 2200 kg. The air drag force acting on
the truck is 600 N and the air drag force acting on the crate is 100 N. If the driving force applied to the
front wheels of the truck is 5000N, find the acceleration of the truck and the horizontal static friction force
acting on the crate. Also find the minimum coefficient of static friction between the truck and the crate
to ensure that the crate does not slide.

Relevant Equations

Acceleration of the truck?
Horizontal static friction force acting on the crate?
Minimum coefficient of static between truck and crate?

Hi everyone is able to help solve this question for my assignment in university?
I've draw a free body diagram for each component of the question but now i am stuck.

[Mentor Note -- Poster has been reminded to show their work when starting a new schoolwork thread]

 

[BvU]

Hi,

Please show your work ! Including the diagrams.
See https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/
Where is it precisely, that you are stuck ?

$\$

 

[Steve4Physics]

Hi @Franklie001. The rules here require that you show us your work before we can offer guidance. So post your free-body diagram(s).

If I were answering this problem, I'd have three diagrams: one for the truck, one for the crate and one for whole system (truck+crate).

Edit: Drats. @BvU beat me to it!

 

[Franklie001]

 

[Franklie001]

Thanks everyone for the quick reply, appreciated your help.

As you can see from the pictures I've managed to draw the free body diagrams for each components and also tried to apply Newton second law to find the friction forces Ff1 and Ff2 but the coefficient hasn't been given.
What am i suppose to do now to find the acceleration and frictions?

 

[PeroK]

Thanks everyone for the quick reply, appreciated your help.

As you can see from the pictures I've managed to draw the free body diagrams for each components and also tried to apply Newton second law to find the friction forces Ff1 and Ff2 but the coefficient hasn't been given.
What am i suppose to do now to find the acceleration and frictions?

If the truck is moving normally, then there is no kinetic friction force acting on the wheels (there is no $F_{f1}$). That's the point of having wheels! They are very efficient and are not subject to kinetic friction, because they are rolling not sliding. There will be some rolling resistance, but as that's not mentioned, you may assume that only the drag forces are resisting the truck's acceleration.

 

[Franklie001]

If the truck is moving normally, then there is no kinetic friction force acting on the wheels (there is no $F_{f1}$). That's the point of having wheels! They are very efficient and are not subject to kinetic friction, because they are rolling not sliding. There will be some rolling resistance, but as that's not mentioned, you may assume that only the drag forces are resisting the truck's acceleration.

Hi PeroK, thanks for the help.

First of all i want so say that the friction force Ff1 is intended to indicate the force that oppose the motion of the truck from the wheels.
Even if moving normally as the ground would be in perfect condition so the coefficient would be zero and this is not the case.

 

[Franklie001]

Hi @Franklie001. The rules here require that you show us your work before we can offer guidance. So post your free-body diagram(s).

If I were answering this problem, I'd have three diagrams: one for the truck, one for the crate and one for whole system (truck+crate).

Edit: Drats. @BvU beat me to it!

Hi what do you think about this problem?

 

[Franklie001]

Hi PeroK, thanks for the help.

First of all i want so say that the friction force Ff1 is intended to indicate the force that oppose the motion of the truck from the wheels.
Even if moving normally as the ground would be in perfect condition so the coefficient would be zero and this is not the case.

If the truck is moving normally also the acceleration would be take into account

 

[PeroK]

Hi PeroK, thanks for the help.

First of all i want so say that the friction force Ff1 is intended to indicate the force that oppose the motion of the truck from the wheels.
Even if moving normally as the ground would be in perfect condition so the coefficient would be zero and this is not the case.

In that case, why can't you find the acceleration from Newton's second law?

 

[Franklie001]

I am thinking at what you said but unsure that's the right answer.
There's indeed something i m missing here

 

[PeroK]

I am thinking at what you said but unsure that's the right answer.
There's indeed something i m missing here

Okay, but it stands to reason that if you don't believe I know anything about physics, then I can't help you!

 

[Franklie001]

Okay, but it stands to reason that if you don't believe I know anything about physics, then I can't help you!

Sorry Pero i am not saying you don't know anything about physics i was just expressing my thought.

So assuming the friction force of the truck is zero. How do i find the friction force Ff2 of the crate?

 

[Franklie001]

And the acceleration of the truck?

 

[PeroK]

Sorry Pero i am not saying you don't know anything about physics i was just expressing my thought.

So assuming the friction force of the truck is zero. How do i find the friction force Ff2 of the crate?

No problem! I think you're supposed to assume that the crate does not slide.

 

[Steve4Physics]

I'll add this, since I'd already drafted it, though it repeats some of what @PeroK has already said...

There are a few mistakes. For example:

A diagram shows force $F_{f1}$. I’m not clear what this is. The driving force, $F_D$ is meant to the total frictional force between the wheels and ground. (Imagine what happens if you tried to accelerate on perfectly smooth/slippery ice.) In this problem, the wheels are rolling and we are meant to take $F_D$ as the only horizontal force between the wheels and the ground.

[On a more advanced note, we are treating the wheels as having negligible mass, so we are ignoring the torque needed to give the non-driving wheels angular acceleration. But ignore this comment if you don't understand it!]

$F_{f2}$ appears to be the force of friction from the truck’s platform, acting to the left on the crate. Fom Newton’s 3rd law, that means there is also an equal magnitude force of friction from the crate, acting to the right on the truck’s platform. But this is missing from your diagram.

Weight is a force, measured in Newtons (N). For example a 100kg mass weighs 981N (on earth). Weight is not measured in kg in a scientific context.

You have taken the +x direction to the left. Not a mistake but a bit unconventional!
_______________

One approach is to take a step back. Start with a fresh free body diagram:
Draw a simple picture of ‘truck+crate’ (no ground).
Mark all the external forces, including their values
Work out the reusltant force, F.
Apply F=ma to find the acceleration.

Then draw a free body diagram for the crate alone and take it from there.

With suitable diagrams, each answer should only be a few lines long.

 

[Delta2]

You have the attention of many good helpers and advisors, so I feel my presence here is redundant, however i just want to say this:
I love the truck design at images 1 and 2 of post #4, you sure got some talent in designing things!

 

[Franklie001]

Hi Steve4Physics,

Thank you so much for the big help.

I've managed to find the acceleration of the truck.

But i am still unsure about what you mentioned earlier about the friction force.
Do i need to include into the free body diagram of the crate an opposite force equal to the friction to satisfy Newton's 3rd law?
If that's how do i find the horizontal force of friction?

If i follow what you said, for the free body diagram of the crate, now I've got three force on the x-axis:
-The drag air force acting on the crate Fa
-The friction force acting to the right of the crate, Ff
-And the resultant force equal and opposite in magnitude of the friction force acting to the left of the crate, R

 

[Steve4Physics]

I've managed to find the acceleration of the truck.

Well done. What did you get? Note, the value is also the accleration of the crate, as the crate and truck move together.

But i am still unsure about what you mentioned earlier about the friction force.
Do i need to include into the free body diagram of the crate an opposite force equal to the friction to satisfy Newton's 3rd law?

(The friction force between crate and truck is what accelerates the crate. Imagine there was no friction and the truck accelerated. What would happen to the crate?)

The 2 forces in a Newton's 3rd law (N3L) pair, act on different bodies.

There is a friction force (magnitude $F_{friction}$ acting to the left on the crate's base.
A free body diagram (FBD) of the crate alone must include this force.

And from N3L, there is a friction force, also magnitude $F_{friction}$, acting to the right on the truck's platform. A FBD of the truck alone must include this force. Though we don't actually need the FBD here.

If that's how do i find the horizontal force of friction?

If i follow what you said, for the free body diagram of the crate, now I've got three force on the x-axis:
-The drag air force acting on the crate Fa
-The friction force acting to the right of the crate, Ff
-And the resultant force equal and opposite in magnitude of the friction force acting to the left of the crate, R

The resultant force is not a separate force.

The resultant force on the crate (say $F_{crate}$) is not a separate force; it is the name we use for the vector-sum of the actual forces on the crate.

The vertical forces on the crate cancel. There are only two horizontal forces on the crate to consider: air resistance to the right and $F_{friction}$ to the left. So can you write an equation expressing $F_{crate}$ in terms of the air resistance force and the friction force?

You know the crate's mass and acceleration. So can you now apply 'F=ma' to the crate taking the crate to be an isolated object with some resultant force acting on it? If it helps, you are actually using $F_{crate} =m_{crate} a_{crate}$.

Edit - typo's corrected.

 

[Franklie001]

Thanks for the clarification.

Solving for the acceleration I've got 0.55 m/s^2 using the Newton's second law.
So using the acceleration found I've found a force of friction Ff = 1310N
And its coefficient of friction u= 0.06.

Am i right?

 

[PeroK]

Thanks for the clarification.

Solving for the acceleration I've got 0.55 m/s^2 using the Newton's second law.

Am i right?

No. That's not right. How did you get that?

 

[Franklie001]

Applying Newton's Second Law for the truck in the x axis:
Driving force(Fd) - Air force resistance on the truck (Fa1) / mass of the truck (m1) = 0.55 m/s^2

 

[Franklie001]

which is numerically: 5000-600 / 8000 = 0.55 m/s^2

 

[PeroK]

Applying Newton's Second Law for the truck in the x axis:
Driving force(Fd) - Air force resistance on the truck (Fa1) / mass of the truck (m1) = 0.55 m/s^2

The truck has a crate on board!

 

[Franklie001]

Sorry PeroK, I am a bit confused now.
What am i supposed to do?
The question asks for the acceleration of the truck therefore i thought that applying Newton second law in the x-axis would be enough.
What am i missing?

 

[PeroK]

Sorry PeroK, I am a bit confused now.
What am i supposed to do?
The question asks for the acceleration of the truck therefore i thought that applying Newton second law in the x-axis would be enough.
What am i missing?

The mass of the crate, which the truck is carrying. The truck must be slowed by the mass of payload it is carrying?

Are there any forces you are missing?

 

[Franklie001]

Is that maybe the reaction friction force?

 

[Franklie001]

I mean the reaction friction force acting to the right of the truck's platform?

Again thank you for the patience you are having in explain this problem, it's just I've always struggled with physics problems

 

[PeroK]

Is that maybe the reaction friction force?

I'm not sure what that is. The mass of the crate is holding the truck back. If we increase the payload by adding another crate, for example, then that must affect the rate at which the truck is able to accelerate.

Also, as long as the crate doesn't slide, then it's no different from any other part of the truck and must be accelerated with the truck. That might give you a clue to a quick way to solve this problem. But, assuming we continue with your approach:

You have an unknown friction force (third law pair) between the truck and crate. That will 1) hold the truck back and b) accelerate the crate.

That gives you two Newton's second law equations: one for the truck and one for the crate, both with an unknown force.

To solve these equations for the unknown force you need a constraint, which in this case is that the acceleration of the truck and crate must be the same.

Time for some simultaneous equations. I bet you never thought want you learned in algebra would be useful!

 

[Steve4Physics]

I mean the reaction friction force acting to the right of the truck's platform?

Yes, you are missing the friction force acting to the right on the truck's platform (as described in Post #19).

But since you don't know the value of this force (yet) it won't help you!

Think of the truck+crate as a single object ‘O’. Try drawing a free body diagram of O. Note, internal forces (e.g. friction between truck and crate) are not included in the diagram - why?

You should be able to calculate the acceleration of O directly using 'F=ma'. The answer is, of course, the same as the truck's acceleration and the crate's acceleration.

 

[Franklie001]

Ok so do i need to consider the mass of the crate on the truck as whole so (mass of the truck + mass of the crate) what about the air forces (Fa1 and Fa2) do i need to add them too, or are they considered as internal forces?

 

[PeroK]

Ok so do i need to consider the mass of the crate on the truck as whole so (mass of the truck + mass of the crate) what about the air forces (Fa1 and Fa2) do i need to add them too, or are they considered as internal forces?

Here's a clue:

 

[Franklie001]

Thanks

 

[Steve4Physics]

@Franklie001, if you want to do it using simultaneous equations, here are the steps...

’1’ is the truck and ‘2’ is the crate.

1. Find the resultant force on the truck, $F_1$, the sum of all the forces on the truck.
Hey, I'll even do that one for you:
Reading-off the truck's free body diagram gives:
$F_1 =5000 - 600 - F_{friction} = 4400 - F_{friction}$.

2. Find the resultant force on the crate, $F_2$, the sum of all the forces on the crate.
You do that one yourself.

3. Apply $F_1=m_1a$ to the truck to get equation 1.
I'll do that one for you:
$4400 - F_{friction} = 8000a$ (equation 1)

4. Apply $F_2=m_2a$ to the crate to get equation 2.
You do that one yourself.

5. You now have 2 simultaneous equations with unknowns $a$ and $F_{friction}$. You solve them.

If you don't like simultaneous equations, see the suggested method in my previous posts.

 

[Franklie001]

I've used the simultaneous equations and i 've got for the acceleration ax= 0.39 m/s^2 and for the force of friction f= -880N
Is that right?