Newton's 3rd law: motorized cart pushes a heavy cart

[TheWonderer1]

Hi there,

Just to map out my question, I am thinking about cart 1 which is a motorized cart and cart 2 -- a cart with a heavy object on it. There is an acceleration rightward.

I understand that Fc1onc2 = -Fc2onc1 which means that the forces do not cancel out as they are applied on different objects. My question pertains to the fact that I'm aware of the push of the wheels on the ground backwards and the ground applies a force on the wheels in forward direction. This allows for the acceleration forward but I'm stuck since the force the ground applies - force of cart 2 applies to cart 1 = net force on cart 1. However, isn't the push from cart 1 created by the force applied forward by the ground on the wheels? Shouldn't that mean cart 1 applies this same force on cart 2?

It's a little hard to explain my question but you could also say the same with a horse and cart, what makes the force of the ground applied to wheel/hooves different than the force the horse/cart 1 applies to cart 2?

I know that these are two different pairs of course:
1. The ground on the wheels and wheels on the ground (opposite direction)
2. cart1 on cart 2 and cart2 on cart1 (opposite direction)

 

[Dale]

It's a little hard to explain my question

I think if you draw the free body diagrams for the card then it will be clear.

 

[TheWonderer1]

Well, the only two forces on the cart 1 in the horizontal direction are the reaction force applied to it by cart 2 and the force applied by the ground. Thus if cart 1 is accelerating then there is a net force in the forward direction.

 

[russ_watters]

Well, the only two forces on the cart 1 in the horizontal direction are the reaction force applied to it by cart 2 and the force applied by the ground. Thus if cart 1 is accelerating then there is a net force in the forward direction.

I'm having trouble visualizing this. I think if you draw the free body diagram it will become clear.

 

[A.T.]

... isn't the push from cart 1 created by...Shouldn't that mean ...

Cause-effect reasoning is irrelevant and only causing confusion here.

 

[TheWonderer1]

I wanted to take some time to think a bit and decided to give more details.

It’s easiest to see the force applied by horse’s foot vs. it’s back will be different because of the control a horse has over it’s muscles. With inanimate objects like a car, it seems to me that the force applied by the bumper of a car should always be the same as the force applied by the wheels on the ground. While I understand this isn’t the case, I’m just interested in this for some reason because I think it suggests a misunderstanding of some fundamental concept.

 

[Dale]

DRAW THE FREE BODY DIAGRAMS!

 

[A.T.]

... it suggests a misunderstanding of some fundamental concept.

Yes, see post #5.

 

[CWatters]

What Dale said.

The whole point of FBD diagrams is to help you visualise and understand the forces that act on each body.

 

[osilmag]

Try summing the forces. F2=m2a2 = F1on2 - friction2.

There is also a change in momentum so you could understand it that way too.

 

[TheWonderer1]

Newton's first law: An object at rest remains at rest, or if in motion, remains in motion at a constant velocity unless acted on by a net external force.I may have found my confusion. If the action-reaction pair is an internal force within the system that is the cars. Then, the external force is the ground pushing against the cars (car 1 pushes on the ground). In the free body diagram, you will not include the action-reaction pair of the two cars and only the ground pushing against the car subtracted by whatever friction.

Fgc - Ff = net f

 

[A.T.]

Newton's first law:

Try 2nd for all bodies, and 3rd for all interactions between bodies. Write down those equations and solve them.

In the free body diagram, you will not include the action-reaction pair of the two cars

You would.

 

[Dale]

In the free body diagram, you will not include the action-reaction pair

I would. I would also take off points for any student that didn’t.

 

[jbriggs444]

If the action-reaction pair is an internal force within the system that is the cars

Then you may have drawn the system boundary in the wrong place.

 

[TheWonderer1]

I made this up and solved this problem:
I’m going to change the carts to cars so now we have car 1 and car 2.

Car 1’s mass is 1,300 Kg and car 2’s mass is 1,700 kg. There is a coefficient of friction of 0.3.

Car 1 is applying a force on the ground and the ground applies a force on car 1 equal to 10,000N. There’s friction on both cars 1 and 2. That will be equal to the normal force x coefficient of friction.
Here’s what is known:
m1 = 1,300 kg
m2 = 1,700 kg

Fg1 = 1,300 kg x 9.8 m/s^-2= -12740N
Fn1 = 12740N
Ff1 = 12740*0.3=3822 N
Fg2 = -16660 N
Fn2 =16660
Ff2=16660 * 0.3 = 4998
Fgc1 = 10,000 N
Fnetsys = (m1+m2)a = Fgc - Ff1 - Ff2 - F21 + F12
Simply (F12 =-F21 so it cancels):
Fnetsys = Fgc - Ff1 -Ff2
Fnetsys = 10,000N - (3822N + 4998N) = 10,000N - 8820N = 1180N
Fnetsys = (m1+m2)a = 1180N
Fnetsys = 3000kg * a = 1180N
a = 0.3933333 m/s^-2

Finding F21 and F12
Fnet1 = 1,300kg * 0.3933333 m/s^-2 = 511.3333 N
511.33333 N= Fgc -F21 -Ff = 10,000N -3822N - F21
511.333333 N= 6178N - F21
-F21 = 6178-511.333= 5666.67 N
F21 = -5666.67 N
Fnet2 = 1,700 kg * 0.393333 m/s^-2
Fnet2 = 668.66777N = F12 - 4998
F21 = 5666.67

I think this works. I may have skipped a few steps.

 

[jbriggs444]

Car 1 is applying a force on the ground and the ground applies a force on car 1 equal to 10,000N.

There’s friction on both cars 1 and 2. That will be equal to the normal force x coefficient of friction.

There is just the one force pair here between the tires of car 1 and the ground? Not two forces. One.
The forward force of ground on car 1's drive wheels is limited by the coefficient of static friction.
For freely rolling tires such as those on car 2, the coefficient of friction is irrelevant. That's the point of using wheels rather than skids.

 

[TheWonderer1]

There is just the one force pair here between the tires of car 1 and the ground? Not two forces. One.
The forward force of ground on car 1's drive wheels is limited by the coefficient of static friction.
For freely rolling tires such as those on car 2, the coefficient of friction is irrelevant. That's the point of using wheels rather than skids.

Okay so I shouldn’t have included friction in either case or only car 2 has friction included?
I believe that’s called the traction force (ut*N at maximum) and yes, I made a mistake. Therefore,
Fnetsys = Fgc - F21 - Ff2 + F12 or Fnetsys = Fgc - F21 + F12

 

[TheWonderer1]

I realized that my free body diagram comment was misstated. The F21 and F12 would cancel out within the system Fnet Force equation because they are equal and opposite (however, acting on different objects). You would still include it when describing the forces on each object in their free body diagrams.

 

[osilmag]

You might be making this more complicated than it should. My above post explains F2. I didn't bother to read your made up problem.

The force on cart 1 would be: F1 = m1a1=F2on1 - friction1.

You would be given/be able to measure/be able to calculate these values.

 

[CWatters]

You haven't said where the friction you mention occurs so we have no way of knowing if you should include this or not and how.

Did you mean air resistance/drag? Rolling resistance? Friction in engine or drive train? Friction between tyres an road? Some combination?

Edit: your equations suggest your cars don't have wheels or the wheels are locked, because you calculate the friction force by multiplying by the normal force.

 

[Dale]

I made this up and solved this problem

This is bizarre. You clearly are not lazy. So why are you so reluctant to draw a free body diagram? You have gone through all this extra effort, why not the little effort that was actually requested by multiple people.

I am done. Best of luck.

 

[osilmag]

You haven't said where the friction you mention occurs so we have no way of knowing if you should include this or not and how.

Did you mean air resistance/drag? Rolling resistance? Friction in engine or drive train? Friction between tyres an road? Some combination?

Edit: your equations suggest your cars don't have wheels or the wheels are locked, because you calculate the friction force by multiplying by the normal force.

I think the OP means rolling friction.

 

[TheWonderer1]

I apologies, I used a YouTube video to figure out what forces I should include and friction was added into their equations. I wanted to add it to be safe but I made a mistake adding them without explanation. Friction between the tires and the road is what I meant. Once again, I assumed wrongly that the video included all the information needed. Here’s a poor resolution free body diagram and I excluded the friction forces:

The force from the ground on the car is difficult to see and I’m sorrry about that. I can update it further but I’m assuming that system is the two cars with the ground applying an external force. I show both car 1 and car 2 but removed the wheels bc I’m not an artist (obviously!). Is it accurate to say the force of the ground on the car or on the wheel? It’s the wheels technically that are applying the force.

 

[CWatters]

Friction between the tires and the road is what I meant.

In which case you were wrong to subtract friction forces in post #15. Friction between tires and road is a good thing and doesn't slow a car down. What it does do is limit the maximum value of Fgc. You wouldn't normally show this type of friction force AND the driving force Fgc on the FBD of a car while it's driving along normally. You might show it if (for example) the car was skidding.

If you had said you meant rolling resistance then you would have been correct to subtract it in #15. You would normally show rolling resistance (and perhaps aerodynamic drag) on a FBD.

 

[CWatters]

Is it accurate to say the force of the ground on the car or on the wheel? It’s the wheels technically that are applying the force.

I don't think it makes a difference.

 

[TheWonderer1]

In which case you were wrong to subtract friction forces in post #15. Friction between tires and road is a good thing and doesn't slow a car down. What it does do is limit the maximum value of Fgc. You wouldn't normally show this type of friction force AND the driving force Fgc on the FBD of a car while it's driving along normally. You might show it if (for example) the car was skidding.

If you had said you meant rolling resistance then you would have been correct to subtract it in #15. You would normally show rolling resistance (and perhaps aerodynamic drag) on a FBD.

This great and super helpful. I appreciate it! I think after thinking about the rolling resistance and aerodynamic drag my answer is pretty well answered.