The Physics of Truck Braking: Why Shorter Distances When Sliding?

In summary: The problem requires you to use the work-energy theorem applied to the truck. Personally, I don't find using this theorem to be the best way to arrive at the answer. It seems to me that the answer is much easier to obtain if you just use concepts of force and acceleration. (Also, see post #6 by @phinds to get an intuitive feel for the answer.) But I might be overlooking something.Anyway, can you tell us your opinion about the following statement?The equation 1/2mv^2 = (Fb+Fk)d is, in fact, misleading if you interpret as saying that Fb and Fk act in the same direction.
  • #1
Andrew1234
18
1
Homework Statement
The problem is attached.
Relevant Equations
1/2mv^2 = (Fb+Fk)d
The truck stops in a shorter distance if the crate slides but why is this the case considering that friction on the crate does positive work on the truck, since it points left for the crate but right for the truck thereby opposing the braking force?
I understand why using the equation
1/2mv^2 = (Fb+Fk)d, where Fb is the braking force and Fk the spring force, d is shorter if sliding occurs but not why Fk is directed opposite the motion given that it points opposite the crate's velocity, left, but must be balanced on the truck's free body diagram by an equivalent force that points to the right.
 

Attachments

  • Screenshot_20200218-152611.png
    Screenshot_20200218-152611.png
    39 KB · Views: 167
Physics news on Phys.org
  • #2
Do you think the centre of mass of the truck-crate system might be relevant?
 
  • #3
Just to be clear, is this your question: Why is the force of friction that acts on the truck (from the crate) in the direction of motion of the truck?
 
  • #4
TSny said:
Just to be clear, is this your question: Why is the force of friction that acts on the truck (from the crate) in the direction of motion of the truck?
My question is why the equation used to solve the problem that leads to the correct answer suggests that the force of friction that acts on the truck (from the crate) opposite the direction of motion of the truck, although if the internal friction of forces are balanced I think the force of friction that acts on the truck (from the crate) is in the direction of motion of the truck.
 
  • #5
PeroK said:
Do you think the centre of mass of the truck-crate system might be relevant?
The center of mass should be displaced by the same distance since the external horizontal forces are equal.
 
  • #6
Try looking at the problem the other way 'round. Is the sliding more likely or less likely under the conditions of the braking force causing:
  1. truck stops abruptly
  2. truck stops slowly
 
  • #7
Andrew1234 said:
My question is why the equation used to solve the problem that leads to the correct answer suggests that the force of friction that acts on the truck (from the crate) opposite the direction of motion of the truck, although if the internal friction of forces are balanced I think the force of friction that acts on the truck (from the crate) is in the direction of motion of the truck.

The equation 1/2mv^2 = (Fb+Fk)d is, in fact, misleading if you interpret as saying that Fb and Fk act in the same direction.

It would be better to write the work-energy principle for the truck as ##\Delta K_{\rm truck}= (\vec F_b+\vec f) \cdot \vec d## where ##K## is kinetic energy and ##\vec f## is the kinetic or static friction force that the crate exerts on the truck.

It is for you to decide if the two force vectors ##\vec F_b## and ##\vec f## are in the same direction or opposite direction. Also, decide if ##\Delta K_{\rm truck}## is positive or negative.
 
  • #8
TSny said:
The equation 1/2mv^2 = (Fb+Fk)d is, in fact, misleading if you interpret as saying that Fb and Fk act in the same direction.

It would be better to write the work-energy principle for the truck as ##\Delta K_{\rm truck}= (\vec F_b+\vec f) \cdot \vec d## where ##K## is kinetic energy and ##\vec f## is the kinetic or static friction force that the crate exerts on the truck.

It is for you to decide if the two force vectors ##\vec F_b## and ##\vec f## are in the same direction or opposite direction. Also, decide if ##\Delta K_{\rm truck}## is positive or negative.
Thank you for your response.
Doesn't this equation lead to the same conclusion since the braking and friction forces oppose each other? Kinetic energy is positive, so if the friction opposes the braking force d must be larger if the crate slides.
 
  • #9
Andrew1234 said:
Doesn't this equation lead to the same conclusion since the braking and friction forces oppose each other?
You are right that the friction force on the truck (from the crate) is in the opposite direction of the braking force on the truck.

Kinetic energy is positive
Yes, but it's the change in kinetic energy that is important in the work-energy theorem: ##\Delta K = W_{\rm net}##.
Is the change in kinetic energy of the truck positive or negative?

so if the friction opposes the braking force d must be larger if the crate slides.
How do you come to this conclusion?

The problem requires you to use the work-energy theorem applied to the truck. Personally, I don't find using this theorem to be the best way to arrive at the answer. It seems to me that the answer is much easier to obtain if you just use concepts of force and acceleration. (Also, see post #6 by @phinds to get an intuitive feel for the answer.) But I might be overlooking something.

Anyway, can you tell us your line of reasoning for concluding that "d must be larger if the crate slides"?
 
Last edited:
  • Like
Likes PeroK
  • #10
Andrew1234 said:
My question is why the equation used to solve the problem that leads to the correct answer suggests that the force of friction that acts on the truck (from the crate) opposite the direction of motion of the truck, although if the internal friction of forces are balanced I think the force of friction that acts on the truck (from the crate) is in the direction of motion of the truck.
You have a moving system: truck + load.
The brakes of the truck are trying to consume all that kinetic energy of the whole system, transforming it into heat.
The problem is asking about stopping time/distance of center of mass of the truck only, disregarding where the load ends up.

Applying the brakes at the same speed, what happens to that truck stopping distance if:
1) You have solidly anchored that load onto the truck, the load remains in its original place on the bed.
2) The load slides 10 meters before both, truck and load come to a full stop (explain this one if you can).
3) The load is on wheels and never stops (it jumps ahead of the truck and keeps on rolling).

:cool:
 
  • #11
Lnewqban said:
The problem is asking about stopping time/distance of center of mass of the truck only, disregarding where the load ends up.
It's not clear on that point. I would interpret the question perhaps as asking where the front of the truck ends up. That would be the important factor in avoiding a collision, for example. Or, equivalently, the centre of mass of the truck itself; not the truck-crate system.

For example: if truck and crate are of equal mass (or even that the crate is much heavier than the truck), and the friction force is small, then the crate may slide a very long distance before stopping. Perhaps much further than the original stopping distance. In this case, the centre of mass of the system must have traveled further than if the crate was secured. But, what about the front of the truck? Does it travel further when the crate slides?
 
  • #12
Andrew1234 said:
Thank you for your response.
Doesn't this equation lead to the same conclusion since the braking and friction forces oppose each other? Kinetic energy is positive, so if the friction opposes the braking force d must be larger if the crate slides.

The thing missing from your equations and your reasoning is mass. Force alone does not determine deceleration or stopping distance. The mass of the object being decelerated is also important.

Alternatively, when the crate does not slide, does the crate not exert a force of the truck in that case too?
 
  • Like
Likes ehild
  • #13
Less work to be done if you're stopping truck and not truck and crate.
 
  • Like
Likes PeroK
  • #14
The question wants you to use the work-energy theorem as applied to the truck: ##W_{\rm truck}^{\rm net} = \Delta K_{\rm truck}##.

Let ##F_B## and ##f## be the magnitudes of the braking force and the friction force acting on the truck. So, ##F_B## and ##f## are defined here to be positive numbers. The horizontal forces acting on the truck are shown below as the truck moves to the right a distance ##d## and comes to rest.

1582137800588.png
Let ##K_0## be the initial kinetic energy of the truck.

@Andrew1234 : Write out the work-energy theorem in terms of ##F_B##, ##f##, ##d##, and ##K_0##. Be careful with signs. Solve for the distance ##d##.
 
  • #15
TSny said:
The question wants you to use the work-energy theorem as applied to the truck: Wnettruck=ΔKtruckWtrucknet=ΔKtruck.

Let FBFB and ff be the magnitudes of the braking force and the friction force acting on the truck. So, FBFB and ff are defined here to be positive numbers. The horizontal forces acting on the truck are shown below as the truck moves to the right a distance dd and comes to rest.

View attachment 257348Let K0K0 be the initial kinetic energy of the truck.

@Andrew1234 : Write out the work-energy theorem in terms of FBFB, ff, dd, and K0K0. Be careful with signs. Solve for the distance dd.
I think I understand how to solve the problem using the system's free body diagram but not if the truck's and crate's free body diagrams are considered separately.

Let m and s represent the crate's mass and stopping distance and M and S represent the truck's mass and stopping distance.
In terms of FB and f, the work energy principle gives

T1+U=T2

1/2(m + M)v^2 +(-FB)S-fs=0

S = (1/2(m + M)v^2-fs)/FB (1)

However, if I consider the crate and truck separately, then applying the work energy principle to the truck's free body diagram gives

T1+U=T2
1/2(M)v^2 +(-Fb)S+fs=0
S = (1/2(M)v^2+fs)/Fb (2)

And applying the work energy principle to the crate's free body diagram gives

T1+U=T2
1/2(m)v^2 -fs=0 (3)
s= 1/2mv^2/f

Plugging this expression into (2),

S = (1/2(M+m)v^2)/FB

which is the same as the expression for as the expression for S if the cratis were not to slide.

The mistake I am making is failing to consider the crate's relative velocity and displacement but I am not sure how to modify the expression (3) to account for that. If it were not to slide then its relative velocity is 0 so since U and T2 are also 0 this is consistent with the fact that T1 is 0.
 
Last edited:
  • #16
Andrew1234 said:
I think I understand how to solve the problem using the system's free body diagram but not if the truck's and crate's free body diagrams are considered separately.

Let m and s represent the crate's mass and stopping distance and M and S represent the truck's mass and stopping distance.
In terms of FB and f, the work energy principle gives

T1+U=T2

1/2(m + M)v^2 +(-FB)S-fs=0

S = (1/2(m + M)v^2-fs)/FB (1)

However, if I consider the crate and truck separately, then applying the work energy principle to the truck's free body diagram gives

T1+U=T2
1/2(M)v^2 +(-Fb)S+fs=0
S = (1/2(M)v^2+fs)/Fb (2)

And applying the work energy principle to the crate's free body diagram gives

T1+U=T2
1/2(m)v^2 -fs=0
s= 1/2mv^2/f

Plugging this expression into (2),

S = (1/2(M+m)v^2)/FB

which is the same as the expression for S if the crate were not to slide.

I'm not sure any of that is valid. The problem I see is that it doesn't adequately reflect the physical scenario.

You have to think first what happens physically if the crate slides. There are two phases of the motion:

1) The truck is decelerating faster than the crate. The external force on the system is ##F_b##, the maximum braking force.

2) After the truck stops, the crate continues to slide until it comes to rest. The external force on the system is no longer the maximum braking force. The driver must keep the brakes applied, but the maximum braking force is not required simply to stop the truck being dragged forward by the sliding crate.

Any analysis must respect these two separate phases of motion.
 
  • Like
Likes Lnewqban
  • #17
Andrew1234 said:
I think I understand how to solve the problem using the system's free body diagram but not if the truck's and crate's free body diagrams are considered separately.
The problem wants you to show that the stopping time/distance of the TRUCK are shorter if the cart slides on it, than they are when the cart does not slide. You need not worry about the cart.
You know for sure that the truck decelerates faster then the sliding cart. From that, you know something about the force of friction between the truck and cart. When is it bigger? if the cart slides or if it does not slide?
The scenario with the constant braking force apply only till the truck moves. As soon as it stops, the applied force has to change in order to keep it steady. The cart moves further, but it is irrelevant for the problem.
 
  • #18
This problem is a mess, with uncleaar requirements, IMO. Everyone has a different interpretation. And justifiably, too.

Just to add some sanity:
First, the most massive unrestrained loads always go as far forward on the bed as possible. Or are restrained. So they cannot shift on braking. Question #3 results in damage to the truck, the driver, and my sanity. :oldgrumpy:

Being helpful:
For #2, assume the crate dissipates all of it's kinetic energy during braking without hitting the restraining front wall of the cargo area. So, whether or not the crate is unrestrained, the ΔKE (loss) has to be the same for #1 and #2.
For #3, since we have a magic truck that let's cargo fly forward unimpeded, you ignore the mass of the cart.
And, FWIW, I do not see why the cart did not roll off the stern to begin with -- when the truck accelerates forward the very first time. So, #3 results in a shorter stopping distance than #1 or #2 ... regardless of when the carts exits the truck.
Magic, again.
 
  • Skeptical
Likes PeroK
  • #19
jim mcnamara said:
This problem is a mess, with uncleaar requirements, IMO.
NO the problem is quite clear.
1582353450392.png
 
  • Like
Likes jim mcnamara and PeroK
  • #20
Perhaps for the sake of Jim's sanity :wink: we can imagine the truck bed to be friction free but the crate rests against a spring which compresses (due to the crate's inertia) when the truck brakes.
 

1. What is the physics behind truck braking?

The physics behind truck braking involves the principles of friction, inertia, and momentum. When the brakes are applied, the friction between the brake pads and the wheels creates a force that opposes the motion of the truck. This force slows down the truck due to the principle of inertia, which states that an object in motion will continue in motion unless acted upon by an external force. As the truck slows down, its momentum decreases, leading to a shorter stopping distance.

2. How does the weight of the truck affect braking distance?

The weight of the truck plays a significant role in braking distance. A heavier truck will have more momentum and therefore require more force to slow down. This means that a heavier truck will have a longer braking distance compared to a lighter truck with the same braking force applied.

3. Why do trucks slide when braking on wet or icy roads?

Trucks can slide when braking on wet or icy roads due to a decrease in friction between the tires and the road surface. Water or ice on the road reduces the contact between the tires and the road, leading to a decrease in the friction force. This results in a longer stopping distance and can cause the truck to slide as the wheels lock up and lose traction.

4. How do anti-lock braking systems (ABS) affect stopping distance?

Anti-lock braking systems (ABS) use sensors to monitor the speed of each wheel and prevent them from locking up during braking. This allows the wheels to maintain traction and prevents the truck from skidding or sliding. ABS can significantly reduce stopping distance, especially on wet or icy roads, by allowing the driver to maintain control of the truck while braking.

5. Is there a way to calculate the exact braking distance of a truck?

Yes, there is a way to calculate the exact braking distance of a truck using the formula: Braking Distance = Initial Speed² / (2 x Deceleration). The deceleration can be calculated by dividing the braking force by the truck's mass. However, the actual braking distance may vary due to factors such as road conditions, weight of the truck, and the efficiency of the braking system.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
4K
  • Introductory Physics Homework Help
Replies
9
Views
3K
  • Introductory Physics Homework Help
Replies
6
Views
7K
  • Introductory Physics Homework Help
Replies
4
Views
6K
  • Introductory Physics Homework Help
Replies
4
Views
16K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
11K
  • Introductory Physics Homework Help
Replies
1
Views
2K
Back
Top