# Does a truck stop faster if the stack on the back of truck is stable or if it moves forward?

My question in the title might not be very descriptive so I am re-writing it here:

If there is a truck in motion and it has stack of hay (lets suppose) on the back. Now if the truck comes to a sudden stop will it stop faster if the force exerted by the truck on hay had overcome the friction force (another wording: will it be faster if the hay slips forward) or will it stop faster if the hay remains constant.

EDIT:

I do not know where everyone is getting tied down option or not. The hay isn't tied down (sorry for misguiding). There are two situations, it slides on the trucks back or it doesn't slide on trucks back. And in which situation does the truck stop faster?

• "Tied down" in my answer is just short hand for "not sliding". – Floris Apr 29 '15 at 1:29
• If the hay is not tied down then it will slip forward. – slebetman Apr 29 '15 at 3:00
• Short answer: If the stack slides, the truck will stop faster, but the stack will stop slower. Depending on what the stack consist of, that may or may not be desirable. – Ilmari Karonen Apr 29 '15 at 10:29
• Applying physics.stackexchange.com/q/143823/25907 , it depends whether the brakes are applied to the front or rear wheels. – user80551 Apr 29 '15 at 20:58

On the whole, static friction is higher than dynamic friction. This means that if you can brake without your wheels skidding, you will come to a halt more quickly. So let's assume that the truck brakes without skidding, and see where that gets us.

Let's assume that your truck has weight $W = Mg$ with a haystack with additional weight $w = mg$ on top. Coefficient of (static) friction between wheels and road is $\mu_1$, and friction between haystack and truck bed is $\mu_2$.

The normal force on the tires is $W+w$ and the maximum static friction is $\mu_1(W+w)$.

If the wheels stop in a distance $D$ and the hay stack can slide an additional distance $d$ on the truck bed, then the work done to bring the entire assembly to a halt (which must be equal to the kinetic energy we started with) is

$$\mu_1 (W+w) D + \mu_2 w d$$

If there was no movement of the haystack, $d=0$ and the second term would disappear.

So if we have to do the same amount of work to bring the truck to a halt in both cases, we can set the distance of the no-slide truck to $D'$ and we get the expression:

$$\mu_1 (W+w) D + \mu_2 w d = \mu_1(W+w) D'\\ D' = D + \frac{\mu_2 w }{\mu_1(W+w)}d$$

Since the second term is always greater than zero, the stopping distance will be longer for the truck that has the hay stack tied down (or otherwise not sliding).

• You seem to have dropped the third law pair of the hay-truck friction. The truck should have $\mu_1 (W+w) - \mu_2 w$ force horizontally. The $\mu_2 \to \infty$ limit should be the same distance as the fixed case. – alemi Apr 29 '15 at 1:10
• @alemi I thought about that but I think conservation of energy still works. If the coefficient of friction between bed and hay is large, then $d$ will be small. I admit I did not explore the limiting cases carefully - as long as the hay slides the stopping distance is shorter. It is true that $d$ in my expression is the distance that the hay slides - which is less than or equal to the size of the truck bed. I should do the math more carefully... – Floris Apr 29 '15 at 1:15
• True enough. You have the right limit, but the third law pair was crucial for getting the limit right when I was considering the dynamics, which makes me suspect here. – alemi Apr 29 '15 at 1:21
• I'll remember this next time a see a 39-wheeled semitrailer lorry with unsecured hay on the back in my mirror. Even though I shall be crushed to death by wayward oversized roll of breakfast cereal, I shall enjoy a half second more life because the juggernaut managed to stop two centimeters behind my car and didn't kill me. – WetSavannaAnimal Apr 29 '15 at 1:23
• @WetSavannaAnimalakaRodVance: You guys have semitrailers with an odd number of wheels? Or is the steering wheel counted in there too? – dotancohen Apr 29 '15 at 17:42

Not to detract from Floris' answer, but I think this is an instance where it is nice to think in terms of limits.

If the hay is tied down, you're stopping an object with mass (truck + hay).

If the hay isn't tied down, but on a sufficiently sticky surface such that it doesn't move, it should be the same as stopping it if it were fixed, since the outcome is the same.

In the other extreme, if the coefficient of friction between the hay and the truck was zero, we wouldn't care about the hay at all, except that now you are just stopping a truck of mass (truck), with a normal force of (truck + hay), which will stop sooner.

The in between cases will be in between.

• The key being that you are stopping a mass of just the truck, with the normal force of truck plus hay (in other words, with more friction available - in that sense we do care about the hay). With that one proviso, this reasoning is nice and clean. – Floris Apr 29 '15 at 5:11
• So what happens after the truck stops and the hay hits the truck? – Moby Disk Apr 29 '15 at 15:55
• Hopefully the static friction of the truck is high enough to withstand the internal impact and prevent it from moving. – mskfisher Apr 29 '15 at 16:33
• – Mooing Duck Apr 29 '15 at 17:02
• @MobyDisk it would still be intermediate between it being nailed down and it being able to slip off the front. – alemi Apr 29 '15 at 17:08

This is the simplest analogy I could think of.

Imagine a long narrow carpet sliding across a huge ice rink at 1kph. On the rear end of the carpet stands a very fat (200kg) man wearing roller skates. You want to bring him to a standstill. You could grab the man and dig your ice skates into the ice until he eventually stops. Alternatively, you could grab the back edge of the carpet. The carpet essentially weighs nothing so you stop it easily. On the other hand, the fat man will start a comedy journey along the carpet, gradually slowing down as he goes along and eventually rolls to a stop.

In other words, some of the energy of the moving fat man has been lost in friction between his skates and the carpet, not by friction between your skates and the ice.

In your original truck scenario, the load slides forward dissipating some of its kinetic energy as heat as it goes. If it hits the front of the trailer, some of the remaining energy will be transferred back into forward motion of the truck - but hopefully, by that time, the truck will have reached a standstill anyway and will have plenty of spare braking capacity.

• The last paragraph nails it in theoretical physics. However I'd be interested in a trucker's opinion on a load shift causing a skid while breaking, though. Heat lost to the floor is negligible; slamming into the back of the truck isn't (at least before ABS). "Hopefully" (mass) is the defining line between a jostle that can overcome the rolling friction of the wheels or just a miniscule amount of heat being transferred to the floor. – Mazura Apr 29 '15 at 10:25
• @Mazura You're absolutely right. My analogy was deliberately chosen to exaggerate the effect as much possible. In reality, the amount of room the load has to slide will be very small, the load's mass compared to the truck will be small and the friction between, what, pallets and plywood, will also be small. I suppose the only thing we can say is that the effect will be unnoticeable in the real world but having the load unsecured can never be WORSE than being secured - ignoring the damage done to the bulkhead in the collision! – Lefty Apr 29 '15 at 11:02
• @Lefty: I don't think the "never be worse" point is true, given the effects of weight distribution on braking efficiency. As an extreme example, consider a very heavy weight at the back of a lightweight semitrailer with excellent brakes, pulled by a lightweight truck with crummy brakes. If the weight can slide, the truck and trailer will stop almost instantly but the weight won't; once the weight reaches its limit of travel the truck+weight combo will still have most of its former momentum, but the wheels with good brakes will have little weight on them, making them almost useless. – supercat Apr 29 '15 at 17:43
• @supercat Yes, very good point well made. I worded it badly when I said "never be worse" - I was mainly trying to encapsulate the idea that any non-zero amount of kinetic energy dissipated as friction on the truck bed is going to be an amount of energy that doesn't need to go into the braking system - but real-world differences like you've suggested will definitely have an effect. – Lefty Apr 29 '15 at 22:43
• @Lefty: Energy isn't the invariant--momentum (mass times velocity) is. If the road is regarded as stationary, the integrated force on the road throughout the stopping distance must equal the initial velocity times the initial mass. – supercat Apr 29 '15 at 22:50

...a truck in motion and it has stack of hay (lets suppose) on the back. Now if the truck comes to a sudden stop will it stop faster if the force exerted by the truck on hay had overcome the friction force (another wording: will it be faster if the hay slips forward) or will it stop faster if the hay remains constant.

I tried to find a braking vehicle that would produce the situation described in the answers: a truck

or a semitrailer

but I couldn't.

Unless the hay is thrown out of the truck, I believe there can't be any difference because even if the hay slips forward, it will allways hit one side of the truck and be eventually stopped by the brakes, isn't that so?

Wanting to be more precise: it will take only a fraction of a second and the cargo will be secured by the bodywork of the truck and it will be stopping anyway truck+hay (plus an infinitesimal amount of energy turned into heat).

• But how far is the front of the truck from the center of mass in both scenarios? – BowlOfRed Apr 29 '15 at 7:21
• @BowlOfRed, how does that influence the fact that the hay will be stopped anyway? – user78040 Apr 29 '15 at 13:20
• You can imagine that at maximum braking, the center of mass of the system will be stopped by a certain point. But the front of the truck does not have to be a constant distance from that center of mass. – BowlOfRed Apr 29 '15 at 16:13

## protected by Qmechanic♦Apr 29 '15 at 6:46

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