Crate on a Braking Pickup: Three Predictions, One Correct
A 40 kg crate of bottled water stands on the flat steel bed of a 1.5 tonne pickup, 2.5 m behind the headboard; the bed is 6.0 m long. Rubber feet on the crate give μs = 0.35 and μk = 0.30 against steel. The truck is doing 54 km/h (15 m/s) on a level road when the driver sees a speed-breaker and brakes, stopping in 30 m without skidding. The driver says the crate is far too heavy to move. The loader says it will slide straight off the tailboard. A third man says it will slide forward and slam into the headboard. Nobody has yet touched the load; the question is what actually happens, and what the driver should have done differently.
1. Decide between the three predictions with numbers: does the crate slide, in which direction relative to the bed, and where does it end up?
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Truck's deceleration. a = v²/2s = 15²/(2 × 30) = 3.75 m/s².
Can friction keep the crate with the truck? The maximum deceleration friction can give the crate is μs g = 0.35 × 9.8 = 3.43 m/s². Since 3.43 < 3.75, the crate cannot keep up: it slides. Because the crate decelerates less than the truck, in the truck's frame it moves forward - the loader's tailboard prediction is wrong.
Once sliding, the crate decelerates at μk g = 0.30 × 9.8 = 2.94 m/s², so the relative acceleration is 3.75 − 2.94 = 0.81 m/s² forward relative to the bed.
Time to the headboard: 2.5 = ½ × 0.81 × t² → t = √(6.17) = 2.49 s, well before the truck stops at 15/3.75 = 4.0 s. Relative speed at impact = 0.81 × 2.49 = 2.0 m/s, i.e. about ½ × 40 × 2.0² = 81 J of relative kinetic energy to be absorbed by the headboard and the bottles.
Had there been a full 6.0 m bed with no headboard, the relative sliding would have reached ½ × 0.81 × 4.0² = 6.5 m while the truck braked, plus a further 1.8 m as the crate slid on the stopped bed - off the tailboard. The headboard, at 2.5 m, is what saves the load. The third man is right.
2. The driver argues that a heavier crate would have been safer because 'there is more weight pressing down'. Evaluate this.
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Both accelerations are independent of mass, so the argument fails.
The condition for not sliding is ma ≤ μs mg, i.e. a ≤ μs g. The m cancels: a heavier crate has more friction (μs mg) but needs proportionally more force (ma) to be decelerated with the truck. The limiting deceleration is μs g = 3.43 m/s² whichever way the mass goes, so doubling the load changes the sliding outcome not at all.
What the mass does change is the consequence. The relative sliding distance (½ × 0.81 × t²) is mass-independent, but the energy the headboard must absorb, ½m(v_rel)², scales with m: at 80 kg the impact is about 162 J instead of 81 J. A heavier crate also raises the truck's centre of mass and lengthens its stopping distance, so the truck may no longer manage 3.75 m/s² at all.
The correct fix is to reduce the required deceleration (brake earlier, so the stopping distance is longer), raise μs (rubber matting, lashing), or move the crate hard against the headboard so there is no run-up.
3. What is the highest speed from which the driver could brake in 30 m without the crate sliding? State the direction of the friction force on the crate while braking, and explain why it is not 'thrown forward'.
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Maximum speed. Set a = μs g = 3.43 m/s² and use v² = 2as with s = 30 m: v = √(2 × 3.43 × 30) = √205.8 = 14.3 m/s ≈ 51.6 km/h. Braking from 54 km/h is just over the limit, which is why a small excess of speed produces a dramatic slide.
Direction of friction. The crate tends to slide forward relative to the bed, so the friction force exerted by the bed on the crate is forward, in the direction of the truck's motion. Friction here is what slows the crate down; without it the crate would continue at 15 m/s and hit the headboard far harder.
Why 'thrown forward' is wrong. In the ground frame there is no forward force on the crate during braking - only the backward-acting? no: only the forward friction, and it is smaller in magnitude than the backward force needed to match the truck (ma = 150 N versus the available 137 N). In the truck's frame a pseudo-force m(3.75) appears to throw the crate forward, but that is a bookkeeping device. Physically the crate simply fails to decelerate as fast as the vehicle under it.