Timeline for Does a truck stop faster if the stack on the back of truck is stable or if it moves forward?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Apr 29, 2015 at 19:33 | comment | added | binaryfunt | @mbeckish I think the sliding of the hay would dissipate kinetic energy as heat, so even if the hay then hit the cab, the stopping distance would still be shorter than a truck with tied-down hay (unless the truck bed was frictionless) | |
| Apr 29, 2015 at 17:42 | comment | added | dotancohen | @WetSavannaAnimalakaRodVance: You guys have semitrailers with an odd number of wheels? Or is the steering wheel counted in there too? | |
| Apr 29, 2015 at 17:23 | comment | added | mbeckish | This analysis assumes that the moving haystack eventually stops due to friction. If instead the haystack stops due to colliding with the cab of the truck, does that change the answer? Also, would the timing of the collision matter (whether the haystack hits the cab before or after the truck comes to a stop)? | |
| Apr 29, 2015 at 2:37 | vote | accept | user3329118 | ||
| Apr 29, 2015 at 1:29 | history | edited | Floris | CC BY-SA 3.0 |
added 27 characters in body
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| Apr 29, 2015 at 1:23 | comment | added | Selene Routley | 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. | |
| Apr 29, 2015 at 1:21 | comment | added | alemi | 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. | |
| Apr 29, 2015 at 1:15 | comment | added | Floris | @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... | |
| Apr 29, 2015 at 1:10 | comment | added | alemi | 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. | |
| Apr 28, 2015 at 23:46 | history | answered | Floris | CC BY-SA 3.0 |