Physics Help Forum determine pressure applied by truck driving into wall at different speeds

 Kinematics and Dynamics Kinematics and Dynamics Physics Help Forum

 Dec 18th 2015, 03:43 AM #1 Junior Member   Join Date: Dec 2015 Location: newport Posts: 4 determine pressure applied by truck driving into wall at different speeds i'm trying to figure out a formula to calculate the pressure of a truck pressing its bumper against a wall with the engine running at different speeds. this video shows exactly what i'm talking about: the mass of the truck is ~2,500 kg, the area of the bumper is ~3,000 sq cm, and the independent variable is obviously going to be the "speed" so i can calculate different pressures at different speeds. unfortunately that's about all i can figure. i don't know how to do the force since the truck isn't literally accelerating, it's just stuck still. is that where i plug the mph (or m/s) variable into the equation? that being the readout on the speedometer, not the actual movement of the truck? then i have no idea what the friction for a setup like that would be. i guess an educated guess is all i can ask for. it's really just to figure out, say the truck is going at 45 mph or 65 mph (pardon my american), then how many lbs of pressure is being applied through the truck's bumper onto the wall in front of it at each speed.
 Dec 18th 2015, 04:51 AM #2 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,331 There are two ways to approach this problem. If you know the torque that the engine produces you can divide that by the radius of the rear tires to arrive at the force that vehicle tries to put down on the asphalt. Divide that figure by the contact area between the truck and wall to get pressure. However - spinning tires look cool, but the amount of force they apply to the road is significantly less than a non-spinning tire. So given the amount of smoke coming off the tires in the video it's pretty clear that the actual amount of pressure on the wall would be significantly less than this calculation would suggest. A second way to approach this is to consider the force due to friction between the tires and the asphalt. A high quality tire has a coefficient of friction of about 0.9, which means the amount of force they can apply to the road is equal to the weight supported by the drive wheels (about half the truck's weight) times 0.9. So - if the rear tires are carrying, say, 2000 pounds of weight on them, they would produce about 1800 pounds of force. Again, divide by the contact area to determine the pressure on the wall, and deduct some amount (at least half I would guess) to account for loss of friction due to the spinning tires. Last edited by ChipB; Dec 18th 2015 at 07:55 AM.
 Dec 18th 2015, 07:11 PM #3 Junior Member   Join Date: Dec 2015 Location: newport Posts: 4 wow, thanks for the quick response! ok, so i've blown pretty much my entire afternoon screwing around with this, didn't get as far as i'd hoped. first off, to fill out variables i couldn't figure on my own, i decided to use a ford f350 for reference. stats are: weight = 6,000 lbs / 2,750 kg (per tire = 1,500 lbs / 688 kg) tire = diameter: 35 in / 89 cm, radius: 17.5 in / 44.5 cm, circumference: 110 in / 280 cm bumper = 1,260 sq in / 8,150 sq cm torque = 405 @ 4,500 rpm thing about torque is, the specs also have a horsepower measurement with 368 @ 5,500 rpm. to compare, i plugged the 405 @ 4,500 rpm into a torque to hp conversion (hp = rpm x torque ÷ 5252) and got 347 hp @ 4,500 rpm. just to check if there was some kind of ratio consistency, i did 368 ÷ 5,500, then 347 ÷ 4,500. first was 0.0669, second was 0.0771. looks like it's some kind of exponential difference, which throws a wrench into what i thought would be a quick formula i could plug the speed variable into. so, with your first point, i used the 4,500 rpm spec to do 405 Nm ÷ 0.445 m to get 910 N. then did 910 N ÷ 0.815 sq m (8,150 sq cm) to get pressure of 1,116.5 Pa, used a Pa to psi converter to get 0.162 psi, then multiplied by 1,260 sq in (bumper surface) to get 204 lbs of pressure from the bumper. divided by the suggested 0.5 to get 102 lbs of pressure on the wall at 4,500 rpm. seems surprisingly low, but even moreso when trying to convert rpm to mph. default formula seems to be diameter (in) x pi x rpm x 60 ÷ 5,280, which gives ... 481 mph. too crazy. further research shows gear ratio plays into it. specs are 1st gear (1-15 mph) = 3.97, 2nd gear (16-30 mph) = 2.31, 3rd gear (31-45 mph) = 1.51, 4th gear (46-60 mph) = 1.14, 5th gear (61 - 75 mph) is 0.85. found an online converter, used 3.97 differential to get 118 mph... but then 2.31 to get 202 mph, and it just got wackier going up from there. so now i think my attempt to figure the speed variables is screwed, because 102 lbs of pressure at those speeds in a 6,000 lb vehicle just doesn't make sense to me, even accounting for tire burnout. moving to the second way, i forget exactly everything i did, but i got 6,000 lbs (4 wheel drive) x 0.9 friction coefficient for 5,400 lbs and somehow got it converted to 24020.4 N, divide by surface area 0.815 sq m to get 29,473 Pa, converted to 4.30 psi, multiplied that by surface area 1,260 sq in to get... 5,418 lbs. just went in a circle, and didn't even account for different speeds. this was just supposed to be an extra credit project of my own choice, but geez. might've bit off more then i can chew ha. so, any idea what i'm screwing up? it can't possibly be this complicated.
 Dec 19th 2015, 04:37 AM #4 Senior Member     Join Date: Apr 2008 Location: Bedford, England Posts: 668 Friction can be a notoriously difficult thing to calculate sensibly, there are just so many factors that can influence it. Friction is highest while the contact is firm, as soon as the contact starts to slip, the friction drops dramatically. The rubber will get stickier as it gets hot (and friction will rise), but if it melts, it will start to lubricate the contact point (and friction will fall). And so on and so forth... Think about people driving on icy roads, if they drive smoothly and the wheels are not slipping, everything is ok. As soon as the wheels start slipping, it makes absolutely no difference how fast they spin the wheels, they go nowhere (or in a straight line at the speed and direction the were traveling in before the slip started). __________________ You have GOT to Laugh !
 Dec 19th 2015, 05:51 AM #5 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,331 I must apologize, but the first technique that I described is missing a key component that is critical to the calculations - the ratio of the turning speed in RPM of the engine to the RPM of the wheels. The effect of the transmission is to reduce the rotation velocity of the wheels which in turn increases the torque that the wheels can drive the vehicle with. The calculation is: Torque at the wheels = engine torque times transmission gear ratio times final drive ratio. If you don't have the data on gear ratios and final drive, a ball park figure might be 3.8 for 1st gear times 4.0 for the final drive. Where did you get the engine torque value of 405? That sounds to me like it's in units of ft-lbs, not N-m, and correcting that would make a huge difference in your calculations. Also, you originally asked about pressure, which is force per unit area. But it seems from this that you end goal is actually to calculate the force of the truck against the wall, not the pressure. Which is it? Once you calculate the force by taking torque at the wheels and dividing by radius of the tires you're done - no need to mess around with surface areas at all. Last edited by ChipB; Dec 19th 2015 at 06:08 AM.
 Dec 19th 2015, 03:03 PM #6 Junior Member   Join Date: Dec 2015 Location: newport Posts: 4 ok, looks i might not have picked the technically correct words, sorry if i got something wrong. using the words loosely: what i was trying to figure was the overall force of the truck (pressing) against the overall wall, not any specific area of the wall. another way to ask the question might be, how much force does the wall need to generate, to be able to match the force of the truck pressing against it at different speeds, without collapsing under the different pressures (again, forget specific area, just take the wall as one unit). maybe the simplest comparison i can make: i've got a bathroom scale laying horizontal on the floor, and i step on it. the force (pressure?) of my foot on the surface makes it read 180 lbs. now say i take that scale, flip it vertical, put it against the wall and use it to measure the lbs of the truck bumper pressing on it. that's all i was trying to get at. see, i only just took intro physics, my first high school class got off to a rough start, now i'm doing some extra work to make up. it uses some stuff we covered in class, but i figured i'd take it one step ahead. we did force, but we didn't cover pressure, so i tried to learn that on my own. maybe i got the technical meaning mixed up with the 'everyday' meaning. i said 'pressure' meaning the truck pressing against the wall... because even though we get vertical acceleration from gravity, i was trying to measure horizontal force, and since the truck was stuck still, i didn't know what to put for 'a' even though i knew it couldn't just be 0. still, since i'm only doing overall force and not force per specific area, i guess i'm not doing pressure anyway, just plain old force. figures. anyway, i know there's no sensible way to figure the friction here. the equipment for that is like professional-lab-level. for this exercise, i just need to show i've accounted for the variable in a way that the guy would consider an 'acceptable ballpark average', so the half-off thing'll be good enough to show i understand it makes a significant difference. the gradual burnout of the tire and the rate of decreasing friction is way advanced for us now. as for engine torque, gear ratios and all that, i dug up the ford f350 specs page. first section has horsepower and torque for the '6.2L V8 gas engine' at different rpms, then over halfway down, transmission specs has the different gear ratios... which is too complicated for this exercise, so i'm going to use the 3.9 average. i read that torque was measured in Nm (not J) so i assumed that's what the specs page meant, but maybe it's ft-lbs. can't tell. either way, i've got the engine torque of 405 at 4,500 rpm and the gear ratio at 3.9, but what's the drive ratio? i don't see that on the specs page. is that where i can plug in the speed variable? or do i have to somehow convert the engine rpms into mph? mph in this case being how far you'd push the pedal to attain that speed if you weren't against a wall. because the point of the exercise is to use an independent variable, so the end result means i need a formula for x mph = y lbs, so i can do a variables chart showing 45 mph = ?? lbs, 55 mph = ?? lbs, etc., meaning i have to find a way to work the different speeds into it and not just the spec'd 405 engine torque @ 4,500 rpm. ok i hope this is clearer, thanks again for helping sort through all this.
 Dec 19th 2015, 04:15 PM #7 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,331 Indeed you are asking about force, not pressure. Remember that pressure is force per unit area - for example if you measure the pressure in the tore od a car and it's 30 PSI that means 30 pounds per square inch. Since this is an American car, I really think the 405 figure is ft-lbs, not N-m. An engine that can put out only 405 N-m, which is equivalent to The final drive ratio depends on the gearing of the differential - it's the ratio of the turning speed of the drive shaft (in RPM) to the axles that connect to the wheels. A typical value for a vehicle like this is 4.0. Another way to determine the gearing ratios is to compare the engine RPM versus road speed. For example, if the tires have an overall diameter of, say, 30 inches, that means they turn 30 x pi = 94 inches per revolution, or about 7.85 ft/revolution. At 15 MPH, which is equal to 22 ft/sec, that means the tires turn at 22/7.85 = 2.8 revs/second, or about 170 RPM. So ... if the engine RPM in 1st gear at 15 MPH is, say, 3000 RPM, the ratio for the entire gear train is 3000/170 = 17.9. Finally - you say you're looking to develop a calculation of force exerted versus an independent variable such as speed. Speed of what? The trick's speed = 0 m so that's not going to help. Engine speed perhaps? You can do a rough analysis as we've already outlined, but keep in mind that the more smoke the spinning tires create the more of the engine's power is going into heating rubber as opposed to developing the force that you want. About the best you can do is calculate an upper limit of force, ignoring the effects of the spinning tires.
 Dec 21st 2015, 10:18 PM #8 Junior Member   Join Date: Dec 2015 Location: newport Posts: 4 yeah that's it, engine output. the truck not moving was what stumped me at the accel part of the force equation, so now i'm using 'speed' in reference to the result of the pedal being pushed. meaning, if i push it down to the 45 mph level, the engine will still put out the same horsepower and the tires will still spin at the same rpm, regardless of if the vehicle's using full traction or slipping on ice or stuck against a wall. it's only to get the final rpm of the tires. as for torque/power, i found this chart detailing the engine specs. you were right, they're measuring torque in lb-ft. that's a heck of a 'curve'. no way i could've calculated that. the top-off in torque seems weird, but i'll get to that in a second. current data for engine (e) and tire (t) reads: 30 mph = 1350 rpm (e) = 625 lb-ft (e) = 160 hp (e) = 360 rpm (t) = 2950 lb-ft (t) [Fc = 52 Gs] [ac = 905 g / Fc = 5.4e6 lb-ft] 40 mph = 1800 rpm (e) = 740 lb-ft (e) = 250 hp (e) = 480 rpm (t) = 3500 lb-ft (t) [Fc = 92 Gs] [ac = 1610 g / Fc = 9.7e6 lb-ft] 50 mph = 2275 rpm (e) = 740 lb-ft (e) = 325 hp (e) = 600 rpm (t) = 3500 lb-ft (t) [Fc = 143 Gs] [ac = 2575 g / Fc = 1.5e7 lb-ft] 60 mph = 2725 rpm (e) = 740 lb-ft (e) = 375 hp (e) = 720 rpm (t) = 3500 lb-ft (t) [Fc = 206 Gs] [ac = 3690 g / Fc = 2.2e7 lb-ft] 70 mph = 3175 rpm (e) = 575 lb-ft (e) = 350 hp (e) = 840 rpm (t) = 2725 lb-ft (t) [Fc = 280 Gs] [ac = 5010 g / Fc = 3.0e7 lb-ft] 80 mph = 3625 rpm (e) = 100 lb-ft (e) = 125 hp (e) = 960 rpm (t) = 475 lb-ft (t) [Fc = 366 Gs] [ac = 6530 g / Fc = 3.9e7 lb-ft] now for the weird torque, that last column for tire torque seems a bit off. i did your 'engine torque x gear ratio (1.43) x drive-axle ratio (3.31)' routine, but should it be staying the same for 40-60 mph, then getting less as it goes up? is this where the horsepower plays into it? i also got centrifugal force (Fc) measurements from 2 rpm calculators i used, plus centrifugal acceleration (ac) from the second, which i added in brackets. should this be figured in somewhere as well? the first measurement in Gs was done using speed (mph) and tire radius with no regard for mass. the second two were done using speed (rpm) and tire radius while accounting for mass. first one didn't tell me what Gs means, but i know it's not the same as g. well, hopefully there's enough data there to be able to plug it into a straightforward formula to get the final forces of the truck on the wall and be done. i'm just not sure what that formula would be, and what data is relevant. your idea of the upper force limit is all i need. then i'll just do the divide-by-half idea and say friction's been accounted for. it's good enough for this.
Dec 22nd 2015, 08:15 AM   #9
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Join Date: Jun 2010
Location: Morristown, NJ USA
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 Originally Posted by DeltBoy so now i'm using 'speed' in reference to the result of the pedal being pushed. meaning, if i push it down to the 45 mph level, the engine will still put out the same horsepower and the tires will still spin at the same rpm, regardless of if the vehicle's using full traction or slipping on ice or stuck against a wall. it's only to get the final rpm of the tires.
Wrong! Torque values as published are determined by determining how much resistance the engine can overcome at each value of RPM. If there is little or no resistance the engine produces very little torque, and since HP = torque times RPM the HP output is small, regardless of speed. For example, if you are going 45 MPH up a steep hill with the gas pedal mashed to the floor the engine produces a lot more torque (and HP) than if you are coasting down a hill at 45 MPH with your foot off the gas. In the case of the spinning tires we know some energy is being expended - otherwise the tires wouldn't smoking - but we really don't know how much.

 Originally Posted by DeltBoy current data for engine (e) and tire (t) reads: ... now for the weird torque, that last column for tire torque seems a bit off. i did your 'engine torque x gear ratio (1.43) x drive-axle ratio (3.31)' routine, but should it be staying the same for 40-60 mph, then getting less as it goes up?
Yes, per the curve you referenced torques is constant throughout a wide range of RPM.

 Originally Posted by DeltBoy is this where the horsepower plays into it?
Yes, HP = torque times RPM, so for this range of RPM HP is linear with respect to RPM.

 Originally Posted by DeltBoy i also got centrifugal force (Fc) measurements from 2 rpm calculators i used, plus centrifugal acceleration (ac) from the second, which i added in brackets. should this be figured in somewhere as well?
No - centripetal acceleration of the tire has nothing to do with the force that the tire exerts to the pavement.

One last comment: you keep referencing horsepower, but I hope you are aware that HP is only vaguely related to the force the force that the tires exert top the road or the acceleration of the truck. It's torque that causes force and acceleration; HP is just a mathematical calculation or torque times RPM. My suggestion is forget HP and focus solely on torque values.

Last edited by ChipB; Dec 22nd 2015 at 08:19 AM.

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