hi to all problem: a truck with a total weight of 5000kN and with 4 wheels will be placed on 4 glass cups each under one wheel. if each tire has a internal air pressure of 2MPa. what will be the load on each cup? and if the truck was standing on the road with out the cups under wheels what will be the load on the road.
if the tires were solid then i can exactly say that the load per tire transfering down to road is 5000/4 = 1250 kN but the tires are like pneumatic pillows. so some of the load should be carried by the tire surface.
the load effecting to road should be the surface of the interface between the tire and road * tire internal pressure.
So if we, say, let out half the tire air (and, therefore, reduced pressure by half), the load on the road would reduce by half? AM
no because if we let the half of air out. then the touching surface area to the road of the tire will increase and the load will remain the same.
The pressure inside the tires is balanced by the forces holding the tire to the wheel. It contributes no net force to the road. AM
got it so if i have a truck of total weight is a N on 4 tires with a tire pressure of p Pa then the surface area touching the road per tire should be [tex] N/(4*p) [/tex] so there force to the road will be pressure times the surface area so N again.
ok lets change the subject. i got a basketball it stands still on the road. then i come and start pressing on the ball with my one foot. the ball will be deformed from the point that i press. and also the area of the ball touching the road will be deformed and increase. i think that because of the volume decrease in the ball the pressure will increase and then this pressure will transfer my weight to the road with it's increased area touching the road. so there reaction force from the road will increase. true?
You are making it too complicated. The tire pressure does not affect the load on the road. At best, it changes the distribution of the load so changing air pressure may change the pressure on the road somewhat. But that is all. The pressure in the tires has no bearing on the total force on the road. AM
:) ok. last one. what will happen if i increase the load (add more things to truck) there must be a limit for that if i put more weight and suppose that the tire surface is strong enough. the air will go upwards and the truck will touch the bottom. but if i put more pressure on it will lift the truck again. so there must be a connection, hmm dunno may be alittle mixed up
If you have a deflated tire and pump it up, the truck will rise as the air pressure builds up inside the tire. The air in the tire does work as it expands. This work raises, slightly, the centre of mass of the truck and increases its gravitational potential energy. But (when the truck stops moving up) the force exerted on the road is exactly the same as it was with the deflated tire. Draw a diagram showing the forces on the truck and on each wheel. Since the truck is not accelerating in the vertical direction, the vertical forces sum to 0. Those forces are the weight of the truck (down) and the normal force of the road on the tires (up). The air in the tire is exerting a force in both directions but is balanced by the mechanical forces holding the tire onto the truck. It would be exactly the same if the truck was sitting on four vertical springs. AM
so can we clearly say that the weight of truck is transfered to pressured tire by [tex] surface area 1*pressure [/tex] and it transfers the same load to the road by [tex] surface area 2*pressure[/tex] surface area 1 : area of the rigid part touching to the tire surface area 2 : area of the tire touching to road
The pressure in the tire does not determine the pressure on the road or the force on the road. A flat tire with 0 air pressure will still exert a force and pressure on the road. The pressure on the road x area of contact with the road would give you the force on the road, provided the pressure was evenly distributed (which it probably is not). But that is not the same as the tire pressure. Don't get hung up on how the weight is mechanically borne by the tire because there are too many variables that affect it. The issue is not pressure. The issue is force. The weight on each tire is 1/4 of the weight of the vehicle regardless of how much pressure is in the tires. AM
I kept myself busy with calculation of tire-pressure and made spreadsheets for it. The formula the tire-manufacturers use , has a power in it but the next article about another way of calculating , I concluded of, that a part of the load is bare by the construction of the tire. This is for smal tires with low pressure about 20% of the max load,so about 60 kg still, and for large truck tires with high pressure only 2%, but still about 120kg. http://cid-a526e0eee092e6dc.office.live.com/self.aspx/.Public/tirepressureatreducedloads.pdf I myself concluded that at a sertain supression of the side wall of the tire, the construction of the tire bares a sertain weight , and this gives a sertain calculatable surface on the ground. The rest of the load is bare by the simple pressure in kgf/sqcm * calculated surface. Not the issue now but , for every speed there is a sertain supression of the tire-side-wall permissable, and so a sertain surface on the ground. Mayby you can do something with my information.