A)How much sress is exerted on a road by an 18-wheel track which has a mass of 15 x 10^5 kg anda footprint area of each tire of .04m^2

B)A 1200 kg auto is supported by 4 tires each with an area of .0015m^2. How much stress does this cause(on the road, not on you)?

C)Compare the two stresses and discuss

D)If the trucker from part A raised 4 of the wheels so they wouldn't touch the ground, this would reduce tire wear. However what would the new stress on the road ?

Question

A)How much sress is exerted on a road by an 18-wheel track which has a mass of 15 x 10^5 kg anda footprint area of each tire of .04m^2 

B)A 1200 kg auto is supported by 4 tires each with an area of .0015m^2. How much stress does this cause(on the road, not on you)?

C)Compare the two stresses and discuss

D)If the trucker from part A raised 4 of the wheels so they wouldn't touch the ground, this would reduce tire wear. However what would the new stress on the road ?

anonymous · Answer

non-related sorry, but dude, what kind of math is this? CSI?

anonymous · Answer

$$Stress= \frac{F\perp}{A}$$

18 wheels means a divided  stress between 18 wheels.    / 18(if all wheels are the same)

$$Stresss=\frac{mg}{A}*18$$
$$stress=\frac{15*10^5kg *9.81 \frac{m}{s^2}}{.04m^2}/18=2.04 *10^5 Pa$$

anonymous · Answer

this is elasticity/ or physics math. fluid dynamics also relates to this sort of which i'm studying for my quiz.. I do my problems online because i find it faster

anonymous · Answer

@saadman a 610 kg racehorse is able to accelerate from rest to a speed of 14 m/s in only 1.1 s

a) Assume tat the acceleration of the racehorse is constant. what is the distance required by the horse to reach this speed?

b) What is the minimum amount of energy required by the horse to reach this speed?

c) how much power does the horse have during acceleration?
2 minutes ago  Block  

a) use $$x=\frac{1}{2}at^2 +v_{0x}t+S_{0x}$$
taking the derivative of this
$$v=at+v_{0x}$$
at when v=14, t=1.1
$$14=1.1a+0$$ (velocity equal zero at inital *at rest)

anonymous · Answer

$$\frac{14}{1.1}=a=12.73$$

anonymous · Answer

so now you use that acceleration

$$x=\frac{1}{2}(12.73)(1.1^2)+(0)(1.1)+0$$
\[x.7.7m

anonymous · Answer

$$x=7.7m$$

anonymous · Answer

$$Stress = \frac{F}{A}=1200*9.81/.0015*4=1.962 *10^5 kg$$

anonymous · Answer

C) the stress per each unit on the tires is quite close to the larger truck with larger tires. this is due to the stress being even distributed on the tires.

anonymous · Answer

The smaller tires are smaller therefore more stress is put on each unit of tire

anonymous · Answer

D) not calculated bu it would be considerably bigger