What is the area of contact each tire makes with the road? (answer must be in $c{m}^{2}$) What gauge pressure is required to give an area of contact of $125c{m}^{2}$ for each tire? (answer must be in $lb/i{n}^{2}$)

Trace Glass
2022-10-17
Answered

The weight of your 1400 kg car is supported equally by its four tires, each inflated to a gauge pressure of 35.0 $lb/i{n}^{2}.$

What is the area of contact each tire makes with the road? (answer must be in $c{m}^{2}$) What gauge pressure is required to give an area of contact of $125c{m}^{2}$ for each tire? (answer must be in $lb/i{n}^{2}$)

What is the area of contact each tire makes with the road? (answer must be in $c{m}^{2}$) What gauge pressure is required to give an area of contact of $125c{m}^{2}$ for each tire? (answer must be in $lb/i{n}^{2}$)

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Jimena Torres

Answered 2022-10-18
Author has **20** answers

Given

$W=1400\times 9.8=13720N$

gauge pressure $=35.0lb/i{n}^{2}.$

in Pa =$35\times 6875Pa=\phantom{\rule{0ex}{0ex}}P=241.325Pa=241325P$a

Force on each type $F=\frac{13720}{4}=3430N$

$P=\frac{F}{A}$

Hence area of Contact

$A=\frac{F}{P}=\frac{3430}{241325}=0.014213{m}^{2}$

$A=1.4213\times {10}^{2}c{m}^{2}\phantom{\rule{0ex}{0ex}}A=142.13c{m}^{2}$

b) If $A=125c{m}^{2}=0.0125{m}^{2}\phantom{\rule{0ex}{0ex}}P=\frac{F}{A}=\frac{3430}{0.0125}=274400Pa\phantom{\rule{0ex}{0ex}}P=\frac{274410}{6895}=39.797lb/i{n}^{2}$

$W=1400\times 9.8=13720N$

gauge pressure $=35.0lb/i{n}^{2}.$

in Pa =$35\times 6875Pa=\phantom{\rule{0ex}{0ex}}P=241.325Pa=241325P$a

Force on each type $F=\frac{13720}{4}=3430N$

$P=\frac{F}{A}$

Hence area of Contact

$A=\frac{F}{P}=\frac{3430}{241325}=0.014213{m}^{2}$

$A=1.4213\times {10}^{2}c{m}^{2}\phantom{\rule{0ex}{0ex}}A=142.13c{m}^{2}$

b) If $A=125c{m}^{2}=0.0125{m}^{2}\phantom{\rule{0ex}{0ex}}P=\frac{F}{A}=\frac{3430}{0.0125}=274400Pa\phantom{\rule{0ex}{0ex}}P=\frac{274410}{6895}=39.797lb/i{n}^{2}$

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