A steel cable with a radius of 1.5 ; cm suppo | vedclass

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(B) The formula for stress is given by $	ext{Stress} = \frac{F}{A}$,where $F$ is the force (load) and $A$ is the cross-sectional area.Given radius $r

Vedclass · Accepted Answer

$7.065 	imes 10^{4} \; N$

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(B) The formula for stress is given by $	ext{Stress} = \frac{F}{A}$,where $F$ is the force (load) and $A$ is the cross-sectional area.Given radius $r = 1.5 \; cm = 0.015 \; m$.The cross-sectional area $A = \pi r^2 = \pi (0.015)^2 \; m^2$.Given maximum stress $\sigma_{max} = 10^8 \; N m^{-2}$.Maximum load $F_{max} = \sigma_{max} 	imes A$.$F_{max} = 10^8 	imes \pi 	imes (0.015)^2$.$F_{max} = 10^8 	imes 3.14159 	imes 0.000225$.$F_{max} = 7.0685 	imes 10^4 \; N \approx 7.065 	imes 10^4 \; N$.Thus,the cable can support a maximum load of $7.065 	imes 10^4 \; N$.

$A$ steel cable with a radius of $1.5 \; cm$ supports a chairlift at a ski area. If the maximum stress is not to exceed $10^{8} \; N m^{-2}$,what is the maximum load the cable can support?

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