A truck pulls boxes up an incline plane. The truck can exert a maximum force of $6,800 N$ . If the plane's incline is $(5 pi )/8$ and the coefficient of friction is $4/5$ , what is the maximum mass that can be pulled up at one time?

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Answer 1 · 2017-06-29T12:55:54

The mass is $=1123.3kg$

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Resolving in the direction parallel to the plane $↗^+$

Let the force exerted by the truck be $=FN$

Let the frictional force be $=F_rN$

The coefficient of friction $mu=F_r/N$

The normal force is $N=mgcostheta$

Therefore,

$F=F_r+mgsintheta$

$=muN+mgsintheta$

$=mumgcostheta+mgsintheta$

$m=F/(g(mucostheta+sintheta))$

$=6800/(9.8(4/5*cos(5/8pi))+sin(5/8pi))$

$=1123.3kg$

A truck pulls boxes up an incline plane. The truck can exert a maximum force of 6,800 N6,800 N. If the plane's incline is (5 pi )/8 (5 pi )/8 and the coefficient of friction is 4/5 4/5 , what is the maximum mass that can be pulled up at one time?