A truck pulls boxes up an incline plane. The truck can exert a maximum force of $4,500 N$ . If the plane's incline is $(2 pi )/3$ and the coefficient of friction is $7/6$ , what is the maximum mass that can be pulled up at one time?

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Answer 1 · 2017-12-30T15:15:58

The mass is $=1624.3kg$

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

Let the force exerted by the truck be $F=4500N$

Let the frictional force be $=F_rN$

The coefficient of friction $mu=F_r/N=7/6$

The normal force is $N=mgcostheta$

The angle of the plane is $theta=2/3pi$

The acceleration due to gravity is $g=9.8ms^-2$
Therefore,

$F=F_r+mgsintheta$

$=muN+mgsintheta$

$=mumgcostheta+mgsintheta$

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

$=4500/(9.8(7/6*cos(2/3pi)+sin(2/3pi))$

$=1624.3kg$

A truck pulls boxes up an incline plane. The truck can exert a maximum force of 4,500 N4,500 N. If the plane's incline is (2 pi )/3 (2 pi )/3 and the coefficient of friction is 7/6 7/6 , what is the maximum mass that can be pulled up at one time?