A truck pulls boxes up an incline plane. The truck can exert a maximum force of $4, 500 N$ $4,500 N$ . If the plane's incline is $\frac{7 π}{8}$ $(7 pi )/8$ and the coefficient of friction is $\frac{5}{3}$ $5/3$ , what is the maximum mass that can be pulled up at one time?

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Answer 1 · 2017-07-23T08:57:25

The mass is $= 396.8 k g$ $=396.8kg$

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

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

Let the frictional force be $= F_{r} N$ $=F_rN$

The coefficient of friction $μ = \frac{F_{r}}{N}$ $mu=F_r/N$

The normal force is $N = m g cos θ$ $N=mgcostheta$

Therefore,

$F = F_{r} + m g sin θ$ $F=F_r+mgsintheta$

$= μ N + m g sin θ$ $=muN+mgsintheta$

$= μ m g cos θ + m g sin θ$ $=mumgcostheta+mgsintheta$

$m = \frac{F}{g (μ cos θ + sin θ)}$ $m=F/(g(mucostheta+sintheta))$

$= \frac{4500}{9.8 (\frac{5}{3} \cdot cos (\frac{7}{8} π)) + sin (\frac{7}{8} π)}$ $=4500/(9.8(5/3*cos(7/8pi))+sin(7/8pi))$

$= - 396.8 k g$ $=-396.8kg$

A truck pulls boxes up an incline plane. The truck can exert a maximum force of 4,500N4,500 N. If the plane's incline is 7π8(7 pi )/8 and the coefficient of friction is 535/3 , what is the maximum mass that can be pulled up at one time?