A truck pulls boxes up an incline plane. The truck can exert a maximum force of $2, 500 N$ $2,500 N$ . If the plane's incline is $\frac{7 π}{8}$ $(7 pi )/8$ and the coefficient of friction is $\frac{1}{3}$ $1/3$ , what is the maximum mass that can be pulled up at one time?

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Answer 1 · 2017-11-23T13:37:14

The maximum mass is $= 3413.9 k g$ $=3413.9kg$

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

Let the force exerted by the truck be $F = 2500 N$ $F=2500N$

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

The coefficient of friction $μ = \frac{F_{r}}{N} = \frac{1}{3}$ $mu=F_r/N=1/3$

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

The angle of the plane is $θ = \frac{7}{8} π$ $theta=7/8pi$

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

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

$= μ N + m g sin θ$ $=muN+mgsintheta$

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

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

$= \frac{2500}{9.8 (\frac{1}{3} \cdot cos (\frac{7}{8} π) + sin (\frac{7}{8} π))}$ $=2500/(9.8(1/3*cos(7/8pi)+sin(7/8pi))$

$= 3413.9 k g$ $=3413.9kg$

A truck pulls boxes up an incline plane. The truck can exert a maximum force of 2,500N2,500 N. If the plane's incline is 7π8(7 pi )/8 and the coefficient of friction is 131/3 , what is the maximum mass that can be pulled up at one time?