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 $\frac{π}{8}$ $pi/8$ and the coefficient of friction is $\frac{6}{5}$ $6/5$ , what is the maximum mass that can be pulled up at one time?

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Answer 1 · 2016-05-20T09:07:06

$m \leq 309.11 k g$ $m<=309.11 kg$

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$T \geq m \cdot g \cdot sin α + u_{k} \cdot m \cdot g \cdot cos α$ $T>=m*g*sin alpha +u_k*m*g*cos alpha$

$4500 = m \cdot g \cdot sin (\frac{π}{8}) + \frac{6}{5} \cdot m \cdot g \cdot cos (\frac{π}{8})$ $4500=m*g*sin(pi/8)+6/5*m*g*cos(pi/8)$

$sin (\frac{π}{8}) = 0.38$ $sin(pi/8)=0.38$

$cos (\frac{π}{8}) = 0.92$ $cos(pi/8)=0.92$

$4500 = m \cdot g (0.38 + \frac{6}{5} \cdot 0.92)$ $4500=m*g(0.38+6/5*0.92)$

$4500 = m \cdot g (0.38 + 1.104)$ $4500=m*g(0.38+1.104)$

$4500 = m \cdot g \cdot 1.484$ $4500=m*g*1.484$

$g = 9.81 \frac{N}{k g}$ $g=9.81 N/(kg)$

$4500 = 9.81 \cdot m \cdot 1.484$ $4500=9.81*m*1.484$

$4500 = 14.55804 \cdot m$ $4500=14.55804*m$

$m = \frac{4500}{14.55804}$ $m=4500/(14.55804)$

$m \leq 309.11 k g$ $m<=309.11 kg$

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 π8pi/8 and the coefficient of friction is 656/5 , what is the maximum mass that can be pulled up at one time?