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

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

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Answer 1 · 2017-04-19T07:01:03

The mass is $= 224.1 k g$ $=224.1kg$

Explanation:

Let the mass be $m k g$ $m kg$

Force of truck is $F = 3500 N$ $F=3500N$

The coefficient of friction is $μ$ $mu$

$μ = \frac{F_{r}}{N}$ $mu=F_r/N$

$N = m g cos θ$ $N=mgcostheta$

$F_{r} = μ \cdot N = μ \cdot m g cos θ$ $F_r=mu*N=mu*mgcostheta$

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

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

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

$= \frac{3500}{9.8 (\frac{7}{4} \cdot cos (\frac{3}{8} π) + sin (\frac{3}{8} π))}$ $=3500/(9.8(7/4*cos(3/8pi)+sin(3/8pi)))$

$= 224.1 k g$ $=224.1kg$

Answer 2 · 2017-04-19T08:06:38

$m = 224.4 k g$ $m=224.4 " "kg$

Explanation:

enter image source here

$Firstly,look into the animation several times to catch the situations that is given below.$ $"Firstly,look into the animation several times to catch the situations that is given below."$

the vector of weight is perpendicular to the horizontal surface.
The vector of weight can be splitted into two component
The component that is perpendicular to the inclined plane causes a Frictional force.
The yellow vector represent the react force that it has applied inclined plane.
The component that is parallel to the inclined plane causes sliding down..
A Tension occurs on rope (T)

enter image source here

$Weight of object (the red vector): G = m \cdot g$ $color(red)("Weight of object (the red vector):"G=m*g)$
$the vertical component that is perpendicular to the inclined plane(the blue vector) : G_{y} = m g \cdot cos θ$ $color(blue)("the vertical component that is perpendicular to the inclined plane(the blue vector) :"G_y=mg*cos theta )$
$the horizontal component that is parallel to the inclined plane(the green vector) F_{x} = m g \cdot sin θ$ $color(green)("the horizontal component that is parallel to the inclined plane(the green vector)"F_x=mg*sin theta)$

$The Frictional Force: F_{f} = μ_{k} \cdot G_{y}$ $"The Frictional Force: "F_f=mu_k*G_y$
$F_{f} = μ_{k} \cdot m g \cdot cos θ$ $F_f=mu_k*mg*cos theta$

$T \geq F_{f} + F_{x}$ $T>=F_f+F_x$

$T \geq μ_{k} \cdot m g \cdot cos θ + m g \cdot sin θ$ $T>=mu_k*mg*cos theta+mg*sin theta$

$3500 = m g (μ_{k} \cdot cos θ + sin θ)$ $3500=mg(mu_k*cos theta+sin theta)$

$m = \frac{3500}{g (μ_{k} \cdot cos θ + sin θ)}$ $m=3500/(g(mu_k*cos theta +sin theta))$

$m = \frac{3500}{9.81 (\frac{7}{4} \cdot 0.383 + 0.924)}$ $m=3500/(9.81(7/4*0.383+0.924))$

$m = \frac{3500}{9.81 \cdot 1.59}$ $m=3500/(9.81*1.59)$

$m = \frac{3500}{15.5970}$ $m=3500/(15.5970)$

$m = 224.4 k g$ $m=224.4" "kg$

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

2 Answers

Explanation:

Explanation:

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A truck pulls boxes up an incline plane. The truck can exert a maximum force of 3,500 N3,500N3,500 N. If the plane's incline is (3 pi )/8 3π8(3 pi )/8 and the coefficient of friction is 7/4 747/4 , what is the maximum mass that can be pulled up at one time?

2 Answers

Explanation:

Explanation:

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Impact of this question

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