Question

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

Answer 1

The mass is `=224.1kg`

Explanation:

Let the mass be `m kg`

Force of truck is `F=3500N`

The coefficient of friction is `mu`

`mu=F_r/N`

`N=mgcostheta`

`F_r=mu*N=mu*mgcostheta`

Resolving in the direction patrallel to the plane `↗^+`

`F=mumgcostheta+mgsintheta`

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

`=3500/(9.8(7/4*cos(3/8pi)+sin(3/8pi)))`

`=224.1kg`

Answer 2

`m=224.4 " "kg`

Explanation:

`"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)

`color(red)("Weight of object (the red vector):"G=m*g)`
`color(blue)("the vertical component that is perpendicular to the inclined plane(the blue vector) :"G_y=mg*cos theta )`
`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=mu_k*G_y`
`F_f=mu_k*mg*cos theta`

`T>=F_f+F_x`

`T>=mu_k*mg*cos theta+mg*sin theta`

`3500=mg(mu_k*cos theta+sin theta)`

`m=3500/(g(mu_k*cos theta +sin theta))`

`m=3500/(9.81(7/4*0.383+0.924))`

`m=3500/(9.81*1.59)`

`m=3500/(15.5970)`

`m=224.4" "kg`