Question

A truck pulls boxes up an incline plane. The truck can exert a maximum force of 4,000N. If the plane's incline is `(5π)/8` and the coefficient of friction is `9/8`, what is the maximum mass that can be pulled up at one time? Thanks

Answer 1

Consider the physical situation, pictured in the free-body diagram,

Given,

`mu_k = 1.125`

`theta = 112.5°`

To figure out the maximum mass the truck can pull, we set,

`SigmaF_x = 0`

Hence, the truck can pull,

`SigmaF_y = F_"N" - mgcostheta = 0`

`=> F_"N" = mgcostheta`

`SigmaF_x = F_"P" - mgsintheta - mu_kmgcostheta = 0`

`=> m = F_"P"/(gsintheta - mu_kgcostheta) approx 422"kg"`

is the maximum mass the truck can pull up the ramp at once.

By extension, any mass `< m` can be pulled up the ramp by the truck.

Answer 2

`color(blue)(827.3kg)`

Explanation:

If we view this as the mass is about to slip up the plane, then:

The frictional force `muR` is acting downwards.

Resolving:

Perpendicular: `\ \ \ \ \ \ R=mgcos((5pi)/8)`

Parallel: `\ \ \ \ \4000=Fr+mgsin((5pi)/8)`

Limiting equilibrium: `Fr=muR`

Substituting:

`Fr=mumgcos((5pi)/8)`

`4000=mumgcos((5pi)/8)+mgsin((5pi)/8)`

`4000=m(mugcos((5pi)/8)+gsin((5pi)/8))`

`m=4000/(g(mugcos((5pi)/8)+sin((5pi)/8))`

`mu=9/8` , `g=9.8`

`m=4000/(9.8(9/8cos((5pi)/8)+sin((5pi)/8)))=827.3121256`

Maximum mass that can be pulled by the truck:

`color(blue)(827.3kg)`

Answer 3

This is what I get

Explanation:

Normally text books state angle of incline as less than `90^@`. This is peculiar case of an angle of incline which is given as greater than `90^@`.
Another thing to be noted is that `mu` is given as greater than `1`.
Now as per convention angles are defined with respect to positive `x`-axis, in anti clockwise direction. As such in the figure `angletheta=(3pi)/8`.

Maximum pulling force of truck `=4000\ N`
Downward pulling force along the incline`=mgsintheta+mumgcostheta`

Equating both for `m_max`, we gat

`m_max g(sintheta+mucostheta)=4000`
`=>m_max=4000/(9.81(sin67.5^@+9/8xxcos67.5^@))`
`=>m_max=301\ kg`