Question

A truck pulls boxes up an incline plane. The truck can exert a maximum force of `2,200 N`. If the plane's incline is `(7 pi )/8` and the coefficient of friction is `1/3`, what is the maximum mass that can be pulled up at one time?

Answer 1

`m_"max" = 325` `"kg"`

Explanation:

We're asked to find the maximum mass that the truck is able to pull up the ramp, given the truck's maximum force, the ramp's angle of inclination, and the coefficient of friction.

NOTE: Ideally, the angle of inclination should be between `0` and `pi/2`, so I'll choose the corresponding first quadrant angle of `pi/8`.

The forces acting on the boxes are

• the gravitational acceleration (acting down the incline), equal to `mgsintheta`

• the friction force `f_k` (acting down the incline because it opposes motion)

• the truck's upward pulling force

In the situation where the mass `m` is maximum, the boxes will be in equilibrium; i.e. the net force acting on them will be `0`.

We thus have our net force equation

`sumF_x = overbrace(F_"truck")^"upward"- overbrace(f_k - mgsintheta)^"downward" = 0`

The expression for the frictional force `f_k` is given by

`ul(f_k = mu_kn`

And since the normal force `color(green)(n = mgcostheta`, we can plug that in above:

`sumF_x = F_"truck" - overbrace(mu_kcolor(green)(mgcostheta))^(f_k = mu_kcolor(green)(n)) - mgsintheta = 0`

Or

`sumF_x = F_"truck" - mg(mu_kcostheta + sintheta) = 0`

Therefore,

`ul(F_"truck" = mg(mu_kcostheta + sintheta))color(white)(aaa)` (net upward force`=`net downward force)

And if we rearrange this equation to solve for the maximum mass `m`, we have

`color(red)(ulbar(|stackrel(" ")(" "m = (F_"truck")/(g(mu_kcostheta + sintheta))" ")|)`

The problem gives us

• `F_"truck" = 2200` `"N"`

• `mu_k = 1/3`

• `theta = (pi)/8`

• and `g = 9.81` `"m/s"^2`

Plugging in these values:

`m = (2200color(white)(l)"N")/((9.81color(white)(l)"m/s"^2)(1/3cos[(pi)/8] + sin[(pi)/8])) = color(blue)(ulbar(|stackrel(" ")(" "325color(white)(l)"kg"" ")|)`