A truck pulls boxes up an incline plane. The truck can exert a maximum force of 2,400 N2,400N. If the plane's incline is (2 pi )/3 2π3 and the coefficient of friction is 5/8 58, what is the maximum mass that can be pulled up at one time?

1 Answer
Aug 20, 2017

m_"max" = 208mmax=208 "kg"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 00 and pi/2π2, so I'll choose the corresponding first quadrant angle of pi/3π3.

The forces acting on the boxes are

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

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

  • the truck's upward pulling force

In the situation where the mass mm is maximum, the boxes will be in equilibrium; i.e. the net force acting on them will be 00.

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" = 2400 "N"

  • mu_k = 5/8

  • theta = (pi)/3

  • and g = 9.81 "m/s"^2

Plugging in these values:

color(blue)(m) = (2400color(white)(l)"N")/((9.81color(white)(l)"m/s"^2)(5/8cos[(pi)/3] + sin[(pi)/3])) = color(blue)(ulbar(|stackrel(" ")(" "208color(white)(l)"kg"" ")|)