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
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
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
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
ul(f_k = mu_kn
And since the normal force
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
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"" ")|)