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 #(5 pi )/8 # and the coefficient of friction is #3/7 #, what is the maximum mass that can be pulled up at one time?

1 Answer
Aug 2, 2017

Answer:

#m_"max" = 328# #"kg"#

Explanation:

We're asked to find the maximum mass the truck can pull up the incline at a time.

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NOTE: The friction force in this situation actually acts DOWN the incline, because the object (the boxes) is being moved upward.

I'd also like to point out that the incline ideally should have an angle of inclination between #0# and #pi/2#, so I'll choose the corresponding first-quadrant angle of #ul((3pi)/8#.

Becuase the maximum mass allowed to pul would give a net force of zero, the equation for the net horizontal force #sumF_x# in this situation is

#ul(sumF_x = F_"truck" - mgsintheta - f_s = 0#

where

  • #f# is the static friction force, equal to

#f_s = mu_sncolor(white)(aa)# (maximum static friction force)

#f_s = mu_smgcostheta#

Plugging this into the above equation, we have

#ul(sumF_x = F_"truck" - mgsintheta - mu_smgcostheta = 0#

Now what we do is solve for the mass, #m#:

#F_"truck" = mgsintheta + mu_smgcostheta#

Divide all terms by #m#:

#(F_"truck")/color(red)(m) = color(green)(gsintheta + mu_sgcostheta)#

Swap the terms #color(red)(m# and #color(green)(gsintheta + mu_sgcostheta#:

#ulbar(|stackrel(" ")(" "m = (F_"truck")/(g(sintheta + mu_scostheta))" ")|)#

We now plug in the variables

  • #F_"truck" = 3500color(white)(l)"N"#

  • #mu_s = 3/7#

  • #theta = (3pi)/8#:

#m = (3500color(white)(l)"N")/((9.81color(white)(l)"m/s"^2)[sin((3pi)/8) + 3/7cos((3pi)/8)]) = color(blue)(ulbar(|stackrel(" ")(" "328color(white)(l)"kg"" ")|)#