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

1 Answer
May 10, 2018

Answer:

The truck will be able to pull up to 144 kg.

Explanation:

This calculation is neglecting any friction or gravity acting on the truck, so it's net force is 2400 N up the slope. Also, I'll take theta as #pi/3# or 60˚. This is because #(2pi)/3# is 120˚ and takes you beyond a vertical slope and back around to 60˚.

The parallel component of the gravitational force on the box is given by

#F_(g|\|)=mgsin(theta)#

This is the force that gravity exerts on the box down the slope.

The normal force acting on the box is given by the perpendicular component of the gravitational force

#N=mgcos(theta)#

The frictional force is then given by

#F_f=mu_sN=mu_smgcos(theta)#

#F_(g|\|)# and #F_f# are the opposing forces so if we set the sum of these equal to 2400 N, that will allow us to solve for the maximum mass

#F_(g|\|)+F_f=2400#

#rArrmgsin(theta)+mu_smgcos(theta)=2400#

#rArrm(gsin(theta)+mu_sgcos(theta))=2400#

#rArrm=2400/(gsin(theta)+mu_sgcos(theta))#

#m=2400/(8.487+8.167)=144" kg"#

Where

#g=9.80" m/s"#

#mu_s=5/3#

#theta=pi/3#