A truck pulls boxes up an incline plane. The truck can exert a maximum force of 4,500 N4,500N. If the plane's incline is pi/8 π8 and the coefficient of friction is 4/3 43, what is the maximum mass that can be pulled up at one time?

1 Answer
Jan 15, 2018

The mass is =284.4kg=284.4kg

Explanation:

![http://spmphysics.onlinetuition.com.my/2013/06/http://inclined-plane.html](https://useruploads.socratic.org/3SLPG4SgRIu153RhGHJq_inclined%20plane.jpg)

Resolving in the direction parallel to the plane ↗^++

Let the force exerted by the truck be F=4500NF=4500N

Let the frictional force be =F_rN=FrN

The coefficient of friction mu=F_r/N=4/3μ=FrN=43

The normal force is N=mgcosthetaN=mgcosθ

The angle of the plane is theta=1/8piθ=18π

The acceleration due to gravity is g=9.8ms^-2g=9.8ms2
Therefore,

F=F_r+mgsinthetaF=Fr+mgsinθ

=muN+mgsintheta=μN+mgsinθ

=mumgcostheta+mgsintheta=μmgcosθ+mgsinθ

m=F/(g(mucostheta+sintheta))m=Fg(μcosθ+sinθ)

=4500/(9.8(4/3*cos(1/8pi)+sin(1/8pi))=45009.8(43cos(18π)+sin(18π))

=284.4kg=284.4kg