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 #(3 pi )/8 # and the coefficient of friction is #7/4 #, what is the maximum mass that can be pulled up at one time?

2 Answers
Apr 19, 2017

Answer:

The mass is #=224.1kg#

Explanation:

Let the mass be #m kg#

Force of truck is #F=3500N#

The coefficient of friction is #mu#

#mu=F_r/N#

#N=mgcostheta#

#F_r=mu*N=mu*mgcostheta#

Resolving in the direction patrallel to the plane #↗^+#

#F=mumgcostheta+mgsintheta#

#m=F/(g(mucostheta+sintheta))#

#=3500/(9.8(7/4*cos(3/8pi)+sin(3/8pi)))#

#=224.1kg#

Apr 19, 2017

Answer:

#m=224.4 " "kg#

Explanation:

enter image source here

#"Firstly,look into the animation several times to catch the situations that is given below."#

  • the vector of weight is perpendicular to the horizontal surface.
  • The vector of weight can be splitted into two component
  • The component that is perpendicular to the inclined plane causes a Frictional force.
  • The yellow vector represent the react force that it has applied inclined plane.
  • The component that is parallel to the inclined plane causes sliding down..
  • A Tension occurs on rope (T)

enter image source here

#color(red)("Weight of object (the red vector):"G=m*g)#
#color(blue)("the vertical component that is perpendicular to the inclined plane(the blue vector) :"G_y=mg*cos theta )#
#color(green)("the horizontal component that is parallel to the inclined plane(the green vector)"F_x=mg*sin theta)#

#"The Frictional Force: "F_f=mu_k*G_y#
#F_f=mu_k*mg*cos theta#

#T>=F_f+F_x#

#T>=mu_k*mg*cos theta+mg*sin theta#

#3500=mg(mu_k*cos theta+sin theta)#

#m=3500/(g(mu_k*cos theta +sin theta))#

#m=3500/(9.81(7/4*0.383+0.924))#

#m=3500/(9.81*1.59)#

#m=3500/(15.5970) #

#m=224.4" "kg#