A truck pulls boxes up an incline plane. The truck can exert a maximum force of 4,000N. If the plane's incline is #(5π)/8# and the coefficient of friction is #9/8#, what is the maximum mass that can be pulled up at one time? Thanks

3 Answers
Apr 30, 2018

Consider the physical situation, pictured in the free-body diagram,

puu.sh

Given,

#mu_k = 1.125#

#theta = 112.5°#

To figure out the maximum mass the truck can pull, we set,

#SigmaF_x = 0#

Hence, the truck can pull,

#SigmaF_y = F_"N" - mgcostheta = 0#

#=> F_"N" = mgcostheta#

#SigmaF_x = F_"P" - mgsintheta - mu_kmgcostheta = 0#

#=> m = F_"P"/(gsintheta - mu_kgcostheta) approx 422"kg"#

is the maximum mass the truck can pull up the ramp at once.

By extension, any mass #< m# can be pulled up the ramp by the truck.

Apr 30, 2018

#color(blue)(827.3kg)#

Explanation:

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If we view this as the mass is about to slip up the plane, then:

The frictional force #muR# is acting downwards.

Resolving:

Perpendicular: # \ \ \ \ \ \ R=mgcos((5pi)/8)#

Parallel: # \ \ \ \ \4000=Fr+mgsin((5pi)/8)#

Limiting equilibrium: #Fr=muR#

Substituting:

#Fr=mumgcos((5pi)/8)#

#4000=mumgcos((5pi)/8)+mgsin((5pi)/8)#

#4000=m(mugcos((5pi)/8)+gsin((5pi)/8))#

#m=4000/(g(mugcos((5pi)/8)+sin((5pi)/8))#

#mu=9/8# , #g=9.8#

#m=4000/(9.8(9/8cos((5pi)/8)+sin((5pi)/8)))=827.3121256#

Maximum mass that can be pulled by the truck:

#color(blue)(827.3kg)#

May 1, 2018

This is what I get

Explanation:

www.texasgateway.org
Normally text books state angle of incline as less than #90^@#. This is peculiar case of an angle of incline which is given as greater than #90^@#.
Another thing to be noted is that #mu# is given as greater than #1#.
Now as per convention angles are defined with respect to positive #x#-axis, in anti clockwise direction. As such in the figure #angletheta=(3pi)/8#.

Maximum pulling force of truck #=4000\ N#
Downward pulling force along the incline#=mgsintheta+mumgcostheta#

Equating both for #m_max#, we gat

#m_max g(sintheta+mucostheta)=4000#
#=>m_max=4000/(9.81(sin67.5^@+9/8xxcos67.5^@))#
#=>m_max=301\ kg#