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 )/12 # and the coefficient of friction is #7/12 #, what is the maximum mass that can be pulled up at one time?

1 Answer
Jan 17, 2017

Maximum mass is # 438.24 \ kg# (2dp)

Explanation:

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For our diagram, #m="mass "kg#, #theta=(5pi)/12 #

If we apply Newton's Second Law up perpendicular to the plane we get:

# R - mg cos theta = 0 #
# :. R = mg cos((5pi)/12) \ \ N#

With a maximum driving force of #3500\ N# upwards (along the plane) #D=3500#. If we Apply Newton's Second Law up parallel to the plane we get:

# D + F - mg sin theta = 0 #
# :. 3500 +F - mg sin ((5pi)/12) = 0#
# :. F = mg sin ((5pi)/12) - 3500#

And the friction is related to the Reaction (Normal) Force by

# F le mu R #
# :. mg sin ((5pi)/12) - 3500 le 7/12 (mg cos((5pi)/12) )#
# :. 12mg sin ((5pi)/12) - 42000 le 7mg cos((5pi)/12) #
# :. 12mg sin ((5pi)/12) - 7mg cos((5pi)/12) le 42000 #
# :. mg(12 sin ((5pi)/12) - 7 cos((5pi)/12)) le 42000 #
# :. m le 42000/((12 sin ((5pi)/12) - 7 cos((5pi)/12))g) #
# :. m le 438.2400291 ... #