Two crates of mass 65 kg and 125 kg are in contact and at rest on a horizontal surface (figure below). A force of 650 N is exerted on the 65 kg crate. If the coefficient of kinetic friction is 0.18, calculate:
a) the acceleration of the system and
b) the force that each crate exerts on the other. Draw a FBD (full body diagram). Show all work.
Answer:
Gravitational acceleration = g = 9.81 m/s2
Mass of the first crate = m1 = 65 kg
Mass of the second crate = m2 = 125 kg
Force acting on the crates = F = 650 N
Coefficient of kinetic friction = \mu = 0.18
Acceleration of the system = a
Force exerted by the crates on each other = F12
Normal force on the first crate = N1
Friction force on the first crate = f1
Normal force on the second crate = N2
Friction force on the second crate = f2
From the free body diagram,
In the vertical direction for the first crate,
m1g = N1
f1 = \muN1
f1 = \mum1g
In the vertical direction for the second crate,
m2g = N2
f2 = \muN2
f2 = \mum2g
In the horizontal direction for the first crate,
m1a = F - F12 - f1
F12 = F - f1 - m1a
In the vertical direction for the second crate,
m2a = F12 - f2
m2a = F - f1 - m1a - f2
(m1 + m2)a = F - \mum1g - \mum2g
(m1 + m2)a = F - \mug(m1 + m2)
(65 + 125)a = 650 - (0.18)(9.81)(65 + 125)
a = 1.655 m/s2
F12 = F - f1 - m1a
F12 = F - \mum1g - m1a
F12 = 650 - (0.18)(65)(9.81) - (65)(1.655)
F12 = 427.648 N
a) Acceleration of the system = 1.655 m/s2
b) Force that crates exert on each other = 427.648 N
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