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:

Question

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 1

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