Two crates, of mass m1 = 70 kg and m2 = 110 kg, are in contact and at rest on a horizontal surface (Fig. 4-54)

Two crates, of mass m1 = 70 kg and m2 = 110 kg, are in contact and at rest on a horizontal surface (Fig. 4-54). A 620 N force is exerted on the 70 kg crate. The coefficient of kinetic friction is 0.15.



Figure 4-54



(a) Calculate the acceleration of the system.

m/s2 (to the right)

(b) Calculate the force that each crate exerts on the other.

N

(c) Repeat with the crates reversed. acceleration m/s2

crate force N

Two crates, of mass m1 = 70 kg and m2 = 110 kg, are in contact and at rest on a horizontal surface (Fig. 4-54)
(a)It is not clear if the crate m2 is on the right of the crate m1 nor it clear in which direction is the force being exerted. Let’s assume that the force if pushing both masses to the right.

F=f + F’ (the force applied equal to the force of friction (f) plus the force due to inertia (F’) )



f = u(N1+N2) = u(m1g +m2g)

f= u(m1 +m2)g

F’=m1 a + m2 a=(m1+m2)a

a=F’/(m1+m2)

F’=F-f= F- u(m1 +m2)g

Finally

a= (F- u(m1 +m2)g)/ (m1+m2)

a=(620 - 0.15(70 + 110) 9.81)/(70+110)

a=1.97m/s^2



(b) m1 is pushing m2 as m2 returns the favor

F=m2 a=110 (1.97) = 217 N

(c) If we reverse the crates the acceleration remains unchanged since the system still has the same force, mass and kinetic friction.

However now the force between the crates is different

F=m1a=70 (1.97)= 137.9N