Question

A force of magnitude 7.50 N pushes three boxes with masses m1= 1.30 kg,m2= 3.20 kg, and m3= 4.90 kg, as shown in the figure. (Ignore friction.)

Find the magnitude of the contact force between boxes 1 and 2

Find the magnitude of the contact force between boxes 2 and 3.

Answer 1

1: 6.46 N,

(b)The contact force between boxes 2 and 3 is
F2 = F1 - m2 * a

Answer 2

2: 3.91 N.

Answer 3

(a)The acceleration of the three boxes is
a = (F/m1 + m2 + m3)
where F = 7.50 N ,m1 = 1.30 kg,m2 = 3.20 kg and m3 = 4.90 kg
The contact force between boxes 1 and 2 is
F1 = F - m1 * a

Answer 4

a = F/m = 7.5/(1.3+3.2+4.9) = 7.5/9.4 = 0.798 m/s^2

The force between 1.3 kg and 3.2 kg boxes is a*m = 0.798*1.3 = 1.037 N

The force between 3.2 kg and 4.9 kg boxes is a*m = 0.798*(1.3+3.2) = 3.591 N