Questions : Two blocks of masses m and 2m are placed in contact on a smooth horizontal surface. Two forces of magnitudes 10 N and 16 N respectively are applied on system as shown in figure. Normal contact force between the blocks is _____

Doubt by Divya
Solution :
F1=+10 N
F2=-16 N
F2=-16 N
Net Force on the system
F=F1+F2
F=10-16
F=-6 N [Towards Left]
F=F1+F2
F=10-16
F=-6 N [Towards Left]
(m1+m2)a=-6
(m+2m)a=-6
(3m)×a=-6
a=-6/3m
a=-2/m — (1)
(m+2m)a=-6
(3m)×a=-6
a=-6/3m
a=-2/m — (1)
-ve sign shows that the acceleration is acting towards the left side.
Let the normal contact force between the two given blocks be N
Considering the first block only
F1+N = m1(a)
+10+(-N)=m(-2/m)
10-N=-2
10+2=N
12=N
N=12 N
Considering the first block only
F1+N = m1(a)
+10+(-N)=m(-2/m)
10-N=-2
10+2=N
12=N
N=12 N
If we consider the second block only
F2+N = m2(a)
-16+N=2m(-2/m)
-16+N=-4
-16+4=-N
-12=-N
N=12 N
-16+N=2m(-2/m)
-16+N=-4
-16+4=-N
-12=-N
N=12 N
Hence, the normal contact force between the block is 12 N.