Two blocks of masses 3 kg and 2 kg are placed beside each other contact with each other on a rough horizontal surface. A horizontal force of 20N is applied on 3 kg. the coefficient of friction between blocks and the surface is 0.1 and . The force of contact between the two blocks is
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Total friction force acting on two blocks will be
F_f = \mu*m_1g + \mu*m_2gF
f
=μ∗m
1
g+μ∗m
2
g
F_f = 0.1*(2+3)*10 = 5 NF
f
=0.1∗(2+3)∗10=5N
now using II law of Newton
F_{net} = F - F_fF
net
=F−F
f
F_{net} = 20 - 5 = 15 NF
net
=20−5=15N
as we know that
F_{net} = m*aF
net
=m∗a
15 = (2+3)*a15=(2+3)∗a
a = 3 m/s^2a=3m/s
2
now using the force diagram of 3 kg block we can say
F_{net} = m*aF
net
=m∗a
F_n - F_f = m*aF
n
−F
f
=m∗a
F_n - 0.1*3*10 = 3* 3F
n
−0.1∗3∗10=3∗3
F_n = 3 + 9 = 12 NF
n
=3+9=12N
so the normal force or contact force between two blocks will be 12 N
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