Question

(Newton's laws/gravitation force/normal force/friction force, 4/ea.) A 60-N force pushes two objects next to each other with masses m_1 = 8 kg, and m_2 = 4 kg, across a horizontal solid surface. As shown below, the force is directed 30 degree below the horizontal. The coefficient of kinetic friction is 0.15, same for both objects. What is the normal force on m_1? What is the normal force on m_2? What is the acceleration of the masses? What is the force the m1 exerts on m_2? What is the force the m2 exerts on m_1?

Answer 1

Free body diagram of m₁ and m₂

Fcos300 N1 Fsin300 f2 fı m1g

Applying Newton's law of motion along vertical direction for mass m₁

N₁= m₁g + Fsin30⁰-----------(1)

f₁ = 0.15*N₁

Applying Newton's law of motion along horizontal direction for mass m₁

Fcos30⁰ - N - f₁ =  m₁a

=> Fcos30⁰ - N - 0.15*(m₁g + Fsin30⁰) =  m₁a-----------(2)

Applying Newton's law of motion along vertical direction for mass m₂

N₂= m₂g -----------(3)

f₂= 0.15*N₂

Applying Newton's law of motion along horizontal direction for mass m₂

N - f₂ =  m₂a

=>N - 0.15*(m₂g ) =  m₂a-----------(4)

From equation 2 and 4

a = [Fcos30⁰ - 0.15*(m₁g +m₂g + Fsin30⁰)]/[m₁ +m₂ ]

From equation 4.

N = 0.15*(m₂g ) + m₂a

So answers are

(a) N₁= m₁g + Fsin30⁰

(b) N₂= m₂g

(c) a = [Fcos30⁰ - 0.15*(m₁g +m₂g + Fsin30⁰)]/[m₁ +m₂ ]

(d) N = 0.15*(m₂g ) + m₂a

(e) N = 0.15*(m₂g ) + m₂a