Magnitude of applied force equation

Question

 Two blocks connected by a string are pulled across a horizontal surface by a force applied to one of the blocks, as shown to the right. The mass of the left block m₁ = 1.1 kg and the mass of the right block m₂ = 4.1 kg. The angle between the applied force and the horizontal is θ = 64°. The coefficient of kinetic friction between the blocks and the surface is μ = 0.18. Each block has an acceleration of a = 3.6 m/s² to the right.

Write an equation for the magnitude of the applied force F acting on the second block. Use variables from the problem statement as well as using g for the gravitational acceleration.

Answer 1

FBD of the second block

Sum force X = m2a

Fcos θ - T -μN2 = m2a   (1)

Sum force Y = 0

-m2g + N2 + Fsin θ = 0

N2 = m2g -Fsin θ   (2)

Sub (2) into (1)

Fcos θ - T -μ(m2g -Fsin θ )=m2a  

Fcos θ = m2a + T+ μ(m2g -Fsin θ )   (3)

FBD of the first block

Sum force X = m1a

T - μN1 = m1a   (4)

Sum force Y = 0

N1 - m1g = 0

N1 = m1g   (5)

Sub (5) into (4)

T - μ*m1g= m1a

T = m1a + μ*m1g

T = m1 (a + μ*g)   (6)

Sub (6) into (4)

Fcos θ = m2a + m1 (a + μ*g) + μ(m2g -Fsin θ )

Fcos θ +μ(Fsin θ )= m2a + m1 (a + μ*g) + μ(m2g)

F(cos θ +μsinθ )= m2(a+ μg) + m1 (a + μg)

Plug knows into this equation :)

F = (m2+m1)(a+ μg)/(cos θ +μsinθ )