Three packing crates of masses, M1 = 6 kg, M2 = 2 kg and M3 = 8 kg are connected by a light string of negligible mass that passes over the pulley as shown. Masses M1 and M3 lies on a 30o incline plane which slides down the plane. The coefficient of kinetic friction on the incline plane is 0.28. A. Draw a free body diagram of all the forces acting in the masses M1 and M2. B. Determine the tension in the string that connects M2 and M3.

Let's start with the free body diagram for crate #2. Let's set the positive direction on top and the negative direction on the bottom. We can see that the forces acting on crate #2 are in the y-direction, so let's use Newton's 2nd Law to write this equation:

∑Fᵧ = maᵧ
T₁ - m₂g = m₂aᵧ

Note that the tension and acceleration are constant throughout the system since the string has a negligible mass. Therefore, we don't really need to write the subscripts under T and a, but I am doing so just so there is no confusion.

Let's solve for T in the equation...

T₁ = m₂aᵧ + m₂g
T₁ = m₂(a + g)

We'll come back to this equation later. Now let's go to the free body diagram for crate #1.

We want to solve for the forces in the x-direction now. Let's set the leftwards direction to be positive and the rightwards direction to be negative.

∑Fₓ = maₓ

F_f - F_g sinΘ = maₓ

The normal force is equal to the x-component of the force of gravity.

(F_n · μ_k) - m₁g sinΘ = m₁aₓ

(F_g cosΘ · μ_k) - m₁g sinΘ = m₁aₓ
[m₁g cos(30) · 0.28] - [m₁g sin(30)] = m₁aₓ
[(6)(9.8)cos(30) · 0.28] - [(6)(9.8)sin(30)] = (6)aₓ
[2.539595871] - [-58.0962595] = 6aₓ
60.63585537 = 6aₓ
aₓ = 10.1059759 m/s²

Now let's go back to this equation:

T₁ = m₂(a + g)

We have 3 known variables and we can solve for the tension force.

T = 2(10.1059759 + 9.8)
T = 2(19.9059759)
T = 39.8119518 N

The tension force is the same throughout the string, therefore, the tension in the string connecting M2 and M3 is 39.81 N.