Question

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.

Answer 1

Answer:

39.81 N

Explanation:

I attached an image of the free body diagrams I drew of crate #1 and #2.  

Using these diagram, we can set up a system of equations for the sum of forces in the x and y direction.

∑Fₓ = maₓ

∑Fᵧ = maᵧ

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.