The elevator may be moving, but if it is moving at a constant velocity, then the observer viewing the mass-rope system is in an inertial reference frame (non-accelerating) and Newton's laws of motion will apply in this reference frame.
A) Choose the point where the ropes intersect (the black dot above m₁) and set up equations of static equilibrium where the forces are acting on that point:
We'll assume that, because rope 3 is oriented vertically, T₃ also acts vertically.
Sum up the vertical components of the forces acting on the point. We will assign upward acting components as positive and downward acting components as negative.
∑Fy = 0
Eq 1: T₁sin(θ₁) + T₂sin(θ₂) - T₃ = 0
Sum up the horizontal components of the forces acting on the point. We will assign rightward acting components as positive and leftward acting components as negative.
∑Fx = 0
Eq 2: T₂cos(θ₂) - T₁cos(θ₁) = 0
T₃ is caused by the force of gravity acting on m₁ which is very easy to calculate:
T₃ = m₁g
m₁ = 3.00kg
g is the acceleration due to earth's gravity, 9.81m/s²
T₃ = 3.00×9.81
T₃ = 29.4N
Plug in known values into Eq. 1 and Eq. 2:
Eq. 1: T₁sin(38.0) + T₂sin(52.0) - 29.4 = 0
Eq. 2: T₂cos(52.0) - T₁cos(38.0) = 0
We can solve for T₁ and T₂ by use of substitution. First let us rearrange and simplify Eq. 2 like so:
T₂cos(52.0) = T₁cos(38.0)
T₂ = T₁cos(38.0)/cos(52.0)
T₂ = 1.28T₁
Now that we have T₂ isolated, we can substitute T₂ in Eq. 1 with 1.28T₁:
T₁sin(38.0) + 1.28T₁sin(52.0) - 29.4 = 0
Rearrange and simplify, and solve for T₁:
T₁(sin(38.0) + 1.28sin(52.0)) = 29.4
1.62T₁ = 29.4
T₁ = 18.1N
Recall from our previous work:
T₂ = 1.28T₁
Plug in T₁ = 18.1N and solve for T₂:
T₂ = 1.28×18.1
T₂ = 23.2N
B) We'll assume that, because rope 2 is horizontally oriented, T₂ also acts horizontally.
Again, choose the point where the ropes intersect and write equations of static equilibrium involving the forces acting at that point:
Sum up the vertical components of the forces
∑Fy = 0
Eq. 3: T₁sin(θ₃) - T₃ = 0
Sum up the horizontal components of the forces
∑Fx = 0
Eq. 4: T₂ - T₁cos(θ₃) = 0
Right away we can solve for T₃, which is the force of gravity acting on m₂:
T₃ = m₂g, m₂ = 6.00kg, g = 9.81m/s²
T₃ = 6.00×9.81
T₃ = 58.9N
Plug in known values into Eq. 3:
T₁sin(61.0) - 58.9 = 0
We can solve for T₁ now that is is the only unknown value in this equation
0.875T₁ = 58.9
T₁ = 67.3N
Plug in known values into Eq. 4:
T₂ - 67.3cos(61.0) = 0
We can solve for T₂ now that it is the only unknown value in this equation
T₂ = 67.3cos(61.0)
T₂ = 32.6N
