In the figures, the masses are hung from an elevator ceiling. assume the velocity of the elevator is constant. find the tensions in the ropes (in n) for each case. note that: θ1 = 38.0°, θ2 = 52.0°, θ3 = 61.0°, m1 = 3.00 kg, and m2 = 6.00 kg.

(a)
t1 =
t2 =
t3 =

(b)
t1 =
t2 =
t3 =

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

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