Answer :
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
The forces acting on a system are in equilibrium, when they exactly balanced each other and the net force acting is zero
The forces acting as tension in the ropes are;
(a) T₁ = 18.12 N
T₂ = 23.2 N
T₃ = 29.43 N
(b) T₁ ≈ 67.3 N
T₂ ≈ 32.63 N
T₃ = 58.86 N
The reason why the above values are correct are as follows;
Known parameters:
θ₁ = 38.0°
θ₂ = 52.0°
θ₃ = 61.0°
m₁ = 3.00 kg
m₂ = 6.00 kg
(a) The weight of the mass m₁, and the tension in T₃
T₃ = W₁ = m·g
Where;
g = The acceleration due to gravity ≈ 9.81 m/s²
∴ T₃ = 3.00 kg × 9.81 m/s² = 29.43 N
T₁·sin(θ₁) + T₂·sin(θ₂) = T₃
T₁·cos(θ₁) = T₂·cos(θ₂)
Which gives;
T₁·sin(38.0°) + T₂·sin(52.0°) = 29.43...(1)
T₁·cos(38.0°) - T₂·cos(52.0°) = 0...(2)
Solving the simultaneous equation with a graphing calculator gives;
T₁ = 18.12 N
T₂ = 23.2 N
Alternatively, making T₁ the subject of equation (2), and substituting the value in equation (1) gives;
[tex]T_1 = \dfrac{T_2 \cdot cos(\theta _2)}{cos(\theta _1) }[/tex]
[tex]\dfrac{T_2 \cdot cos(52^{\circ})}{cos(38.0^{\circ}) }\cdot sin(38.0^{\circ}) + T_2 \times sin(52.0^{\circ}) = 29.43[/tex]
[tex]T_2 = \dfrac{29.43}{\dfrac{cos(52^{\circ})}{cos(32.0^{\circ}) }\cdot sin(38.0^{\circ}) + sin(52.0^{\circ}) } = 23.19[/tex]
T₂ ≈ 23.19 N
[tex]T_1 = \dfrac{23.19 \times cos(52^{\circ})}{cos(38.0^{\circ}) } \approx 18.12 \[/tex]
T₁ ≈ 18.12 N
(b) T₁·cos(θ₃) = T₂
T₁·sin(θ₃) = T₃
T₃ = W₁ = m₁·g
T₃ = 6.0 kg × 9.81 m/s² = 58.86 N
T₁·sin(61.0°) = 58.86 N
[tex]T_1 = \dfrac{58.86 \ N}{sin(61.0^{\circ})} \approx 67.3 N[/tex]
T₁ ≈ 67.3 N
T₂ ≈ 67.3 N × cos(61.0°) ≈ 32.63 N
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