To find the tension in each rope, we can use the concepts of vertical and horizontal components of forces.
Let's start by finding the horizontal component of the tension in each rope. The horizontal component of the tension in the first rope can be calculated using the formula:
T₁⋅cos(θ₁) = T₁x
Where T₁ is the tension in the first rope and θ₁ is the angle of the first rope above the horizontal.
Similarly, for the second rope, the horizontal component of the tension in the second rope can be calculated using the formula:
T₂⋅cos(θ₂) = T₂x
Where T₂ is the tension in the second rope and θ₂ is the angle of the second rope above the horizontal.
Now let's find the vertical component of the tension in each rope. The vertical component of the tension in the first rope can be calculated using the formula:
T₁⋅sin(θ₁) = T₁y
Similarly, for the second rope, the vertical component of the tension in the second rope can be calculated using the formula:
T₂⋅sin(θ₂) = T₂y
Now, let's set up the equations to solve for the tensions.
Since the hammock is in equilibrium, the vertical components of the tensions in both ropes must add up to the weight of the person in the hammock:
T₁y + T₂y = m⋅g
Where m is the mass of the person in the hammock (60 kg) and g is the acceleration due to gravity (9.8 m/s²).
Let's substitute the equations we derived for the vertical components of the tension:
(T₁⋅sin(θ₁)) + (T₂⋅sin(θ₂)) = m⋅g
Now, let's consider the horizontal components of the tensions. The horizontal components of the tensions in both ropes must balance each other out:
T₁x + T₂x = 0
Which implies:
T₁⋅cos(θ₁) = -T₂⋅cos(θ₂)
We can solve this equation for T₂x:
T₂x = -((T₁⋅cos(θ₁)) / cos(θ₂))
Now, we have the equations and can solve for the tensions. Let's combine the equations for T₁y and T₂y to solve for T₁:
T₁⋅sin(θ₁) + T₂⋅sin(θ₂) = m⋅g
T₁⋅sin(θ₁) = m⋅g - T₂⋅sin(θ₂)
T₁ = (m⋅g - T₂⋅sin(θ₂)) / sin(θ₁)
Finally, let's substitute the value for T₁ into our equation for T₂x:
T₂x = -((T₁⋅cos(θ₁)) / cos(θ₂))
Now, we have the values of T₁ and T₂x, and we can solve for T₂y by rearranging the equation for T₂y:
T₂⋅sin(θ₂) = m⋅g - T₁⋅sin(θ₁)
T₂ = (m⋅g - T₁⋅sin(θ₁)) / sin(θ₂)
Now we have both T₁ and T₂. Plugging in the given values of θ₁ = 19.4°, θ₂ = 35.3°, m = 60 kg, and g = 9.8 m/s², we can calculate the tensions in each rope.