Find the tension in each of the two ropes supporting a hammock if the first one is at an angle of 19.4° above the horizontal and the second is at an angle of 35.3° above the horizontal. The person sleeping in the hammock (unconcerned about tensions and ropes) has a mass of 60 kg.

To find the tension in each of the two ropes supporting the hammock, we can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration. In this case, the net force is equal to zero, as the person in the hammock is not accelerating vertically.

Let's denote the tensions in the first and second ropes as T1 and T2, respectively.

Step 1: Resolve the forces
We can start by resolving the forces acting on the person in the hammock into horizontal and vertical components.

The horizontal component:
The only horizontal force acting on the person is the tension in the second rope (T2).

The vertical component:
There are two vertical forces acting on the person: the tension in the first rope (T1) and the gravitational force (mg), where m is the mass of the person and g is the acceleration due to gravity (approximately 9.8 m/s^2).

Step 2: Write equations for the horizontal and vertical forces:
Horizontal forces:
T2 = 0 (since there is no horizontal acceleration)

Vertical forces:
T1 - mg = 0

Step 3: Solve the equations:
From the horizontal forces equation, we can see that T2 = 0.

From the vertical forces equation:
T1 - mg = 0
T1 = mg

Substituting the values:
m = 60 kg
g = 9.8 m/s^2

T1 = 60 kg × 9.8 m/s^2
T1 = 588 N

Therefore, the tension in the first rope (at an angle of 19.4° above the horizontal) is 588 N.

Since the tension in the second rope (at an angle of 35.3° above the horizontal) is equal to zero, we don't need to calculate it.

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.

M*g = 60 * 9.8 = 588 N. = Wt. of the person.

Eq1: T1*Cos9.4 + T2*Cos35.3 = 0.
T1*Cos9.4 = -T2*Cos35.3.
T1 = -0.827T2.

Eq2: T1*sin9.4 + T2*sin35.3 =-588[270o].
0.1633T1 + 0.578T2 = 0 + 588
Replace T1 with -0.827T2:
0.1633*(-0.827T2) + 0.578T2 = 588
-0.135T2 + 0.578T2 = 588
0.443T2 = 588.
T2 = 1327 N.

T1 = -0.827T2 = (-0.827)*1327 = -1098 N.