A 3.0m wooden platform is suspended from the roof of a house by ropes attached to its end. A painter whose mass 80kg stands 1.2m from the left-hand end of the platform, whose own mass is 18kg. Find the tension in each rope.
To find the tension in each rope, we need to analyze the forces acting on the wooden platform and the painter in equilibrium.
Let's start by calculating the center of mass of the system. The center of mass is the point where the total mass of the system can be considered to be concentrated.
Given:
- Length of the wooden platform (L) = 3.0m
- Mass of the painter (m1) = 80kg
- Distance of the painter from the left-hand end (d1) = 1.2m
- Mass of the wooden platform (m2) = 18kg
The total mass of the system:
m_total = m1 + m2
m_total = 80kg + 18kg
m_total = 98kg
The center of mass can be calculated using the following formula:
x_cm = (m1 * d1 + m2 * d2) / m_total
Substituting the values:
x_cm = (80kg * 1.2m + 18kg * d2) / 98kg
To maintain equilibrium, the center of mass of the system should be exactly at the mid-point of the platform (since the platform is symmetric).
Therefore, x_cm = L/2
(80kg * 1.2m + 18kg * d2) / 98kg = 1.5m
96kg + 18kg * d2 = 147kg
Solving for d2:
18kg * d2 = 147kg - 96kg
18kg * d2 = 51kg
d2 = 51kg / 18kg
d2 ≈ 2.8m
Now, we can analyze the forces acting on the system.
Since the system is in equilibrium, the sum of the vertical forces and the sum of the horizontal forces must be zero.
Vertical forces:
1. The weight of the painter (m1 * g), acting downwards.
2. The weight of the platform (m2 * g), acting downwards.
3. The tension in the right-hand rope (T1), acting upwards.
Sum of vertical forces:
T1 - (m1 * g) - (m2 * g) = 0
Horizontal forces:
1. The tension in the left-hand rope (T2), acting towards the left.
2. The tension in the right-hand rope (T1), acting towards the right.
Sum of horizontal forces:
T2 - T1 = 0
From the horizontal forces equation, we can conclude that T2 = T1.
Substituting T2 = T1 into the vertical forces equation:
T1 - (m1 * g) - (m2 * g) = 0
T1 = (m1 * g) + (m2 * g)
Substituting the values:
T1 = (80kg * 9.8m/s²) + (18kg * 9.8m/s²)
T1 = 784N + 176.4N
T1 ≈ 960.4N
Since T2 = T1, the tension in each rope is approximately 960.4N.