To haul a boat out of the water for the winter, a worker at the storage facility uses a wide strap with cables operating at the same angle (measured from the horizontal) on either side of the boat. Determine the tension in each cable if the boat has a mass of 510 kg and the angle of each cable is 46 ∘ from the horizontal and the boat is being momentarily held at rest. Compare this to the tension when the boat is raised and held at rest so the angle becomes 31 ∘.
2 * T * sin(Θ) = m g
T is tension , m is mass of boat
To determine the tension in each cable, we can start by setting up a free body diagram of the boat.
When the boat is being momentarily held at rest with the angle of each cable at 46 degrees from the horizontal (θ = 46°), we can analyze the forces acting on the boat as follows:
1. The weight of the boat acts vertically downward and can be calculated using the mass (m) and acceleration due to gravity (g).
Weight = m * g = 510 kg * 9.8 m/s^2 = 4998 N (rounded to 4 significant figures).
2. The tension in each cable is directed upwards, inclined at the same angle (θ = 46°) on either side of the boat.
To find the tension in each cable, we can resolve the weight into its vertical and horizontal components.
The vertical component of the weight (W_vert) is equal to W * sin(θ), and the horizontal component (W_horiz) is equal to W * cos(θ).
Using the given angle of 46°, the vertical component becomes:
W_vert = 4998 N * sin(46°) = 4998 N * 0.71934 ≈ 3594 N (rounded to 4 significant figures).
Since the cables are pulling upwards to counteract the weight, the tension in each cable is equal to half of the vertical component (Tension = W_vert/2). Hence, the tension in each cable is approximately:
Tension = 3594 N / 2 ≈ 1797 N (rounded to 4 significant figures).
Now let's consider the situation when the boat is raised and held at rest, causing the angle of each cable to change to 31°.
Using the same approach as before, we can calculate the new vertical component of the weight:
W_vert_new = 4998 N * sin(31°) = 4998 N * 0.51504 ≈ 2574 N (rounded to 4 significant figures).
Again, the tension in each cable is equal to half of the vertical component, so the tension in each cable becomes:
Tension_new = 2574 N / 2 ≈ 1287 N (rounded to 4 significant figures).
Comparing the tensions in the two scenarios:
- When the boat is held at rest with the angle of each cable at 46°, the tension in each cable is approximately 1797 N.
- When the boat is held at rest with the angle of each cable at 31°, the tension in each cable is approximately 1287 N.
Therefore, the tension in each cable decreases as the angle between the cables and the horizontal decreases.
To determine the tension in each cable, we can use the concepts of equilibrium in a system.
When the boat is being held at rest, the forces acting on it must be balanced. The weight of the boat acts downwards, and the tension in the cables acts upwards at an angle to the horizontal.
Let's start by calculating the weight of the boat. The weight is given by the formula:
Weight = mass * gravitational acceleration
Weight = 510 kg * 9.8 m/s^2 (taking gravitational acceleration as 9.8 m/s^2)
Weight = 5009 N (approx.)
Now, let's analyze the forces acting on the boat when the angle is 46°. We'll use the following diagram:
T
________
/ \
| |
| |
———————————————
Weight (5009 N)
In this case, we have two cables at the same angle of 46° from the horizontal. We know that the vertical component of the tension will balance the weight, while the horizontal components of the forces in the cables cancel each other out.
Let's calculate the vertical component of the tension, which is responsible for balancing the weight:
Vertical component of T = T * cos(angle)
Vertical component of T = T * cos(46°)
Now, let's set up an equation for vertical equilibrium:
Vertical component of T = Weight
T * cos(46°) = 5009 N
Now, solve for T:
T = 5009 N / cos(46°)
T ≈ 7065 N (approx.)
Therefore, each cable has a tension of around 7065 N when the boat is raised and held at rest with an angle of 46°.
Now, let's analyze the forces acting on the boat when the angle is 31°. Again, using the concept of equilibrium, we'll set up an equation:
Vertical component of T = Weight
T * cos(31°) = 5009 N
Solve for T:
T = 5009 N / cos(31°)
T ≈ 5900 N (approx.)
Therefore, each cable has a tension of around 5900 N when the boat is raised and held at rest with an angle of 31°.
To compare the tension in each cable, we can see that it decreases as the angle decreases. When the angle is 46°, the tension is around 7065 N, whereas when the angle is 31°, the tension reduces to around 5900 N.