A crane with a movable pulley system on a horizontal arm is moving a large container. The 355 kg container is hanging from a cable that is 6.15 m out on the 7.50 m arm. The arm has a mass of 345 kg. A cable that is attached to its end makes and angle 32.0 degrees with the horizontal arm.
1. What is the tension in the cable supporting the arm?
2. What force does the vertical part of the crane exert on the boom at the point of contact?
8000
To find the tension in the cable supporting the arm, we can first calculate the torque applied to the arm by the weight of the container hanging from the cable.
1. Find the torque (τ) applied to the arm:
The torque is given by the equation: τ = F × d × sin(θ), where F is the force applied perpendicular to the arm, d is the distance from the pivot point to the point of application of the force, and θ is the angle between the force and the arm.
In this case, the force applied is the weight of the container, which is given by F = m × g, where m is the mass of the container and g is the acceleration due to gravity.
Calculating the torque:
τ = (m × g) × d × sin(θ)
= (355 kg × 9.8 m/s^2) × 6.15 m × sin(32°)
2. Find the torque (τ_a) applied to the arm by its own weight:
The torque applied to the arm by its own weight can be calculated using the equation τ_a = m_a × g × (d_a/2), where m_a is the mass of the arm, g is the acceleration due to gravity, and d_a is the distance between the pivot point and the center of mass of the arm.
Calculating the torque applied by the arm:
τ_a = 345 kg × 9.8 m/s^2 × (7.50 m/2)
3. Find the net torque (τ_net) on the arm:
Since the system is in equilibrium (not rotating), the net torque on the arm is zero. So we have:
τ_net = τ - τ_a = 0
Therefore, we can set up the equation:
(m × g) × d × sin(θ) = m_a × g × (d_a/2)
Now we can solve for the tension in the cable supporting the arm:
1. Rearrange the equation to solve for the tension (T):
T = (m_a × g × (d_a/2)) / (d × sin(θ))
2. Substitute the given values:
m_a = 345 kg (mass of the arm)
g = 9.8 m/s^2 (acceleration due to gravity)
d_a = 7.50 m (distance between the pivot point and the center of mass of the arm)
d = 6.15 m (distance from the pivot point to the point where the container is hanging)
θ = 32.0° (angle between the cable and the horizontal arm)
Now, you can plug in these values into the equation to find the tension (T) in the cable supporting the arm.
To find the force exerted by the vertical part of the crane on the boom at the point of contact:
1. Resolve the tension force (T) into horizontal component (T_h) and vertical component (T_v).
T_h = T × cos(θ)
T_v = T × sin(θ)
2. The force exerted by the vertical part of the crane on the boom is equal to the vertical component of the tension force (T_v).
So, the force exerted by the vertical part of the crane on the boom is T_v.
Now, you can plug in the calculated tension (T) value and solve for the force exerted on the boom at the point of contact.