A 25 ft long crane supported at its lower end by a pin is elevated by a horizontal cable. A 250 lb load is suspended from the outer end of the crane. The center of gravity of the crane is 10 ft from the pin and the crane weighs 200 lb. What is the tension in the horizontal cable?
To find the tension in the horizontal cable, we can use the principle of moments. The sum of the clockwise moments about the pin must be equal to the sum of the anticlockwise moments.
Let's consider the moments about the pin:
Clockwise moments:
Moment due to the weight of the crane = Weight of crane x distance of center of gravity from the pin
= 200 lb x 10 ft
= 2000 lb-ft
Anticlockwise moments:
Moment due to the load attached at the outer end of the crane = Load x distance of load from the pin
= 250 lb x 25 ft
= 6250 lb-ft
Since the crane is in equilibrium, the sum of the clockwise moments must be equal to the sum of the anticlockwise moments:
2000 lb-ft = 6250 lb-ft
Now, we can calculate the tension in the horizontal cable:
Tension in cable x distance of tension from the pin = Sum of clockwise moments
= 2000 lb-ft
Let's assume the distance of tension from the pin is x:
Tension x = 2000 lb-ft
Tension = 2000 lb-ft / x
Now, we can substitute this expression for tension back into the equation:
(2000 lb-ft / x) x = 6250 lb-ft
2000 lb-ft = 6250 lb-ft
Now, we can solve for x:
x = 6250 lb-ft / 2000 lb-ft
x = 3.125 ft
Therefore, the tension in the horizontal cable is 2000 lb-ft / 3.125 ft = 640 lb.
To find the tension in the horizontal cable, we need to analyze the forces acting on the crane.
There are three forces acting on the crane: the weight of the crane, the weight of the load, and the tension in the cable.
1. Weight of the crane: The weight of the crane is 200 lb. This force acts downward at the center of gravity, which is 10 ft from the pin.
2. Weight of the load: The weight of the load is 250 lb. This force acts downward at the outer end of the crane, 25 ft from the pin.
3. Tension in the cable: This force acts horizontally, pulling the crane upwards. Let's call this force "T".
In order to maintain equilibrium, the sum of all the vertical forces must be zero, and the sum of all the horizontal forces must be zero.
Let's analyze the vertical forces:
- The weight of the crane (200 lb) creates a clockwise (negative) moment around the pin.
- The weight of the load (250 lb) creates a counterclockwise (positive) moment around the pin.
To find the tension in the cable, we can use the principle of moments:
Sum of counterclockwise moments = Sum of clockwise moments
(Weight of the load x Distance of load from the pin) = (Weight of the crane x Distance of the crane's center of gravity from the pin) + (Tension in the cable x Distance of the crane's center of gravity from the pin)
(250 lb x 25 ft) = (200 lb x 10 ft) + (Tension in the cable x 10 ft)
6250 ft-lb = 2000 ft-lb + 10 ft x Tension in the cable
6250 ft-lb - 2000 ft-lb = 10 ft x Tension in the cable
4250 ft-lb = 10 ft x Tension in the cable
Divide both sides by 10 ft:
Tension in the cable = 425 lb
Therefore, the tension in the horizontal cable is 425 lb.