What is the Tension in the Horizontal Cable Supporting a Crane?

[jrrodri7]

crane static problem...

Homework Statement

A 25-ft. 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 lbs. What is the Tension in the horizontal cable?

Homework Equations

t = I (alpha)
F = ma
Sums of F = 0
Sums of t = 0
r x F = t
Magnitude of t = Fdsin(theta)

The Attempt at a Solution

I have no idea, I tried doing a basic way of doing the load force mechanics but i realized it's not only that there's something with torque and the load proportion I don't know about.

 

[PhanthomJay]

For a system in equilibrium, there is no acceleration, so scratch out your first 2 relevant equations. Try summing torques about the lower pivot, and note also that the weight forces act through their center of mass, and that tension forces act along the longitudinal axis of the cable (horizontally in this case). You must show some work before we can be of further assistance.

 

[Moetasim]

I would approach this problem by first drawing a free body diagram of the crane. This will help to identify all the forces acting on the crane, including the tension in the horizontal cable. From the given information, we know that there are three forces acting on the crane: the weight of the crane, the weight of the load, and the tension in the cable.

Next, I would apply the principle of equilibrium, which states that the sum of all forces and torques acting on a body must equal zero for the body to be in static equilibrium. This means that the sum of the forces in the vertical direction must equal zero, and the sum of the torques about any point must also equal zero.

In this problem, we can choose the pin as the point about which we calculate the torques. Since the crane is in static equilibrium, the sum of the torques about the pin must be equal to zero. This can be expressed as:

Tsin(theta) x 25ft - 200lbs x 10ft = 0

Where T is the tension in the horizontal cable and theta is the angle between the cable and the horizontal.

Solving for T, we get:

T = 200lbs x 10ft/25ftsin(theta) = 80lbs/cos(theta)

Therefore, the tension in the horizontal cable is dependent on the angle between the cable and the horizontal. If the cable is parallel to the ground (theta = 0), the tension would be 200lbs. If the cable is at an angle of 45 degrees to the horizontal (theta = 45 degrees), the tension would be approximately 113lbs.

In summary, the tension in the horizontal cable can be calculated using the principle of equilibrium and the known distances and weights in the problem. The angle of the cable with respect to the horizontal will affect the magnitude of the tension.