Equilibrium of Tensions The cranes in the figure are lifting an object that weighs 18,278 lb. Fi

Equilibrium of Tensions The cranes in the figure are lifting...

Key Concepts

Tension Forces in Cables

Tension forces refer to the pulling force transmitted through a string, rope, or cable when it is subjected to stretching. Understanding tension is crucial in problems where cables or cranes lift objects, as each cable may carry a different portion of the load depending on the geometry and angles involved.

Equilibrium of Moments

Equilibrium of moments, also known as rotational equilibrium, involves balancing the torques about a pivot point such that their sum is zero. This concept complements force equilibrium by addressing the tendency of forces to create rotation, and it is often used to solve for unknown forces when multiple forces are applied at different distances from a reference point.

Equilibrium of Forces

The equilibrium of forces involves ensuring that the total forces acting in any direction cancel each other out. In practical problems, this often involves setting up equations where the sum of the vertical, horizontal, or other directional forces equals zero, which is critical for determining unknown forces like tensions in cables.

Free-Body Diagram

A free-body diagram is a simplified representation of an object where all forces acting on it are represented by arrows. This tool is essential for visualizing the problem, identifying the points of application of forces, and setting up equilibrium equations to solve for unknown quantities.

Static Equilibrium

Static equilibrium is the condition in which an object remains at rest, meaning that the sum of all forces and the sum of all moments acting on the object are zero. This principle is fundamental in statics, ensuring that there is no net acceleration or rotation, allowing one to set up balance equations to solve for unknown forces.

Transcript

00:01 So the problem gives us the following diagram where a mass of 18 ,278 pounds is being held up by two different cranes, each with a different tension and a different direction at which they're pulling at.

00:15 And we're asked to find what these two tensions are.

00:19 So to do this, we are going to have to separate each tension into different components and also use a system of equations.

00:30 To solve for each of the tensions.

00:34 And each equation will have to do with either the horizontal components or the vertical components.

00:42 So, as we've talked about in previous problems, to find the horizontal and vertical components of these tension vectors, we just use trigonometry.

00:53 So, for example, the cosine of 22 .3 degrees should be equal to the adjacent of the which is the horizontal component over the hypotenuse.

01:04 And using that, we can solve for what this adjacent horizontal component is, and in this case, it's going to be t1 times the cosine of 22 .3 degrees.

01:15 So now that we understand this, we can start looking at our equations that we need to come up with.

01:21 So the first thing we should look at is the horizontal components of these tensions.

01:27 Because the mass isn't going to be moving left or right, it's just stationary in the middle, the two horizontal components of the tensions need to cancel out, so they will be equal to each other.

01:42 So the horizontal component for tension 1 is going to be equal to t1 times cosine of 22 .3 degrees.

01:54 And this has to be equal to the horizontal component.

01:57 For tension 2, which is t2 times the cosine of 41 .5 degrees.

02:08 Now, if we look at the vertical components of these tensions, we see that both vertical components need to add together to equal this mass down here so that it can hold the mass up without it either moving up or down because it's once again stationary there...

Please give Ace some feedback