A two-side 'A' shaped ladder and then tension in the wire that holds it together

[IAmPat]

Homework Statement

A ladder is made in the shape of the letter A. Treat the two sides of the ladder as identical uniform rods, each weighing 455 N, with a length of 3.60 m. A frictionless hinge connects the two ends at the top, and a horizontal wire, 1.20 m long, connects them at a distance 1.40 m from the hinge, as measured along the sides. The ladder rests on a frictionless floor. What is the tension in the wire?

Homework Equations

$$\sum$$Fx = 0
$$\sum$$Fy = 0

The Attempt at a Solution

I honestly don't know where exactly to start on this solution.

I know that the system is in equilibrium, so all the forces should sum to 0. Couldn't I just pretend it was just one ladder leaning against a wall, and then double the tension to get the tension of wire in this double ladder?

Where else can I start on this problem?

 

[eczeno]

hello,

the first thing, if you haven't done it already, would be to draw a figure. once you have that, just consider the forces, and torques, on one of the beams.

hope this helps

 

[IAmPat]

hello,

the first thing, if you haven't done it already, would be to draw a figure. once you have that, just consider the forces, and torques, on one of the beams.

hope this helps

So should I consider just one ladder and double the tension in order to get the tension in the 'A' ladder

 

[eczeno]

consider one ladder, but don't double the tension, it will pull in equal but opposite directions on each ladder.

cheers

 

[rwisz]

I would approach this problem by first drawing a free body diagram of the ladder and identifying all the forces acting on it. From the given information, we know that the ladder is in equilibrium, so we can use the equations \sumFx = 0 and \sumFy = 0 to solve for the unknown tension in the wire.

We can start by considering the forces acting on the ladder in the x-direction. Since the ladder is in equilibrium, the sum of all the forces in the x-direction must be equal to 0. The only force acting in the x-direction is the tension in the wire, so we can set \sumFx = T = 0.

Next, we can consider the forces acting in the y-direction. The ladder is resting on a frictionless floor, so the normal force from the floor will cancel out the weight of the ladder. This means that the only force acting in the y-direction is the weight of the ladder, which is given as 455 N for each side. We can set \sumFy = 2(455 N) - T = 0.

Solving these two equations simultaneously, we can find that the tension in the wire is 910 N. This makes sense, as the tension in the wire should be equal to the weight of the ladder to keep it in equilibrium.

In conclusion, by treating the ladder as a system in equilibrium and considering the forces acting on it, we can solve for the unknown tension in the wire.