1. Mar 21, 2006

Swagger

A ladder consists of two legs, AB and BC, that are pinned at joint B and held together by a massless wire between D and E. The two legs are uniform and have the same length (so the ladder forms an isosceles triangle), but the legs have unequal masses. The heavier of the two legs, BC, has a mass of 34.7 kg and is shown shaded gray. The other leg, AB, has a mass of 21.5 kg. The ladder is standing on a smooth floor. The dimensions of the ladder are shown below.

The top of the ladder B is y = 2.45 m above the floor, the distance AC (the base of the ladder) is x = 1.49 m, and the cross brace DE is h = 1.35 m above the floor.

What are the vertical forces on the two legs? (Use "+" if the force component is up, "-" if it is down.) (Ay and Cy)

I know that all of the "internal" forces and torques cancel out if you look at the ladder as a single rigid body. You can then treat it as "center of gravity" problem where two masses are supported on a massless board of length 1.49 m. But I am not sure how to solve it like this? Please Help. Thanks.

Attached Files:

• beam-3.png
File size:
7.7 KB
Views:
116
2. Mar 21, 2006

Staff: Mentor

Since the ladder is in equilibrium, you know that (1) the sum of the forces in any direction must be zero, and (2) the sum of the torques about any point must be zero. Since each leg is uniform, you can treat its weight as acting at its center. What's a good point to pick as your axis for finding torques?

Give it a shot.