The 40-lb ladder AC is leaning on a 10-lb block at B and a frictionles

[wing911]

Homework Statement

The 40-lb ladder AC is leaning on a 10-lb block at B and a frictionless corner at C. Can the system remain at rest in the position shown? Consider both sliding and tipping.
here is the picture of the problem
http://www.flickr.com/photos/19566492@N03/4146098284/

Homework Equations

The Attempt at a Solution

try to do free body but got stuck

 

[tiny-tim]

Welcome to PF!

Hi wing911! Welcome to PF!

First assume that the block is fixed, and check whether the ladder will slip.

The check whether the block will slide, then check whether the block will tip over.

Show us what you get. ​

 

[Mene]

I would first analyze the forces acting on the system and determine if the system can remain at rest in the given position. In this case, there are three forces acting on the system: the weight of the ladder (40 lbs), the weight of the block (10 lbs), and the normal force from the frictionless corner at C.

For the system to remain at rest, the net force and net torque must be equal to zero. Since the ladder is leaning against a frictionless corner, there is no friction force to consider. Therefore, we only need to consider the forces acting along the ladder and the block.

In terms of sliding, the ladder will slide if the force of gravity acting on the ladder (40 lbs) is greater than the normal force from the corner (0 lbs). This means that the system cannot remain at rest in this position.

In terms of tipping, the ladder will tip over if the torque due to the weight of the ladder (40 lbs) is greater than the torque due to the weight of the block (10 lbs). This means that the system also cannot remain at rest in this position.

In conclusion, the system cannot remain at rest in the given position for both sliding and tipping. To prevent sliding, a frictional force would need to be present at the corner. To prevent tipping, the weight of the block would need to be increased.