# Load on ladder leaning against a wall

1. Oct 10, 2005

### engcon

Hello, I got this problem from a book

"A light ladder leans against a perfetly smooth vertical wall at an angle of 30 degrees to the horizontal. A load of 800N is placed 3-quarters of the way up the ladder. If the ladder rests on a rough horizontal surface which prevents slipping, find the magnitude and direction of the reaction between the ladder and the ground"

Since the system is in equilibrium, I resolved vertically and got

R sin(theta) = 800

where R is the reaction force, and theta is the angle of the reaction force with the horizontal.

The problem is I can't get another equation so that I can find R and theta.

Any help is appreciated, thanks!

2. Oct 10, 2005

### mezarashi

Okay, first of all, my definition of "reaction" is the normal force produced by any surface, which I think does not coincide with your definition. In anycase, This problem involves three equations which are easily solved.

1. Fnet = 0 (in the x direction)
2. Fnet = 0 (in the y direction)
3. Mnet = 0 (torque, no rotation)

The frictionless wall will only produce a force in x. The floor will only produce a force in y. The friction at ladder's base will produce a force in -x. The 800 N load will produce a force in the -y.

Now for moments. Take the moments about the ladder's base (you can choose any point, but this point is particularly easy. Then the two forces that will create a torque around this point will be the 800N load (counter clockwise) and the reaction force of the wall (clockwise). They must sum up to zero.

Go get em =)

3. Oct 10, 2005

### engcon

No mention of the ladder length is made, so I cannot use moments

4. Oct 10, 2005

### mezarashi

Are you sure? Try denoting the length of the ladder as L. Leave it as L (you don't need to substitute a number), and see what happens.

5. Oct 11, 2005

### engcon

Solved

Did that, and solved it

Thanks