# Ladder Question - Force of friction required to prevent ladder from slipping

1. Jul 11, 2007

### avsj

1. The problem statement, all variables and given/known data

A uniform 15kg ladder whose length is 5.0 m stands on the floor and leans against a vertical wall, making an angle of 25 degrees with the vertical. Assuming that the friction between the ladder and the wall is negligble, what is the minimum amount of friction between the ladder and the floor that will keep the ladder from slipping?

2. Relevant equations

Torque = Fdistance
Fg = mg
And possibly : Ff= 1/2mgtan(x)

also can Fg = 1/2mg because we're delaing with torques? if so why?

3. The attempt at a solution

I tried making force vectors perpendicular to (torque) and straight down (gravity) and finding horizontal components (bottom of triangle...so friction?) but could not arrive at the correct answer which is 34 N but would really really appreciate it if someone could help me understand this.

Thanks a lot

2. Jul 11, 2007

### Dick

If you have arrived at a wrong answer it would be really useful for you to post the wrong solution. What are the forces you've determined and how did you combine them to get the wrong answer? The more details you show the more people will jump in and point out errors. As it is, you are not giving us much to go on.

3. Jul 11, 2007

### Gear300

Since you're dealing with torque, you're probably also dealing with rotational motion. Its best then to plot out a radius and center. Since the wall is frictionless with the ladder, gravity is pulling it down and so it is up to the friction on the floor to prevent it from moving. Now, since the ladder is not moving because the friction on the floor is preventing it from doing that, the point in which the ladder touches the floor could be considered constant (not moving), so you can call it the center and the ladder the radius. Now apply what you know about torque.

Last edited: Jul 11, 2007
4. Jul 12, 2007

### avsj

Fg= 15 x 9.8 = 147 (straight down)
F torque (perpendicular to the ladder) = 147/sin(25)= 346.8..... or is it 2.5 x 15 x 9.8 = 367.5?

solving the horizontal of this vertical line (fg) and the hypotenuse (torque) to get 314....

This is in the 'brain busters' section of my book which often asks questions for which we havent learned something yet, but I believe the next question which is of the same format gives the formula:

Ff = 1/2mgtan@
If we plug in with @=25, m=15 and g=9.8, we get the correct answer of 34. But i dont see where this equation comes from.

Any ideas?

5. Jul 12, 2007

### Dick

Ff=1/2mgtan@ is the formula for the solution to this particular problem - not a general formula. So figuring out where it comes from is the same as figuring out the problem. You have four forces to deal with gravitational force, normal force and frictional force from the floor and normal force from the wall. All four of those must sum to zero since the ladder is not accelerating. Once you've ensured they sum to zero then you can place the center for computing total torque ANYWHERE. Where ever is convenient. Then set the sum of all of the torques generated by each force equal zero. It is handy to put it at the point of contact with the floor as Gear300 suggests, since then two of the forces have zero torque.