# Homework Help: Statics Problem

1. Sep 15, 2011

### eurekameh

A uniform ladder of mass m = 30.0 kg and length l = 2.50 m leans against a frictionless vertical wall. The ladder forms an angle theta = 75.0 degrees with the rough floor. The coefficient of static friction μ between the floor and the ladder is equal to 0.288. Determine the force Fw exerted on the ladder by the wall, and the frictional force f that the floor exerts on the ladder.

So here's my force diagram: http://imageshack.us/photo/my-images/90/gha.png/

Here are my equilibrium equations:
Fnet,x = Fw - f = 0
Fnet,y = Fv - mg = 0
Tnet,z = -(Fw)(2.4148) + (mg)(0.3235238) = 0

From the sum of the torques, I found Fw:
Fw = [(mg)(0.3235236)] / (2.4148) = 39.4 N.
Fw = f = 39.4 N.
What did I do wrong?

2. Sep 16, 2011

### vela

Staff Emeritus
Your solution looks fine to me. Why do you think it's wrong?

3. Sep 16, 2011

### pabloenigma

I went through your solution casually, but isnt it that you have forgot to take the torque contribution due to the normal reaction of the floor,that you have labelled as Fv in the diagram?

4. Sep 16, 2011

### eurekameh

I took torque around the axis concurrent through both the normal reaction and the friction acting from the floor, so the torque from both the normal and the friction is zero. I think it's wrong because this was on a test and I got no points for it.

5. Sep 16, 2011

### BruceW

You did calculate the torque correctly, I checked myself. (Its the torque around an axis going through the base of the ladder, right?).

6. Sep 16, 2011

### BruceW

If the ladder was stationary (i.e. in equilibrium), then you are right that Fw=f, and you do have the right answer for the frictional force.

So I think you should have gotten full marks for the problem.

But if the ladder was not stationary, ie moving, then the answer might be different. Are you sure this problem is meant to be a statics problem? (I'm just trying to figure out why you didn't get full marks)...

7. Sep 16, 2011

### eurekameh

Yeah, I'm pretty sure the ladder's not moving.
Here's the entire question:
A uniform ladder of mass m = 30.0 kg and length l = 2.50 m leans against a frictionless vertical wall. The ladder forms an angle theta = 75.0 degrees with the rough floor. The coefficient of static friction μ between the floor and the ladder is equal to 0.288. Determine the force Fw exerted on the ladder by the wall, and the frictional force f that the floor exerts on the ladder.
a. Draw the force diagram for the ladder.
b. Write the equilibrium conditions for the ladder.
c. Determine the force Fw exerted on the ladder by the wall.
d. Calculate the frictional force f that the floor exerts on the ladder.
e. Determine the minimum angle theta between the ladder and the floor for which the ladder will not slip.

Now that I think of it, and considering (e), the ladder might have been moving and "slipping" at the angle of 75 degrees. Hmm...
The wording is weird and makes me assume at first that the ladder is not moving.
What do you guys think? Is it moving or not?

8. Sep 16, 2011

### vela

Staff Emeritus
[strike]Yes, that was the problem. I tried checking that last night, but apparently I made a mistake when I did so.[/strike]

What's the maximum force of friction the floor can exert on the ladder?

EDIT: Apparently, I did check it correctly last night. I messed up this morning!

Last edited: Sep 16, 2011
9. Sep 16, 2011

### eurekameh

Isn't that just f,max = 0.288Fv?

10. Sep 16, 2011

### BruceW

I think vela is suggesting you work out the problem under the assumption that the ladder isn't necessarily stationary. Is that right?

11. Sep 16, 2011

### eurekameh

Hmm... How do I go about doing that? In a statics problem, a = 0. I'm not given any accelerations or velocities.

12. Sep 16, 2011

### vela

Staff Emeritus
You can assume it's stationary and work out what the forces have to be, which you've done, but then you need to go back and check if that's physically possible, which is what you need to do now. If it's not, that tells you your assumption was wrong and that the ladder will slide.

13. Sep 16, 2011

### pabloenigma

Of course its static,and you seem to have done it correctly except for the (e) part.Its static for an angle of $\theta$=75 degrees,since the the frictional force required is less than the maximum permissible frictional force,which makes equilibrium possible.Note that the maxumum possible frictional force is constant(=$\mu$Fv=$\mu$mg,since Fv=mg,as you have calculated for equilibrium)
Now,for some values of $\theta$,the frictional force calculated for equilibrium will be more than the maximum permissible value,making the equilibrium physically impossible.You have to find that value of $\theta$,lesser than which equilibrium is not possible

14. Sep 17, 2011

### eurekameh

Tnet,z = -(Fw)(2.5sintheta) + (mg)(1.25costheta) = 0
Fw = [(mg)(1.25costheta)] / (2.5sintheta)
Fw - f = 0
[(mg)(1.25costheta)] / (2.5sintheta) = μsFv = μs(mg)
μs = 1.25costheta / 2.5sintheta = 0.288
theta = 60.1 degrees
Is this right?
Because if so, then this whole problem I did correctly, but got no points for.

15. Sep 18, 2011

### eurekameh

Anyone?

Last edited: Sep 19, 2011
16. Sep 19, 2011

### BruceW

Yep, I got the same. Nice work. It is weird that you got no points for this question...