# Homework Help: Torque to find forces

1. Nov 24, 2008

### veronicak5678

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

A 6 meter long ladder rests against a wall which is 4 m tall. The base of the ladder is 1.5 m away from the base of the wall (in the picture you can see the wall touches the ladder about 3/4 way up the ladder). The wall is frictionless, but the ground is not.
a- Find the normal force on the ladder due to the wall (remember it is perpendicular to the ladder)
b- Find the normal force on the ladder due to the ground.
c- If the ladder is just about to slide, find the coefficient of static friction between the wall and the ground.

2. Relevant equations

torque = r * force * sin (angle between them)

3. The attempt at a solution
a- I put the pivot at the point where the floor meets the ladder. I found the angle to be 69.5
-torque of normal wall - torque of weight = 0
r*normal wall * sin 90 = r*weight * sin 200.5
normal wall = 17.2 N

b- I put the pivot at the point where the wall meets the ladder.
torque of normal floor - torque static friction - torque weight = 0

I dont know normal floor or friction force, so how do I solve this?

2. Nov 24, 2008

### naresh

How are the forces acting on the ladder related? Remember, the ladder is at rest.

3. Nov 24, 2008

### veronicak5678

Can I just say the normal force of the floor = the weight - y component of normal of wall?
Is the y component just sin 45 * normal wall?

4. Nov 24, 2008

### naresh

Yes, this is right.

Hmm, Don't know where you get 45 from. The direction of the normal force due to the wall is given to you. You're even asked to remember it

5. Nov 24, 2008

### veronicak5678

Oops! I meant to say 90. It's been a long day...

So the normal force from the ground is mg-normal wall *sin 90 =

10kg*9.8-17.2*sin 90 = 80.8 N?

6. Nov 24, 2008

### naresh

Well , 90 isn't right either.

"a- Find the normal force on the ladder due to the wall (remember it is perpendicular to the ladder)"

Along which axis are you equating the forces?

7. Nov 24, 2008

### veronicak5678

God, I was looking at the normal force for the wall again. Sorry. So it should be sin 69.5.

8. Nov 24, 2008

### naresh

Good

Edit: I don't know which of the angles is 69.5, so I don't know if you have the sine term right. Looks like you may be off -- I'd check it again

9. Nov 24, 2008

### veronicak5678

Oh! This is the part that always messes me up. Looking again, I think the angle should be 90 + (90-69.5) = 110.5.

So I get normal = 10kg*9.8m/s^2 - 17.2*sin 110.5 = 81.9 N. Does this seem right?

10. Nov 24, 2008

### naresh

I can't really reproduce the 17.2N number, but there is nothing wrong with your approach.

11. Nov 24, 2008

### naresh

1." r*normal wall * sin 90 = r*weight * sin 200.5"

Are the r's the same? What values do they take?

2. "I think the angle should be 90 + (90-69.5) = 110.5. "

Hmm, sin 69.5 = sin 110.5, therefore this is not correct either. You need to look at it a little more.

12. Nov 24, 2008

### veronicak5678

1) I used 4.27m*normal wall*sin90 = (4.27m/2)*weight*sin200.5

2) For this, I really don't know. I just drew a diagram, and now it looks t me like the y component of the normal wall should use sin 20.5 because the original theta (69.5) would make a 90 degree angle with it. I'm not sure though...

13. Nov 24, 2008

### naresh

The ladder is 6 m long I also assume you mean 20.5 degrees

Yes, now it is correct. You should probably practise these diagrams a few times, so that you don't get confused.

14. Nov 24, 2008

### veronicak5678

The r from the normal wall to the pivot on the ground should be 4.27, but the r for the weight should be 6/2 = 3, right? Using all this, I get normal wall = 24.1 N. And yes, I did mean 20.5. Sorry for being so sloppy and confused! This homework makes me flustered.

Last edited: Nov 24, 2008
15. Nov 24, 2008

### naresh

That's fine. I think you know enough to solve this problem now.

16. Nov 24, 2008