What is the Coefficient of Static Friction in a Torque Problem?

[veronicak5678]

Homework Statement

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.

Homework Equations

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

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 don't know normal floor or friction force, so how do I solve this?

 

[naresh]

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

 

[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?

 

[naresh]

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

Yes, this is right.

Is the y component just sin 45 * normal wall?

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

 

[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?

 

[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?

 

[veronicak5678]

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

 

[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

 

[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?

 

[naresh]

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

 

[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.

 

[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...

 

[naresh]

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

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

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...

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

 

[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.

 

[naresh]

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

 

[veronicak5678]

Great. Thank you so much for your help (and your patience!).

 

[veronicak5678]

I thought I had this figured out, but now I am having trouble trying to solve for the coefficient of static friction. I am using friction = coeffiient*normal ground

looking at the diagram, I said -friction = normal of wall * cos 69.5 which gave me friction = -8.44

solving for the coefficient with friction / normal ground is -8.44 / 89.6, which is negative, so I am doing something wrong.