# Homework Help: Easy beginner torque/com question.need help please

1. May 20, 2013

### martinlematre

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 with the vertical. Assuming that the friction between the ladder and the wall
is negligible, what is the minimum amount of friction between the ladder and the floor which will
(34 N)

2. Relevant equations

torque = fperp * distance
3. The attempt at a solution

I keep on writing stuff but never get the right answer (which is 1/2 tan theta * mg)
i have no idea how to get there. i keep mapping and drawing FBD and i cant get there... i think its my understanding of torque and not understanding where the pivot point on this ladder is...

anyone care to share

2. May 20, 2013

### ehild

Can you show your work in detail?

ehild

3. May 20, 2013

### martinlematre

sure 1 sec lemme write neatly

4. May 20, 2013

### martinlematre

okay past this point im not sure what to do..i assume the force wanting it to slip is Force-> (the horizontal ocmponent of MG) so i try to equate Ff (Friction) to The Force -> (Fparallelsintheta) but i get the answer 68 not 34...

http://i44.tinypic.com/2vmvalg.jpg

i realize my problem lays in the fact i dont know what forces are acting on what and my free body diagram is probably wrong

Last edited: May 20, 2013
5. May 20, 2013

### ehild

The ladder is in rest, so the resultant force is zero and the torque is also zero.

You have vertical forces acting on the ladder: mg and FN (from the ground) and horizontal ones: FWN (the normal force from the wall) and the force of static friction, Ff.

How are the magnitudes of these forces related?

The forces have torque. Choose an appropriate point to write it up.

ehild

6. May 20, 2013

### martinlematre

this doesnt help me =[

7. May 20, 2013

### ehild

You do not need Fparallel. Write up the equilibrium for both the vertical and horizontal forces separately.

Last edited: May 20, 2013
8. May 20, 2013

### martinlematre

my problem is i dont know what forces are where

9. May 21, 2013

### ehild

See figure. It is identical to your one.

mg acts at the midpoint of the ladder, vertically down. FWN acts at the upper end of the ladder, horizontal, away from wall. N and Ff act at the lower end of the ladder, N vertically up, Ff horizontally, towards the wall, as it is against slipping the ladder outward.

#### Attached Files:

File size:
7.9 KB
Views:
109
10. May 21, 2013

### martinlematre

2 hr later i am still stuck ty..

11. May 21, 2013

### ehild

The ladder is a rigid body. To be in equilibrium (no motion) the sum of all forces has to be zero both the horizontal and vertical components. Also the torque has to be zero.

Do you see the forces? The ladder can not move vertically, so what forces have to cancel?
It can not move horizontally, so the horizontal forces have to cancel.
It can not rotate about its upper end so the sum of torques is zero.

Can you write up the corresponding equations?

ehild

12. May 21, 2013

### martinlematre

i got it no biggie

13. May 21, 2013

### ehild

So you have solved the problem?

ehild

14. May 21, 2013

### martinlematre

No ive just decided im gonna go into liberal arts instead