# Homework Help: Question about static equilibrium

1. Feb 26, 2015

### bioengi12345

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

A Ladder weighs 20 Lbs. IT rests against a smooth wall at point B. Theta=45. If you pull the ladder at its midpoint, how much force is needed to make it slide at point A? u=.6.
The ladder is 15 ft long, Point b is the top and point A is the bottom. 45 Deg is the bottom angle.

Force of pull= 20lb. Now i think the number worked out this was just cause the angle is 45 degrees and gravity as well as the force of the pull act about the same point, however, this just seems like that it would take to tip the ladder backwards, not make it slide. Im not really sure where to go from here.

2. Relevant equations
(sum of forces in x dir) : Force of pull - force of static friction = 0

(sum of forces in y dir) : -20 + normal force at A = 0

(sum of moments about A) : -20lb(5.3m)-force of pull(5.3m) = 0

3. The attempt at a solution
Force of pull= 20lb. Now i think the number worked out this was just cause the angle is 45 degrees and gravity as well as the force of the pull act about the same point, however, this just seems like that it would take to tip the ladder backwards, not make it slide. Im not really sure where to go from here.

2. Feb 26, 2015

### bioengi12345

Anyone have any ideas?

3. Feb 27, 2015

### ehild

You forgot the normal force from the wall.
To make the ladder just slide, but not tip over, both the resultant force components and the torque have to be zero.

4. Feb 27, 2015

### bioengi12345

So it would be: Force of pull - Force of static friction - normal force of the wall=0? Then i have 3 unknowns. Sorry im normally really good at this stuff, the whole idea of equilibrium gets me sometimes.

5. Feb 27, 2015

### haruspex

Three unknowns is fine since in 2D statics problems you have three equations.
Is the direction of pull given?

6. Feb 27, 2015

### bioengi12345

pulling horizontally away from the wall at the midpoint.

7. Feb 27, 2015

This problem actually has four unknowns, but also four equations. (although only three are needed to solve it)

8. Feb 27, 2015

### haruspex

Ok, so have you drawn a free body diagram? Why do you write "force of pull = 20 lbs"? Isn't that what we're to find?
Create symbols for the known and unknown forces and lengths. Work symbolically until the final step. Write three equations: one for horizontal force balance, one for vertical force balance, and one for torque balance. (Choose a reference axis for the torque.)

9. Feb 27, 2015

### bioengi12345

Im trying to say that is what i got as the answer but i dont think its right. I havent made a whole lot of progress here some direction would be very helpful. i have set up 3 equations. sum of f in x ( fp-fs-nB=0, -20+nA=0, -20(5.3m)+fP(5.3)+nB(10.6)=0 ...... is any of this looking right? i need to get this turned in so some solid hint would be very nice.

10. Feb 27, 2015

### SammyS

Staff Emeritus
When you put all of that together, what do you get for fP ?

11. Feb 27, 2015

### haruspex

First, you had not previously posted that version of your equations; second, you have a sign wrong in the x direction balance; third, you would do well to work entirely symbolically as I suggested (it avoids the units conversion); fourth, you have not used the fact that you are interested in the case where the friction is only just enough to balance the forces.
I presume you no long claim 20lbs as an answer.

12. Feb 27, 2015