Static equilibrium ladder question

[hskrnt8590]

Homework Statement

A nonuniform fire escape ladder is 6.0 long when extended to the icy alley below. It is held at the top by a frictionless pivot, and there is negligible frictional force from the icy surface at the bottom. The ladder weighs 250 , and its center of gravity is 2.0 along the ladder from its bottom. A mother and child of total weight 750 are on the ladder 1.5 from the pivot. The ladder makes an angle theta with the horizontal.

1)Find the magnitude of the force exerted by the icy alley on the ladder?
2)What is the direction of the force exerted by the icy alley on the ladder? (Up/down)
3)Find the magnitude of the force exerted by the ladder on the pivot.
4)Do your answers in parts A and C depend on the angle theta? (Yes/No)

Homework Equations

F=ma
Torque= RxL

The Attempt at a Solution

1) Isn't the magnitude of the force by the icy alley on the ladder just the normal force? Since it doesn't mention the ladder sliding it has to be stationary which means all forces are cancelling out. 250+750 = 1000 Newtons upward.
2) It should be upward to cancel out the weight of the people and ladder
3) It would be the normal force of the wall on the ladder correct? This would be 0 though since friction is negligible I think.
4) The forces shouldn't depend on theta because they are strictly talking forces and not torque.

These are my initial tries, but the first one is wrong when I try the answer, so I am assuming my thinking is flawed. Any help would be appreciated.

 

[PhanthomJay]

Homework Statement

A nonuniform fire escape ladder is 6.0 long when extended to the icy alley below. It is held at the top by a frictionless pivot, and there is negligible frictional force from the icy surface at the bottom. The ladder weighs 250 , and its center of gravity is 2.0 along the ladder from its bottom. A mother and child of total weight 750 are on the ladder 1.5 from the pivot. The ladder makes an angle theta with the horizontal.

1)Find the magnitude of the force exerted by the icy alley on the ladder?
2)What is the direction of the force exerted by the icy alley on the ladder? (Up/down)
3)Find the magnitude of the force exerted by the ladder on the pivot.
4)Do your answers in parts A and C depend on the angle theta? (Yes/No)

Homework Equations

F=ma
Torque= RxL

The Attempt at a Solution

1) Isn't the magnitude of the force by the icy alley on the ladder just the normal force?

yes, you must calculate it

Since it doesn't mention the ladder sliding it has to be stationary which means all forces are cancelling out. 250+750 = 1000 Newtons upward.

what about the force of the pivot on the ladder? Try summing torques .

2) It should be upward to cancel out the weight of the people and ladder

yes it is upward, but it doesn't cancel out the weights

3) It would be the normal force of the wall on the ladder correct? This would be 0 though since friction is negligible I think.

yes, the normal force in the x direction would be 0, but what about forces in the vertical y direction?

4) The forces shouldn't depend on theta because they are strictly talking forces and not torque.

you must use sum of torques = 0 about any point to solve this problem.

 

[hskrnt8590]

Ok, so the only force acting in the x direction is the force of the pivot on the ladder (F_Nx).
Sum of forces in y= 1000N - F_Ny(force of alley on ladder)
Sum of torques around the pivot= (4m)(250N)(cos\vartheta)+(1.5m)(750N)(cos\vartheta) -(6m)(F_Ny)(cos\vartheta)=0

I canceled out the cos\vartheta and got F_Ny = 354N which was correct

I am still having trouble with number 3 though. The answer wasn't 0.

 

[PhanthomJay]

For number 3, can there be any horizontal force at the pivot? Recheck your thinking about the sum of forces in the y direction. Your 4th line above (correct) does not agree with your 2nd line (incorrect). Also look at sum of forces in x direction.

 

[hskrnt8590]

Nevermind I got it. The pivot was exerting a force upwards on the ladder, so had to change my pivot point to the bottom of the ladder and redo the sum of the torques. I then could cancel out the cos and solve for the missing force The answer was 645.8 N. Thanks for the help.

 

[PhanthomJay]

OK, so you also realize that F_Nx is 0. By the way, a belated Welcome to Physics Forums!

 

[hskrnt8590]

Yes I do now haha. Thanks once again.