Force on peg preventing a uniform ladder from slipping

[Ishida52134]

Homework Statement

A uniform ladder is 10m long and weighs 400 N. It rests with its upper end against a frictionless
vertical wall. Its lower end rests on the ground and is prevented from slipping by a peg driven
into the ground. The ladder makes a 30◦ angle with the horizontal. The magnitude of the
force exerted on the peg by the ladder is:

A. zero
B. 200N
C. 400N
D. 470N
E. 670N

Homework Equations

torque = r x F

The Attempt at a Solution

First I identified the forces. So normal force, frictional force, weight, and force from the wall and force from the peg which acts horizantally.
Basically, I just took moments about the base of the ladder. I got -400(5cos30) + 5Fw = 0.
Then I solved for Fw getting 346.
Then I set all x forces equal to each other. Getting 346 = Fpeg + Ffrictional.
And I got stuck lol.

I don't know if there's no frictional force at all and that the peg is acting in place of it or if it's the sum of them together that provides the x component.

 

[Ishida52134]

any ideas

 

[PhanthomJay]

There is no friction at the peg. So calculate the horizontal force that the peg exerts on the ladder. Don't forget the vertical force also.

 

[Ishida52134]

oh thanks.
But I keep getting 529 as the magnitude which isn't any of the answer choices. And the answer is supposed to be 470N

 

[PhanthomJay]

Yeah, that's what I get. looks like a book error to me.