Ladder on a wall held in place by a peg

[Jacobiam]

Homework Statement

Figure attached
A uniform ladder is 10 m 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 degree angle with the horizontal. The magnitude of the force exerted on the peg by the ladder is:
W = 400N
L = 10m
Answer: 470N

Homework Equations

Torque = r x F

The Attempt at a Solution

Calculating torque about the lower end, I was able to find force exerted by the wall
T_weight = T_wall
400N(5cos30) = F_wall(10sin30)
F_wall = 346.4N

the forces I have so far are: normal force, weight and force from the wall.
Are there two other forces from the peg(perpendicular to the ladder going down) and the force of the ladder on the peg(perpendicular to ladder up)?

 

[haruspex]

Are there two other forces from the peg(perpendicular to the ladder going down) and the force of the ladder on the peg(perpendicular to ladder up)?

You can thonk of the force from the peg as a single force or as composed of two forces at right angles to each other; in the latter choice, you could use vertical and horizontal or perpendicular and parallel to the ladder.
I suggest using vertical and horizontal.

 

[ehild]

Homework Statement

Figure attached
A uniform ladder is 10 m 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.

In your figure, the peg looks as part of the ladder and perpendicular to the leg of the ladder. But the peg should stand out from the ground, probably vertical. Was the figure attached to the problem, or have you drawn it yourself?

the forces I have so far are: normal force, weight and force from the wall.
Are there two other forces from the peg(perpendicular to the ladder going down) and the force of the ladder on the peg(perpendicular to ladder up)?

You have to collect the forces acting on the ladder, and their sum should result zero. Determine the force of the peg on the ladder. The force of the ladder on the peg is equal in magnitude and of opposite direction according to Newton's third law.

 

[CWatters]

The usual form of this problem has friction with the ground rather than a peg preventing it slipping. Unless there is additional info in the original problem statement I would solve the peg version the same way. eg assume the peg can only provide a horizontal force.