Ladder resting against wall - find forces

[UrbanXrisis]

A 15m ladder weighing 500N rests against a frictionless wall. The ladder makes a 60 degree angle with the horizontal. I need to find the horizontal and vertical force that the ground exerts on the base of the ladder when an 800N firefighter is 4m from the bottom.

the total mg down is 1300N. I don't know why I can't get the answer if I just did mg for the vertical force and mgtan(theta) for the horizontal force? I know this doesn't work.

 

[marlon]

A 15m ladder weighing 500N rests against a frictionless wall. The ladder makes a 60 degree angle with the horizontal. I need to find the horizontal and vertical force that the ground exerts on the base of the ladder when an 800N firefighter is 4m from the bottom.

the total mg down is 1300N. I don't know why I can't get the answer if I just did mg for the vertical force and mgtan(theta) for the horizontal force? I know this doesn't work.

Write down all forces along the x and y axis.
There is one reaction force at the point where the ladder touches the wall and two reaction forces at the ground...

Then apply :
1) sum of all forces is zero
2) sum of all torques is zero

good luck

regards
marlon

 

[UrbanXrisis]

I understand all of it, just not why my method doesn't work...I'm looking at my book and it shows mg acting downwards... that means the normal force is upward. Force of friction has to equal the force at the wall.

I undersatnd what you're trying to say. I just need more information. Yes, I have everything equal to zero. There's just a mental block I need to get over. I don't understand why mg is not the nornal force. What is the equation for the force of friction and normal force?

 

[whozum]

When you stand on a ladder leaning on a wall you are applying force both to the floor and the wall. Think about what would happen if you took away the wall, your ladder will slam to the ground. Not all the force of gravity is applied to to the floor, some of it is applied to the wall.

 

[UrbanXrisis]

right, so the normal force is mg sin(theta) and the friction force is mg cos(theta)

but that's not it, is it?

 

[whozum]

I think the normal force is actually mg cos(theta). The friction force is just the normal force times the coefficient of friction.

 

[UrbanXrisis]

does the fact that the firefighter is 4 m from the ground make a difference in my calculations?