Horizontal components of force on an object?

Question

A uniform 12 kg, 2.8 m long ladder rests against a vertical frictionless wall at a point 2.4 meters off the floor. Find the vertical end horizontal components of force exerted by the ladder on the floor.


I am very confused on this problem assigned by my physics professor and would like any help possible in finding a way to solve. He gave us the answers of F(Y)=117.6N and F(X) = 55N BUt i don't know how he got those answers so any help would be great!
Thanks

Answer 1

F(Y)=117.6N becasue W=mg

m=12kg
g=9.8m/s/s
12*9.8=117.6N

but i disagree with the second part of f(X)=55N. though i could be wrong i guess

start with some trig.

sin^-1(2.4/2.8)=58.99 deg.

you know your Y component is 117.6(opposite side in a trangle from the angle just found) and you are looking for your X component(adjacent side to your angle) and that tan=opp/adj

so:
117.6/tan(58.99)=70.68N

i'm not saying your teacher is wrong, but if he is just getting the numbers out of the answer book then the book could be wrong. i see this all the time even in my college books.

Answer 2

using trig and Pythagorean theorem

The triangle is a right triangle with the
a=2.3
h=2.8
b=sqrt(2.8^2-2.4^2)

the ladder weighs a total of
w=12*9.8 N
which is the vertical force on the floor of 117.6N

I don't get 55N, though
Using torque at the base,
m*g*(2*tan(angle))
is the reaction force off the wall, which is in the horizontal direction, which is equal to the frictional force.
117.6/(2*(2.4/(sqrt(2.8^2-2.4^2)))
=35N

http://farside.ph.utexas.edu/teaching/301/lectures/node133.html


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