Recent content by iceice655

So currently what is incorrect, and what is correct? Based on what you said, am I right to say ##F_{N (wall on ladder)} = F_{f (ground on ladder)}## so that ##L \cdot F_{N (wall on ladder)} \sin{\theta} = \frac L 2 Mg \cos{\theta}## then I need to figure out how to replace ##F_{N (wall on...

I chose the point where the ladder and floor touch as my pivot point, I thought this is what I have been doing? Don't only forces perpendicular to the lever arm result in torque on the ladder? On another note, looking at just the vertical forces on the FBD, why couldn't I say the ##F_N## of...

So at this point I know ##F_N## of the wall on the ladder is equal to the ##F_f## at the base of the ladder. So my equation is ##F_f \sin{\theta} = \frac 1 2 Mg \cos{\theta}##. Is it right at this step? My next step should be replacing ##F_f## with the given values, but I am stuck.

I assumed the friction force at the base was ##\mu \cdot F_N##. ##F_N## being that of the ground on the ladder and what I am looking for. Is this correct?

So I was thinking in the y-direction, the normal force of the wall on the ladder would equal the friction force at the base of the ladder. (Sorry, touchpads aren't the best writing tools, if there needs to be more clarification in the drawing, I will modify it)

Ok what I tried doing was make ##\sum{\tau} = 0## and ##\sum{F} = 0##. Taking the base as the pivot point, I found the perpendicular (to the ladder) force of M at the center to be ##mg \cos(\theta)##, and the normal force of the wall on the ladder at the top to be ##\mu F_N \sin(\theta)##...

Hello, I was recently tested on finding the normal force at the base of a ladder leaning against a wall as well as its friction force. So this is the question from memory. Given: θ the acute angle between the ground and the ladder μ as the coefficient of friction between the ground and the...