Recent content by 9vswt

Thank you so much for pointing that out ! I've managed to get my answer

Thank you for your help, I've already found the answer

A clearer picture of my free body diagram. Σy=0, F(n)y+F(r)y=981 N (weight of the ladder and the person combined) Σx=0, F(r)x=f(n)x Using principle of moments, take the contact between the ladder and the floor as the pivot. 981 N* 5cos(60°)m=F(n)x*10sin(60°)m [10m being the ladder's length]...

Uhm...so I can only equate Y forces as 0 when I only take the weight of the ladder into account ? So do I equate X when I take both the ladder and the person's weight into account ? Also, what do you mean by rotational equation ? Is it the principle of moments ? If it is, Then I have used it in...

I was trying to use the principle of moments, taking the ground contact as the point of reference

Homework Statement A ladder,10 m long and having a mass of 25 kg, rests on a rough horizontal floor and is supported by a rough vertical wall. The ladder is inclined at 60 degrees as shown and starts to slip when a person having a mass of 75 kg has climbed halfway up it. The coefficient of...