Question

a man climbs the ladder. the ladder rest on a frictionless wall but the bottom end rest on a rough surface. neglect the rotational effect on the system. please help me to draw a free body diagram bout this! Thanks

Answer 1

First, list all of your forces (there are a few of them):

Vertical forces:
Weight of the ladder (down)
Weight of the man (down)
Normal force from the ground onto the ladder (up)

Horizontal forces:
Normal force from the wall (even though it has no friction, it still exerts a normal force!)
Friction from the ground onto the ladder

For your free body diagram, just have the friction force pointing toward the wall (in other words, if the ladder is to the left of the wall, the friction points right, if the ladder is to the right of the wall, friction points left). Everything else should be pretty straight forward.

Answer 2

@dudebobmac yay! thank you so much!! =)

Answer 3

No problem :)

Answer 4

assume the man to be a point mass and the ladder to be inclined at an angle theta
so the dotted arrow represents the force of gravity acting on the man
so by breaking the force into two components
one i.e. parallel to the inclined ladder and the other that is perpendicular to the inclined ladder
and the normal force provided by the ladder to the man be N and t will be equal to mgcos theta so as to balance the force in that direction
and even i have chosen to my co-ordinate system to have it's axes perpendicular and parallel to the inclined ladder
so the fbd of the man looks something like this
i have drawn the force of gravity with a dotted line as i have broken it into two components and now we will take into account the forces that we've got by breaking the force mg and the force represented with a dotted line i.e. mg will not be taken into account while solving a problem
always remember the following points while drawing a fbd:
first of all chose a convinient coordinate system
next, analyze the forces
next break them along the axes of your coordinate system
and then enjoy solving physics:)
one more point for this question
the angle between the dotted line force(mg) and the force perpendicular to the plane i.e.
mgcos theta is theta
if you want some help wth resolution(breaking) of vectors, you can ask for it:)
hope this helps you:)