Optimal Ladder Position for Painting: Friction and Wall Conditions Explained

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Ladder Problem (urgent)

A light of negligible mass and length d is supported on a rough floor and leans against a smooth vertical wall makeing an angle (theta) with the floor. The coefficient of friction between the ladder and floor is M.

If a painter climbs the ladder up distance x, what value of x is when the ladder begins to slip? And how far can he climb is the floor is smooth and the wall is rough?

Attempt at solution:

Have a force diagram with mass of painter Mg downwards, with anti clockwise moment xMgcos(theta) taking base of the ladder as a pivot.

clockwise moment from the top off ladder reaction from the wall R2.
= R2cos(theta)d (i think)

R2 = Mmg (i think)

So both moments ---> Mmgcos(theta)d = mgcos(theta)x

so from this x = dM which doesn't seem right..
 
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Myrddin said:
A light ladder of negligible mass and length d is supported on a rough floor and leans against a smooth vertical wall making an angle θ (theta) with the floor. The coefficient of friction between the ladder and floor is M. (I'd rather use μ .)

If a painter climbs a distance x up the ladder, at what value of x does the ladder begin to slip? And how far can he climb is the floor is smooth and the wall is rough?

Attempt at solution:

Have a force diagram with mass of painter Mg downwards, with anti clockwise moment x·m·g·cos(θ) taking base of the ladder as a pivot.

clockwise moment from the top off ladder reaction from the wall R2.
= R2·cos(θ)·d (i think)
This should be R2·sin(θ)·d .
R2 = μ·m·g (i think)

So both moments ---> μ·m·g·cos(theta)d = m·g·cos(theta)x

so from this x = d·μ which doesn't seem right.
R2 looks OK.

Just fix the anti-clockwise moment.
 


Anti clock wise moment should equal, dR2sin(theta)? so ---> dMmgsin(theta)
this gives anticlock wise = clockwise

So dMmgsin(theta) = xmgcos(theta)
so x = dMmgsin(theta) / mgcos(theta) = dMtan (theta)?
 


x = d·μ·tan(θ) looks good to me.

x is proportional to d and μ. x increases as θ increases.