In this problem i've got a person with mass 80 kilograms climbing a ladder of length 5 meters and
the ladder itself has non-negligible mass that's 15 kilograms it's propped up against a smooth wall
meaning i don't have to worry about friction there i know the angle of elevation is 50 degrees and at
the bottom the ladder feels a static friction coefficient of 0.4 i'm trying to figure out
the maximum height the person can climb before the ladder slips and before i start doing any
calculations i want to figure out why should there be a maximum height they can climb to before it
slips like what's going on physically so i'm going to start looking at some force vectors in here
one of them is this contact point with the wall there's a normal force exerted there
that i'll call nw for wall and then i have gravity pulling down on the center of mass of the ladder
with a magnitude of ml times g and then gravity is pulling down at the center
of mass of the person so i just estimated where that would be in the picture and put mp
times g and then at the contact point with the ground i have the normal force
exerted straight up so i'll call that n f for the floor and then i've got to think about which way
does the static friction force point and i see the normal force exerted by the wall pushing to the
left which just qualitatively thinking about the sum of the forces in the x direction that means
the static friction force must point to the right so that the x forces will sum to zero
and then real quick i'm going to look at a qualitative torque analysis around the
contact point with the ground to try to figure out what's going on
so i have two clockwise torques in my analysis and they're caused by the force of gravity down
on the center of mass of the ladder together with the force of gravity down on the person
well as this person climbs higher and higher and higher up the ladder that torque due
to their weight is increasing that means my counterclockwise torque must increase which means
the normal force exerted by the wall increases and that means my friction force pointing to the
right must increase so i guess what's happening here is we eventually get to the point where that
rightward pointing static friction force is maxed out so i can say that it's mu s
times the normal force all right so now i get like why the ladder might slip when someone
climbs higher and higher on it and i need to get into the quantitative analysis of the problem
so the first thing i'm going to do is the torque analysis about the contact point with the ground
so the sum of all the torques around that axis must vanish that means the
clockwise torques must add up to the same as the counterclockwise torques and i've got some
difficult trig to do here if i draw a perpendicular through the center of mass in
the ladder i've got to figure out what this angle is so i can get the perpendicular component of the
weights well this is exactly the same diagram we've used dozens of times for inclined planes
and it just turned out that angle was the same as the angle of incline and we're going to do
the same thing at the location of the person one last thing to do here is to realize that
my angle of incline is an alternate interior angle with this angle for the normal force exerted by
the wall so that's going to be 50 degrees as well plugging all that in and i'm going to
stay symbolic for a while i have my clockwise torques mass of the ladder times g times the
cosine of 50 degrees i'll just leave it as a theta for now exert it through a lever arm of l over 2
because it's happening right at the center of the ladder then i have the mass of the person times g
cosine of the angle exerted through a lever arm of x and that's the prize here we're trying to find
x that's going to be the height the person is at when the friction force is maxed out
and then i have my counterclockwise torque exerted by the normal force at the wall which is going to
be nw sine of theta it's happening at the end of the ladder so the lever arm there is the entire
length of the ladder l okay so that's nice but there's two unknowns in it
there's x and there's the normal force exerted by the wall so i've got to keep going and the first
thing i'll do is to write down the sum of the forces in the x direction must be equal to zero
that means all the forces pointing to the right must be equal to all the forces pointing to the
left and there's not much going on there i have the friction force pointing to the right
and i have the normal force exerted by the wall pointing to the left so there's my second equation
and i'm making progress here but i did introduce another unknown it was the normal force exerted by
the floor so i need to write down another equation and i'm going to go to the vertical force analysis
i could write it this way all the forces pointing up must be equal to all the forces pointing down
so the normal force exerted by the floor is the only upward pointing force
and then i have the weight of the ladder and the weight of the person as the only force is pointing
down so this last equation allows me to actually write down a number for the normal force exerted
at the floor if i wanted to but i'm still going to keep it symbolic for a second so i plug that
nf in to my second equation and i find out the normal force exerted by the wall is going to be
mu s times nf and i'm going to go ahead and do some factoring when i write down nf it's
g times the sum of the two masses now i can plug that expression for the normal force at the wall
back into my original torque equation and i should be all set to actually put numbers into this thing
so i'm going to go down to the bottom and i get ml g cosine theta times l over 2 plus
mpg cosine theta times x is equal to mu s g times the quantity ml plus mp times l
so i didn't see a lot of advantage to staying symbolic here i just went ahead and plugged in
all the numbers at this step so the mass of the ladder was 15 kilograms g is 9.8 theta
is 50 degrees the length of the ladder is 5 meters the mass of the person was 80 kilograms
my static friction coefficient was 0.4 plug all that stuff in and smash it in your calculator
and you get 236.2 plus 503.9 x equals 1426.4 solving for x
i get 2.36 meters which means the person doesn't even make it halfway up the ladder before it
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