Daniel B. answered • 11/09/20
A retired computer professional to teach math, physics
Please draw a picture of the situation, where
the ground is the x-axis, the wall is the y-axis, and the ladder is the in first quadrant.
Let
L = 7.47 m be the length of the ladder,
Θ be a variable angle between the ground and the ladder,
θ = 60.2° be the value of Θ for which we are to solve the problem,
f be the smallest coefficient of friction (to be calculated) preventing the ladder from sliding when Θ = θ,
x(Θ) = L cos(Θ) be the horizontal distance between the wall and the foot of the ladder,
m1 = 14.5kg be the mass of the ladder,
x1(Θ) = L cos(Θ) / 2 be the distance of the ladder's center of gravity from the wall,
y1(Θ) = L sin(Θ) / 2 be the height of the ladder's center of gravity,
m2 = 41.3 kg be the mass of the woman,
x2(Θ) = L cos(Θ) / 4 be the distance of the woman from the wall,
y2(Θ) = L sin(Θ) 3/4 be the height of the woman,
Fv(Θ) be the vertical force (due to gravity) acting at the foot of the ladder,
Fh(Θ) be the horizontal force (due to friction) acting at the foot of the ladder,
g = 9.8 m/s2 be gravitation acceleration.
We need to calculate Fv and Fh because their ratio is the coefficient of friction f.
There is also a vertical and horizontal force acting on the ladder at the wall,
but we will not need them, because the friction at the wall is 0.
To calculate Fv we use equality of torgues around the point where the ladder touches the wall.
The weights g m1 and g m2 are trying to turn the ladder in clockwise,
and the force of the ground, equal (in magnitude) to Fv, is trying to turn it counter-clockwise.
The friction force Fh plays no role because that would act only if the ladder were moving.
The force Fv acts with torque Fv x,
and the two weights act with torgue g m1 x1 + g m2 x2.
Fv x = g m1 x1 + g m2 x2
Fv = (g m1 L cos(Θ) / 2 + g m2 L cos(Θ) / 4) / L cos(Θ)
= g m1 / 2 + g m2 / 4
Please note that Fv turns out to be independent of Θ.
To calculate Fh we will use conservation of energy.
Imagine that the ladder starts sliding.
Then the potential energy of the ladder and the woman is being converted
into their kinetic energy and the energy dissipated through friction.
At the start of any sliding the speed is 0, and therefore so is their kinetic energy.
Therefore the entire loss of potential energy gets dissipated by friction.
In other words, if the angle Θ changes by a very small amount dΘ then
gravity performs small amount of work to be equal the work performed by friction Fh.
A change of dΘ causes these changes
dx = (dx/dΘ) dΘ = -L sin(Θ) dΘ
dy1 = (dy1/dΘ) dΘ = (L cos(Θ) / 2) dΘ
dy2 = (dy2/dΘ) dΘ = (L cos(Θ) 3/4) dΘ
To understand the meaning of positive and negative signs, as in the value of dx,
imaging that dΘ is positive, i.e., the ladder is spontaneously rising.
That would make y1 and y2 increase (positive change) and x decrease (negative change).
If the ladder is falling then dΘ is negative and the polarities reverse.
The direction of the force Fh is always opposite to the direction of dx.
Then the work of gravity is g m1 dy1 + g m2 dy2, and the work of friction is Fh dx.
The equality of those two:
Fh dx = g m1 dy1 + g m2 dy2
Fh dx/dΘ = g m1 dy1/dΘ + g m2 dy2/dΘ (after dividing by dΘ )
Fh(-L sin(Θ)) = g m1 (L cos(Θ) / 2) + g m2 (L cos(Θ) 3/4) (after substituting the derivatives)
Fh = -g cot(Θ) (m1 / 2 + 3 m2 / 4)
The desired coefficient of friction f is then the ratio between the two forces when Θ = θ
f = Fh(θ)/Fv
= g cot(θ) (m1 / 2 + 3 m2 / 4) / (g m1 / 2 + g m2 / 4)
= cot(θ) (2m1 + 3m2) / (2m1 + m2)
Substituting actual values
f = cot(60.2°) (2 . 14.5 + 3 . 41.3) / (2 . 14.5 + 41.3) = 1.2
