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A) Force of the wall on the ladder: 186.3 N
B) Normal force of the ground on the ladder: 725.2 N
C) Minimum value of the coefficient of friction: 0.257
D) Minimum absolute value of the coefficient of friction: 0.332
Explanation:
a)
The free-body diagram of the problem is in attachment (please rotate the picture 90 degrees clockwise). We have the following forces:
: weight of the ladder, with m = 20 kg (mass) and
(acceleration of gravity)
: weight of the person, with M = 54 kg (mass)
: normal reaction exerted by the wall on the ladder
: normal reaction exerted by the floor on the ladder
: force of friction between the floor and the ladder, with
(coefficient of friction)
Also we have:
L = 4.1 m (length of the ladder)
d = 3.0 m (distance of the man from point A)
Taking the equilibrium of moments about point A:
where
is the component of the weight of the ladder perpendicular to the ladder
is the component of the weight of the man perpendicular to the ladder
is the component of the normal force perpendicular to the ladder
And solving for , we find the force exerted by the wall on the ladder:
B)
Here we want to find the magnitude of the normal force of the ground on the ladder, therefore the magnitude of .
We can do it by writing the equation of equilibrium of the forces along the vertical direction: in fact, since the ladder is in equilibrium the sum of all the forces acting in the vertical direction must be zero.
Therefore, we have:
And substituting and solving for N2, we find:
C)
Here we have to find the minimum value of the coefficient of friction so that the ladder does not slip.
The ladder does not slip if there is equilibrium in the horizontal direction also: that means, if the sum of the forces acting in the horizontal direction is zero.
Therefore, we can write:
And re-writing the equation,
So, the minimum value of the coefficient of friction is 0.257.
D)
Here we want to find the minimum coefficient of friction so the ladder does not slip for any location of the person on the ladder.
From part C), we saw that the coefficient of friction can be written as
This ratio is maximum when N1 is maximum. From part A), we see that the expression for N1 was
We see that this quantity is maximum when d is maximum, so when
d = L
Which corresponds to the case in which the man stands at point B, causing the maximum torque about point A. In this case, the value of N1 is:
And substituting, we get
And therefore, the minimum coefficient of friction in order for the ladder not to slip is
Learn more about torques and equilibrium:
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