We can solve the problem by requiring the equilibrium of the forces and the equilibrium of torques.
1) Equilibrium of forces:

where

is the weight of the person

is the weight of the scaffold
Re-arranging, we can write the equation as

(1)
2) Equilibrium of torques:

where 3 m and 2 m are the distances of the forces from the center of mass of the scaffold.
Using

and replacing T1 with (1), we find

from which we find

And then, substituting T2 into (1), we find
To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.
In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.
By definition we know that the tensile strength is defined as

Where,
Tensile strength
F = Tensile Force
A = Cross-sectional Area
In the other hand we have that the shear strength is defined as

where,
Shear strength
Shear Force
Parallel Area
PART A) Replacing with our values in the equation of tensile strenght, then

Resolving for F,

PART B) We need here to apply the shear strength equation, then



In such a way that the material is more resistant to tensile strength than shear force.
Answer:
Their efforts would be expressed in units of Joules per second
Explanation:
The unit of their efforts can be derived from the formula of power which is given by the product of mass, acceleration and distance (the product is energy with unit joules) divided by time taken to complete the task (unit is seconds)
Therefore, the unit of their efforts would be joules per second
To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.
The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,


M,m = Counts per second
Our radios are given by



Therefore replacing we have that,






Therefore the number of counts expect at a distance of 20 cm is 19.66cps