PLS HELP FOR PHYSICS PLS
1 answer:
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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:
Check attachment for better understanding
Explanation:
Given that
a= 0.75m
b=1.31m
c= 2.2m
Weight of pole is 26.90
Then, Fg = Weight = 26.90
Using Equilibrium of forces
ΣFy = 0
U — D — Fg = 0
U — D = Fg
U — D = 26.9
To calculate U,
We will take moment about point A.
ΣMa = 0
Let the clockwise moment be positive and anti-clockwise be negative
Fg(a+b) — U(a) = 0
26.9(1.31+0.75) —0.75U = 0
26.9(2.06) = 0.75U
0.75U = 55.414
U = 55.414/0.75
U = 73.89 N
To calculate D,
U — D = 26.9
73.89—D =26.9
73.89—26.9 = D
D = 46.99N
