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2 answers:
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Answer:
τ (bc. max) =25.37 MPa
Explanation:
From the question, T = 1.3Kn.m;
(G = 77.2 GPa) and from the image of this solid shaft system i attached;
d(ab) = 50mm; d(bc) = 38mm; L(ab) =0.2m and L(bc) = 0.25m
So ΣT = 0 → Ta + Tc = 1.3Kn.m
So the system is statically indeterminate.
Let's check at the equation that makes it compatible ;
ψ = 0 and ψ(c/b) + ψ(b/a) + ψ(a) = 0
[{T(bc)} / {J(bc)G}] + [{T(ab)} / {J(ab)G}] + 0 =
ΣT = 0 and T(bc) = T(c)
ΣT = 0 and T(ab) = T(c) - 1.3 Kn.m
Now,
[(T(c) x 250mm)/{(π/2)(19^4)}] + [{((T(c) - 1300000 Nmm) x 200mm}/{(π/2)(25^4)}] = 0
So, T(c) = 273374 Nmm = 273.374Nm
T(a) = 1300Nm - 273.374Nm = 1026. 63Nm
From the beginning we saw that;
T(bc) = T(c) = 273.374Nm
Now let's find the maximum shear stress in shaft BC;
τ (bc. max) = {τ (bc) x r(bc)} / J(bc)
τ (bc. max) = (273374Nmm x 19mm)/ {(π/2)(19^4)} = 5194106 / 204707
= 25.37 MPa
Answer:
7.5 x 10⁻⁸N
Explanation:
Given parameters:
Mass 1 = 60kg
Mass 2 = 75kg
Distance between the bodies = 2m
Unknown:
Gravitational fore = ?
Solution:
The gravitational force between the two bodies can be derived using;
F =
G is the universal gravitation constant = 6.67 x 10⁻¹¹m³kg⁻¹s⁻²
Insert the parameters and solve;
F = = 7.5 x 10⁻⁸N