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2 answers:
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Answer:
a) the values of the angle α is 45.5°
b) the required magnitude of the vertical force, F is 41 lb
Explanation:
Applying the free equilibrium equation along x-direction
from the diagram
we say
∑Fₓ = 0
Pcosα - 425cos30° = 0
525cosα - 368.06 = 0
cosα = 368.06/525
cosα = 0.701
α = cos⁻¹ (0.701)
α = 45.5°
Also Applying the force equation of motion along y-direction
∑Fₓ = ma
Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)
525sin45.5° + F + 212.5 - 600 = 27.95
374.46 + F + 212.5 - 600 = 27.95
F - 13.04 = 27.95
F = 27.95 + 13.04
F = 40.99 ≈ 41 lb
Answer:
we have to find out the critical resolved shear stress. As it it given in the question
Ф = 28.1°and the possible values for λ are 62.4°, 72.0° and 81.1°.
a) Slip will occur in the direction where cosФ cosλ are maximum. Cosine for all possible λ values are given as follows.
cos(62.4°) = 0.46
cos(72.0°) = 0.31
cos(81.1°) = 0.15
Thus, the slip direction is at the angle of 62.4° along the tensile axis.
b) now the critical resolved shear stress can be find out by the following equation.
τ = σ
( cosФ cosλ)
now by putting values,
= (1.95MPa)[ cos(28.1) cos(62.4)] = 0.80 MPa (114 Psi) 7.23