During the collision, is the magnitude of the force of asteroid A on asteroid B greater than, less than, or equal to the magnitu
2 answers:
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Solution :
i. Slip plane (1 1 0)
Slip direction -- [1 1 1]
Applied stress direction = ( 1 0 0 ]
τ = 50 MPa ( Here slip direction must be perpendicular to slip plane)
τ = σ cos Φ cos λ
τ = σ cos Φ cos λ
∴
σ = 122.47 MPa
ii. Slip plane --- (1 1 0)
Slip direction -- [1 1 1]
τ = σ cos Φ cos λ
∴
σ = 122.47 MPa
iii. Slip plane --- (1 0 1)
Slip direction --- [1 1 1]
τ = σ cos Φ cos λ
∴
σ = 122.47 MPa
iv. Slip plane -- (1 0 1)
Slip direction ---- [1 1 1]
τ = σ cos Φ cos λ
∴
σ = 122.47 MPa
∴ (1, 0, -1). (1, -1, 1) = 1 + 0 - 1 = 0