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Rashid [163]
4 years ago
13

Sometimes cos ϕ.cosλ in Equation 7.2 is termed the Schmid factor. Determine the magnitude of the Schmid factor for an FCC single

crystal oriented with its [120] direction parallel to the loading axis.The slip occurs on the (111) plane and in the [] direction.

Physics
1 answer:
BARSIC [14]4 years ago
3 0

Answer:

Explanation:

Attached is the solution

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We have the following data:

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Particle A of charge 2.79 10-4 C is at the origin, particle B of charge -5.64 10-4 C is at (4.00 m, 0), and particle C of charge
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Answer:

a) 0 b) 29.9 N c) 21.7 N d) -17.4 N e) -13.0 N f) -17.4 N  g) 16.9 N

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b) Similarly, as Fx = 0, the entire force is directed along the y-axis, and is going upward, due both charges repel each other.

Fyca = k*qa*qc / rac² = (9.10⁹ N*m²/C²*(2.79)*(1.07)*10⁻⁸ C²) / 9.00 m²

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⇒ Fbc = k*qb*qc / rbc² = (9.10⁹ N*m²/C²*(5.64)*(1.07)*10⁻⁸ C²) / 25.0 m²

⇒ Fbc = 21.7 N

d) In order to get the x component of Fbc, we need to get the projection of Fcb over the x axis, taking into account that the force on particle C is attractive, as follows:

Fbcₓ = Fbc * cos (-θ) where θ, is the angle that makes the line of action of the force, with the x-axis, so we can write:

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The angle from the horizontal can be found as follows:

Ф = arc tg (16.9 / 17.4) = 44.2º

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