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dangina [55]
3 years ago
15

Find the Cartesian components of force PP acting in the x, y, and z directions given P=50.0 NP=50.0 N, α=143.5∘α=143.5∘, β= 70.0

∘β= 70.0∘, and γ=60.9∘γ=60.9∘. Recall that αα is the angle between the vector and the x axis, ββ is the angle between the vector and the y axis, and γγ is the angle between the vector and the z axis.
Physics
1 answer:
BlackZzzverrR [31]3 years ago
4 0

Answer:

vec(F) = [ -40.193 i + 17.101 j + 24.317 k ]

Explanation:

Given:

-The force P = 50.0 N

- Angle with x-axis α = 143.5

- Angle with y-axis β = 70.0

- Angle with z-axis γ = 60.9

Find:

Find the Cartesian components of force P acting in the x, y, and z directions,

Solution:

- The Force vector in the Cartesian coordinate system is given by the dot product of the Force P and the unit vector in its direction.

                                    F.unit(u) = vec(F)

- Where, the unit vector is defined as:

                      unit (u) = [ cos(α) i + cos(β) j + cos(γ) k ]

- Using the given unit angles α , β, and γ compute unit (u):

                      unit (u) = [ cos(143.5) i + cos(70) j + cos(60.9) k ]  

                      unit (u) = [ -0.80386 i + 0.34202 j + 0.48634 k ]

- The force vector is:

                      vec(F) =  50.[ -0.80386 i + 0.34202 j + 0.48634 k ]

                      vec(F) = [ -40.193 i + 17.101 j + 24.317 k ]

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Answer:

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Explanation:

1)

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       F_{g} = F_{c} (1)

  • where Fg is the gravitational attraction, that can be written as follows        according Newton's Universal Law of Gravitation:

       F_{g} = G*\frac{m_{1}*m_{2}}{r_{12}^{2}} (2)

  • Fc, due to it is the electrical repulsion between both charged spheres, must obey Coulomb's Law (assuming  we can treat both spheres as point charges), as follows:

       F_{c} = k*\frac{q_{1}*q_{2}}{r_{12}^{2}} (3)

  • since m₁ = m₂ = 0.0024 kg, and  r₁₂ = 0.1m, G and k universal constants, and q₁ = q₂ = Q, we can replace the values in (2) and (3), so we can rewrite (1) as follows:

       G*\frac{(0.0024kg)^{2}}{r_{12}^{2}} = k*\frac{Q^{2}}{r_{12}^{2}} (4)

  • Since obviously the distance is the same on both sides, we can cancel them out, and solve (4) for Q² first, as follows:

       Q^{2} = \frac{6.67e-11*(0.0024kg)^{2}}{9e9Nm2/C2} = 4.27*e-26 C2 (5)

  • Since both charges are the same, the charge on each sphere is just the square root of (5):
  • Q = 2.066* 10⁻¹³ C.

2)

  • Assuming that both spheres were electrically neutral before being charged, the negative charge removed must be equal to the positive charge on the spheres.
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  • n_{e} =\frac{-2.066e-13C}{-1.6e-19C} = 1,291,250 electrons. (6)
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