A particle moves through an xyz coordinate system while a force acts on it. When the particle has the position vector vector r = (2.00 m)ihat - (3.00 m)jhat + (2.00 m)khat, the force is vector F = Fxihat + (7.00 N)jhat - (5.30 N)khat and the corresponding torque about the origin is ô = (1.90 N·m)ihat + (-1.60 N·m)jhat + (-4.30 N·m)khat. Determine Fx.
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To determine Fx, we can use the relationship between force (F), torque (τ), and position vector (r) given by the equation:
τ = r x F
Here, "x" denotes the cross product between vectors. Since torque (τ) is given as ô = (1.90 N·m)ihat + (-1.60 N·m)jhat + (-4.30 N·m)khat, and the position vector (r) is given as r = (2.00 m)ihat - (3.00 m)jhat + (2.00 m)khat, we can substitute these values into the equation to solve for F.
Dividing both sides of the equation by the magnitude of r, we have:
(τ / |r|) = ((r x F) / |r|)
Since the magnitude of the position vector |r| is sqrt((2.00^2) + (-3.00^2) + (2.00^2)) = sqrt(17) m, we can write the equation as:
(τ / sqrt(17)) = ((2.00i - 3.00j + 2.00k) x F) / sqrt(17)
Now, let's expand the cross product:
τ = (2.00i - 3.00j + 2.00k) x F
Using the determinant method for cross products, we can write:
τ = (3.00 • 2.00 - 2.00 • (-1.60))i - (2.00 • -4.30 - 2.00 • 1.90)j + (2.00 • -1.60 - 3.00 • -4.30)k
Simplifying the equation, we get:
τ = (12.00 + 3.20)i - (-8.60 - 3.80)j + (-3.20 + 12.90)k
= 15.20i + (-12.40)j + 9.70k
Comparing this with the given torque, we have:
(τ / sqrt(17)) = ((15.20i + (-12.40)j + 9.70k) / sqrt(17))
Now, we equate the corresponding components of the two sides:
τx / sqrt(17) = 15.20 / sqrt(17)
τy / sqrt(17) = -12.40 / sqrt(17)
τz / sqrt(17) = 9.70 / sqrt(17)
Solving for τx, we have:
τx = 15.20 N·m / sqrt(17)
Thus, Fx can be determined as the x-component of the force vector F:
Fx = τx = 15.20 N·m / sqrt(17)