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What is the magnitude of the torque about the origin of the force (−3i^ 3 j^ 4 k^) n if it is applied at the point whose position vector is r⃗ =(−2i^ 4j^) m.

Sagot :

LammettHashidn

The torque vector τ has magnitude

|τ| = |F| |r| sin(θ)

where θ is the angle made by the force vector F and position vector r.

Recall the dot product identity,

F • r = |F| |r| cos(θ)

and use to find the measure of the angle in question.

F = (-3 i + 3 j + 4 k) N

r = (-2 i + 4 j) m

F • r = (6 + 12 + 0) N•m = 18 N•m

|F| = √((-3 N)² + (3 N)² + (4 N)²) = √34 N

|r| = √((-2 m)² + (4 m)²) = √20 m

So, we have

cos(θ) = (18 N•m) / ((√20 m) (√34 N)) = 9/√170

⇒   θ = arccos(9/√170) ≈ 46.3°

Now just find |τ| :

|τ| = (√34 N) (√20 m) sin(arccos(9/√170))

|τ| = (√680 N•m) sin(arccos(9/√170))

|τ| = (√680 N•m) √(89/170)

|τ| = 2√89 N•m ≈ 18.9 N•m

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