Answered

Westonci.ca is the Q&A platform that connects you with experts who provide accurate and detailed answers. Discover the answers you need from a community of experts ready to help you with their knowledge and experience in various fields. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

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 :

LammettHash

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

We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.