Determine the moment of inertia of the thin ring about the z axis. The ring has a mass m.
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Notes Week 2 LECTURE 3 Position vector from A to B = (Xb – Xa) i, (Ya – Yb)j, (Za – Zb)k STEPS TO FINDING FORCE VECTOR a. Find position vector along 2 points ^^^ b. Find the unit vector r(ab)/r(ab) c. Multiply the unit vector by the magnitude of the force EXAMPLE: r(ac) = (2i + 3j 6k)m and F = 420 N r(ac) = √2^2 + 3^2 + (6)^2 = 7m u(ac) = (2/7i + 3/7j + 6/7k) F(ac) = F(ac) * u(ac) F(ac) = (420N)* (2/7i + 3/7j 6/7k) F(ac) = (120i + 180j – 360k)N DOT PRODUCT The dot product of vectors A and B is defined as A ∙ B = ABcosϴ ϴ is always the smallest angle between A and B 1. The result of the dot product is a scalar 2. The units of the dot product is the product of the units of A and B **Basically asking how much of A lies in the axis of B or vice ver
Textbook: Engineering Mechanics: Statics & Dynamics
Author: Russell C. Hibbeler
Since the solution to 17-3 from 17 chapter was answered, more than 327 students have viewed the full step-by-step answer. The answer to “Determine the moment of inertia of the thin ring about the z axis. The ring has a mass m.” is broken down into a number of easy to follow steps, and 19 words. This full solution covers the following key subjects: ring, mass, inertia, determine, moment. This expansive textbook survival guide covers 22 chapters, and 2358 solutions. Engineering Mechanics: Statics & Dynamics was written by and is associated to the ISBN: 9780133951929. This textbook survival guide was created for the textbook: Engineering Mechanics: Statics & Dynamics , edition: 14. The full step-by-step solution to problem: 17-3 from chapter: 17 was answered by , our top Engineering and Tech solution expert on 11/10/17, 05:20PM.