Page1 / 3

SOL23 - Thevalingam, Donald Homework 23 Due: Mar 21 2007,...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon

SOL23 - Thevalingam, Donald Homework 23 Due: Mar 21 2007,...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Thevalingam, Donald Homework 23 Due: Mar 21 2007, midnight Inst: Eslami 1 This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Three particles are located in a coordinate system as shown in the figure below. The acceleration of gravity is 9 . 8 m / s 2 . The mass 8 kg is located at the origin, the mass 3 kg is located on the x coordinate, and the mass 6 kg is located on the y coordinate. x y 6 kg 8 kg 3 kg 13 m 4 m Find the distance of the center of gravity from the origin. Correct answer: 2 . 69371 m. Explanation: The center of gravity of the x and y compo- nents from the origin respectively are x c = m 0 + m 1 x 1 m + m 1 + m 2 = (3 kg) (13 m) (8 kg) + (3 kg) + (6 kg) = 2 . 29412 m and y c = m 0 + m 2 y 2 m + m 1 + m 2 = (6 kg) (4 m) (8 kg) + (3 kg) + (6 kg) = 1 . 41176 m . Then the distance of the center of gravity from the origin is d = q x 2 c + y 2 c = q (2 . 29412 m) 2 + (1 . 41176 m) 2 = 2 . 69371 m ....
View Full Document

Ask a homework question - tutors are online