1-8(10) - Create assignment, 00223, Homework 8, Nov 04 at...

Info iconThis preview shows pages 1–3. 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: Create assignment, 00223, Homework 8, Nov 04 at 12:57 pm 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. Net Torque on a Wheel 10:07, trigonometry, numeric, > 1 min, wording-variable. 001 A circular shaped object has an inner radius of 12 m and an outer radius of 25 m. Three forces (acting perpendicular to the axis of ro- tation) of magnitudes 11 N, 24 N, and 14 N act on the object, as shown. The force of magnitude 24 N is 31 below horizontal. 11 N 14 N 24 N 11 kg 31 12 m 25 m Find the magnitude of the net torque on the wheel about the axle through the center of the object. Correct answer: 3 . 37 Nm. Explanation: Let : a = 12 m , b = 25 m , F 1 = 11 N , F 2 = 24 N , F 3 = 14 N , and = 31 . F 1 F 3 F 2 M a b The total torque is = a F 2- b F 1- b F 3 = (12 m)(24 N)- (25 m) h (11 N) + (14 N) i =- 3 . 37 Nm k ~ k = 3 . 37 Nm . keywords: Blocks and a Wedge 10:08, trigonometry, numeric, > 1 min, nor- mal. 002 A block of mass 2 kg and one of mass 6 kg are connected by a massless string over a pul- ley that is in the shape of a disk having a radius of 0 . 25 m, and a mass of 9 kg. In addi- tion, the blocks are allowed to move on a fixed block-wedge of angle = 30 , as shown. The coefficient of kinetic friction is 0 . 27 for both blocks. 2 kg 6 k g 30 . 25 m 9 kg Determine the acceleration of the two blocks. The acceleration of gravity is 9 . 8 m / s 2 . Assume: The positive direction is to the right. Correct answer: 0 . 828719 m / s 2 . Explanation: Let : m 1 = 2 kg , m 2 = 6 kg , M = 9 kg , and R = 0 . 25 m . Newtons second low is X ~ F = m~a Create assignment, 00223, Homework 8, Nov 04 at 12:57 pm 2 For m 1 N 1- m 1 g = m 1 a y = 0 N 1 = m 1 g and the force of friction is then f 1 = N 1 = (0 . 27)(2 kg)(9 . 8 m / s 2 ) = 5 . 292 N ....
View Full Document

This note was uploaded on 05/06/2011 for the course PHYS 103N taught by Professor Chiu during the Spring '11 term at University of Texas at Austin.

Page1 / 5

1-8(10) - Create assignment, 00223, Homework 8, Nov 04 at...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online