Homework 17

# Homework 17 - Gozick, Brandon – Homework 17 – Due: Nov...

This preview shows pages 1–3. Sign up to view the full content.

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Gozick, Brandon – Homework 17 – Due: Nov 10 2006, 5:00 pm – Inst: D Weathers 1 This print-out should have 10 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 Two weights attached to a uniform beam of mass 15 kg are supported in a horizontal po- sition by a pin and cable as shown in the figure. The acceleration of gravity is 9 . 8 m / s 2 . 43 kg 19 kg 2 . 5 m 7 . 6 m 42 ◦ 15 kg What is the tension in the cable which sup- ports the beam? Correct answer: 0 . 595278 kN. Explanation: Let : m = 15 kg , M 1 = 43 kg , M 2 = 19 kg , L 1 = 2 . 5 m , L 2 = 7 . 6 m , and θ = 42 ◦ . Basic Concepts: X ~ F = 0 X ~ τ = 0 . Solution: The sum of the torques about the pivot is T L 2 sin θ- mg L 2 2- M 1 gL 1- M 2 gL 2 = 0 Solving for the tension T T = m g L 2 2 + M 1 g L 1 + M 2 g L 2 L 2 sin θ = (15 kg)(9 . 8 m / s 2 ) µ 7 . 6 m 2 ¶ (7 . 6 m)sin θ + (43 kg)(9 . 8 m / s 2 )(2 . 5 m) (7 . 6 m)sin θ + (19 kg)(9 . 8 m / s 2 )(7 . 6 m) (7 . 6 m)sin θ = 0 . 595278 kN . keywords: 002 (part 1 of 1) 10 points A uniform plank of length L = 7 . 7 m and mass 37 kg rests horizontally on a scaffold, with ‘ = 1 . 3 m of the plank hanging over one end of the scaffold. The acceleration of gravity is 9 . 8 m / s 2 . L l x How far can a painter of mass 210 kg walk on the overhanging part of the plank x before it tips? Correct answer: 0 . 449286 m. Explanation: Basic Concepts: In equilibrium, X ~ F = 0 X ~ τ = 0 Solution: Considering the edge of the scaf- fold to be the pivot point, then X τ = 0 m g • L 2- ‘ ‚- M g x = 0 Gozick, Brandon – Homework 17 – Due: Nov 10 2006, 5:00 pm – Inst: D Weathers 2 Therefore, x = m M • L 2- ‘ ‚ = (37 kg) (210 kg) × • (7 . 7 m) 2- (1 . 3 m) ‚ = 0 . 449286 m ....
View Full Document

## This note was uploaded on 09/12/2008 for the course PHYS 1710 taught by Professor Weathers during the Fall '08 term at North Texas.

### Page1 / 5

Homework 17 - Gozick, Brandon – Homework 17 – Due: Nov...

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

View Full Document
Ask a homework question - tutors are online