h27 - hernandez (ejh742) homework 27 Turner (58120) 1 This...

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: hernandez (ejh742) homework 27 Turner (58120) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points A 14 m ladder whose weight is 297 N is placed against a smooth vertical wall. A person whose weight is 486 N stands on the ladder a distance 5 . 4 m up the ladder. The foot of the ladder rests on the floor 8 . 82 m from the wall. 486 N 297 N 5 . 4 m 14 m b 8 . 82 m Note: Figure is not to scale. Calculate the force exerted by the wall. Correct answer: 272 . 54 N. Explanation: Let : = 14 m , d = 5 . 4 m , s = 8 . 82 m , W = 297 N , and W p = 486 N . Pivot b N w F f N f W p W = arccos s = arccos parenleftbigg 8 . 82 m 14 m parenrightbigg = 50 . 9499 In equilibrium summationdisplay vector F = 0 and summationdisplay vector = 0 . summationdisplay : W p d cos + W 2 cos N w sin = 0 , where d is the distance of the person from the bottom of the ladder. Therefore 2 F w sin = 2 W p d cos + W cos F w = 2 W p d + W 2 cos sin = 2 (486 N)(5 . 4 m) + (297 N)(14 m) 2 (14 m) cos 50 . 9499 sin50 . 9499 = 272 . 54 N . 002 (part 2 of 2) 10.0 points Calculate the normal force exerted by the floor on the ladder. Correct answer: 783 N. Explanation: Applying translational equilibrium, summationdisplay F y : N f W p W = 0 N f W p W = 0 . N f = W p + W = 486 N + 297 N = 783 N . hernandez (ejh742) homework 27 Turner (58120) 2 003 (part 1 of 3) 10.0 points An 18 . 3 m , 489 N uniform ladder rests against a frictionless wall, making an angle of 64 with the horizontal. b 825 N 489 N 5 . 25 m 18 . 3 m 64 Note: Figure is not drawn to scale. Find the horizontal force exerted on the base of the ladder by Earth when an 825 N fire fighter is 5 . 25 m from the bottom. Correct answer: 234 . 687 N. Explanation: Let : L = 18 . 3 m , = 5 . 25 m , = 64 , W = 489 N , and W p = 825 N . L sin L cos L 2 cos cos Pivot b N w f N g W p W Applying rotational equilibrium with the pivot at the point of contact with the ground, summationdisplay = W p cos + W L 2 cos N w L sin = 0 2 W p cos + W L cos = 2 N w L sin N w = W p cos L sin + W cos 2 sin = W p L...
View Full Document

This note was uploaded on 06/14/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

Page1 / 7

h27 - hernandez (ejh742) homework 27 Turner (58120) 1 This...

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