{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW27_Solutions - hiarker(srh959 – homework 27 – Turner...

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 Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: hiarker (srh959) – homework 27 – Turner – (58220) 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 13 m ladder whose weight is 253 N is placed against a smooth vertical wall. A person whose weight is 880 N stands on the ladder a distance 5 . 1 m up the ladder. The foot of the ladder rests on the floor 3 . 77 m from the wall. 880 N 253 N 5 . 1 m 13 m b 3 . 77 m Note: Figure is not to scale. Calculate the force exerted by the wall. Correct answer: 142 . 945 N. Explanation: Let : ℓ = 13 m , d = 5 . 1 m , s = 3 . 77 m , W ℓ = 253 N , and W p = 880 N . Pivot b N w F f N f W p W ℓ θ θ = arccos s ℓ = arccos parenleftbigg 3 . 77 m 13 m parenrightbigg = 73 . 142 ◦ In equilibrium summationdisplay vector F = 0 and summationdisplay vector τ = 0 . summationdisplay τ ◦ : W p d cos θ + W ℓ ℓ 2 cos θ − N w ℓ sin θ = 0 , where d is the distance of the person from the bottom of the ladder. Therefore 2 F w ℓ sin θ = 2 W p d cos θ + W ℓ ℓ cos θ F w = 2 W p d + W ℓ ℓ 2 ℓ cos θ sin θ = 2 (880 N)(5 . 1 m) + (253 N)(13 m) 2 (13 m) × cos 73 . 142 ◦ sin73 . 142 ◦ = 142 . 945 N . 002 (part 2 of 2) 10.0 points Calculate the normal force exerted by the floor on the ladder. Correct answer: 1133 N. Explanation: Applying translational equilibrium, summationdisplay F y : N f −W p −W ℓ = 0 N f −W p −W ℓ = 0 . N f = W p + W ℓ = 880 N + 253 N = 1133 N . hiarker (srh959) – homework 27 – Turner – (58220) 2 003 (part 1 of 3) 10.0 points An 11 . 5 m , 552 N uniform ladder rests against a frictionless wall, making an angle of 61 . 2 ◦ with the horizontal. b 884 N 552 N 4 . 63 m 11 . 5 m 61 . 2 ◦ Note: Figure is not drawn to scale. Find the horizontal force exerted on the base of the ladder by Earth when an 884 N fire fighter is 4 . 63 m from the bottom. Correct answer: 347 . 393 N. Explanation: Let : L = 11 . 5 m , δ = 4 . 63 m , α = 61 . 2 ◦ , W ℓ = 552 N , and W p = 884 N . L sin α L cos α L 2 cos α δ cos α Pivot b N w f N g W p W ℓ Applying rotational equilibrium with the pivot at the point of contact with the ground, summationdisplay τ = W p δ cos α + W ℓ L 2 cos α −N w L sin α = 0 2 W p δ cos α + W ℓ L cos α = 2 N w L sin α N w = W p δ cos α L sin α + W ℓ cos α 2 sin α = W p δ...
View Full Document

{[ snackBarMessage ]}

Page1 / 7

HW27_Solutions - hiarker(srh959 – homework 27 – Turner...

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

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