hw28_s - homework 28 YOO, HEE Due: Apr 8 2008, 4:00 am...

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

View Full Document Right Arrow Icon
homework 28 – YOO, HEE – Due: Apr 8 2008, 4:00 am 1 Question 1, chap 14, sect 2. part 1 of 3 10 points Consider a uniform ladder leaning against a smooth wall and resting on a smooth Foor at point P . There is a rope stretched horizon- tally, with one end tied to the bottom of the ladder essentially at P and the other end to the wall. The top of the ladder is at a height is h up the wall and the base of the ladder is at a distance b from the wall. The weight of the ladder is W 1 . Jill, with a weight W 2 , is one-fourth the way p d = 4 P up the ladder. The force which the wall exerts on the ladder is F . P b W 1 W 2 d T θ b h b F Note: ±igure is not to scale. The torque equation about P is given by 1. ( W 1 + W 2 ) h 2 = F b 2. ( W 1 + W 2 ) b 2 = F h 3. h 2 W 2 + hW 1 = F b 4. h 4 W 2 + h 2 W 1 = F b 5. b 2 W 2 + bW 1 = F h 6. b 4 W 2 + b 2 W 1 = F h correct Explanation: b Pivot b F T N f W 2 W 1 θ s F x : T F = 0 , (1) s F y : N f W 2 W 1 = 0 , and (2) s τ P : W 2 d cos θ + W 1 2 cos θ (3) F ℓ sin θ = 0 , where d is the distance of the person from the bottom of the ladder. Therefore 2 F ℓ sin θ = 2 W 2 d cos θ + W 1 cos θ . Since sin θ = h and cos θ = b , the torque equation about P is given by b 4 W 2 + b 2 W 1 = F h . (4) Question 2, chap 14, sect 2. part 2 of 3 10 points Given: W 2 = 3 W 1 = W , h = b . Determine the force F the wall exerts on the ladder. 1. F = 5 12 W correct 2. F = 1 4 W 3. F = 1 12 W
Background image of page 1

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

View Full DocumentRight Arrow Icon
homework 28 – YOO, HEE – Due: Apr 8 2008, 4:00 am 2 4. F = 1 2 W 5. F = 1 6 W 6. F = 5 6 W 7. F = 2 3 W 8. F = 3 4 W 9. F = 1 3 W 10. F = 7 12 W Explanation: For W 2 = 3 W 1 = W and h = b , Eq. 4 gives b 4 W + b 2 W 3 = F b, so that F = W 4 + W 6 = (3 + 2) 12 W = 5 12 W . Question 3, chap 14, sect 2. part 3 of 3 10 points Given: W 2 = 3 W 1 = W , h = b. When Jill has climbed up the ladder such that the rope tension reaches T = W 2 deter- mine Jill’s height y from the ±oor. 1. y = 1 4 b 2. y = 5 12 b 3. y = 2 3 b 4. y = 3 4 b 5. y = 1 6 b 6. y = 5 6 b 7. y = 7 12 b 8. y = 1 3 b correct 9. y = 1 2 b 10. y = 1 12 b Explanation: For rope tension T = W 2 , and by applying Newton’s 2nd law in the horizontal direction (giving T F = 0), we have F = T = W 2 . With the origin at
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 7

hw28_s - homework 28 YOO, HEE Due: Apr 8 2008, 4:00 am...

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