{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

midterm03 - Version 097/ABCAB midterm 03 Turner(58220 1...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Version 097/ABCAB – midterm 03 – Turner – (58220) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 17 . 5 kg person climbs up a uniform 142 N ladder. The upper end of the ladder rests on a frictionless wall. The bottom of the ladder rests on a floor with a rough surface. The angle between the horizontal and the ladder is 43 . The person can climb 35 % up the ladder before the base of the ladder slips on the floor. 17 . 5 kg 142 N d 2 m 43 μ s > 0 μ = 0 Note: Figure is not to scale. What is the coe ffi cient of static friction μ s between the base of the ladder and the floor? The acceleration of gravity is 9 . 8 m / s 2 . 1. 0.611755 2. 0.228951 3. 0.33071 4. 0.259405 5. 0.211536 6. 0.270395 7. 0.448189 8. 0.516792 9. 0.279139 10. 0.299018 Correct answer: 0 . 448189. Explanation: P ivot N w f N f m g W θ Let : g = 9 . 8 m / s 2 , θ = 43 , L = 2 m , d = 0 . 35 L , W = 142 N , and m = 17 . 5 kg . Applying translational equilibrium hori- zontally, F x : f - N w = 0 f = N w τ : m g d cos θ + W L 2 cos θ - N w L sin θ = 0 where d is the distance of the person from the bottom of the ladder 2 f L sin θ = 2 m g d cos θ + W L cos θ f = 1 2 W tan θ + d L m g tan θ The ladder may slip when f = f max = N w , so f max μ ( W + m g ) 1 2 W tan θ + d L m g tan θ μ 1 ( W + m g ) tan θ 1 2 W + d L m g 1 [142 N + (17 . 5 kg)(9 . 8 m / s 2 )] tan43 × 1 2 (142 N)
Background image of page 1

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

View Full Document Right Arrow Icon
Version 097/ABCAB – midterm 03 – Turner – (58220) 2 +(0 . 35)(17 . 5 kg)(9 . 8 m / s 2 ) 0 . 448189 . 002 10.0 points A mobile consisting of four weights hanging on three rods of negligible mass. 24 N W 1 W 2 W 3 5 m 9 m 7 m 4 m 4 m 8 m Find the weight of W 3 . 1. 155.048 2. 19.5556 3. 45.3968 4. 29.3333 5. 60.8395 6. 62.3333 7. 23.8333 8. 34.0476 9. 63.0667 10. 116.111 Correct answer: 29 . 3333 N. Explanation: Let : W = 24 N , r 1 = 9 m , l 1 = 5 m , r 2 = 4 m , l 2 = 7 m , r 3 = 8 m , and l 3 = 4 m . We can apply the balance condition τ = 0 successively, starting with the lowest part of the mobile, to find the value of each unknown weight. Apply τ = 0 about an axis through the point of suspension of the lowest part of the mobile, we obtain l 1 W - r 1 W 1 = 0 W 1 = l 1 W r 1 = (5 m) (24 N) 9 m = 13 . 3333 N . Apply τ = 0 about an axis through the point of suspension of the middle part of the mobile, we obtain l w W 2 - r 2 [ W + W 1 ] = 0 W 2 = r 2 [ W + W 1 ] l w = (4 m) (24 N + 13 . 3333 N) 7 m = 21 . 3333 N . Apply τ = 0 about an axis through the point of suspension of the top part of the mobile, we obtain l e [ W + W 1 + W 2 ] - r 3 W 3 = 0 W 3 = l 3 [ W + W 1 + W 2 ] r 3 = 4 m 8 m (24 N + 13 . 3333 N +21 . 3333 N) = 29 . 3333 N . keywords: 003 10.0 points
Background image of page 2
Version 097/ABCAB – midterm 03 – Turner – (58220) 3 The puck in the figure has a mass of 0 . 145 kg. Its original distance from the center of rota- tion is 51 . 9 cm, and the puck is moving with a speed of 75 . 7 cm / s. The string is pulled downward 15 . 1 cm through the hole in the frictionless table. 51 . 9 cm 75 . 7 cm / s 0 . 145 kg What is the magnitude of the work was done on the puck? Treat the hockey puck as a point mass. 1. 0.0460518 2. 0.0410899 3. 0.0650961 4. 0.0963719 5. 0.0580828 6. 0.055173 7. 0.0345982 8. 0.127202 9. 0.0349704 10. 0.039917 Correct answer: 0 . 0410899 J.
Background image of page 3

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

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}