Midterm03 - Version 097/ABCAB midterm 03 Turner(58220 1 This print-out should have 19 questions Multiple-choice questions may continue on the next

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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 – fnd all choices beFore answering. 001 10.0 points A17 . 5kgpersonclimbsupauniForm142N ladder. The upper end oF the ladder rests on aFr ict ion lesswa l l . Thebottomo Ftheladder rests on a ±oor 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 ±oor. 17 . 5kg 142 N d 2m 43 μ s > 0 μ =0 Note: ²igure is not to scale. What is the coe³cient oF static Friction μ s between the base oF the ladder and the ±oor? The acceleration oF gravity is 9 . 8m / 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: Pivot N w f N f mg W θ Let : g =9 . / s 2 , θ =43 , L =2m , d . 35 L, W =142N , and m =17 . . Applying translational equilibrium hori- zontally, ± F x : f - N w f = N w ± τ : mgd cos θ + W L 2 cos θ - N w L sin θ where d is the distance oF the person From the bottom oF the ladder 2 fL sin θ =2 cos θ + W L cos θ f = 1 2 W tan θ + d L tan θ The ladder may slip when f = f max = N w , so f max μ ( W + ) 1 2 W tan θ + d L tan θ μ 1 ( W + )tan θ ² 1 2 W + d L ³ 1 [142 N + (17 . 5kg)(9 . / s 2 )] tan43 × ´ 1 2 (142 N)
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Version 097/ABCAB – midterm 03 – Turner – (58220) 2 +(0 . 35)(17 . 5kg)(9 . 8m / s 2 ) ± 0 . 448189 . 002 10.0 points Amobileconsistingoffourweightshanging on three rods of negligible mass. 24 N W 1 W 2 W 3 5m 9m 7m 4m 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 =24N , ) r 1 =9m , ) l 1 =5m , ) r 2 =4m , ) l 2 =7m , ) r 3 =8m , and ) l 3 . We can apply the balance condition ² = 0successively ,startingwiththelowestpartof the mobile, to ±nd the value of each unknown weight. Apply ² =0aboutanax isthroughthe 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) =13 . 3333 N . Apply ² 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) =21 . 3333 N . Apply ² point of suspension of the top part of the mobile, we obtain ) l e [ W + W 1 + W 2 ] - ) r 3 W 3 W 3 = ) l 3 [ W + W 1 + W 2 ] ) r 3 = (24 N + 13 . 3333 N +21 . 3333 N) = 29 . 3333 N . keywords: 003 10.0 points
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Version 097/ABCAB – midterm 03 – Turner – (58220) 3 The puck in the fgure has a mass oF 0 . 145 kg. Its original distance From the center oF rota- tion is 51 . 9cm ,andthepuckismov ingw ith aspe edo F75 . 7cm / s. The string is pulled downward 15 . 1cmth roughth eho l einth e Frictionless table. 51 . 75 . / s 0 . 145 kg What is the magnitude oF the work was done on the puck? Treat the hockey puck as apointmass. 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. Explanation: Given : m p =0 . 145 kg , r i =51 . 9cm=0 . 519 m , v i =75 . / s=0 . 757 m / s , and Δ r =15 . 1cm=0 . 151 m . r v F c m The kinetic energy is KE = 1 2 2 = 1 2 ( mr 2 ) ± v r ² 2 = 1 2 mv 2 Angular momentum is conserved, so I i ω i = I f ω f ( i 2 ) v i r i = ( f 2 ) v f r f v f = v i r i r f Thus the work done on the puck is W = 1 2 2 f - 1 2 2 i = 1 2 m ³ v i r i r f ´ 2 - 1 2 2 i = 1 2 m µ r 2 i r 2 f - 1 v 2 i = 1 2 (0 . 145 kg) ³ (0 . 519 m) 2 (0 . 368 m) 2 - 1 ´ × (0 . 757 m / s) 2 . 0410899 N
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This note was uploaded on 10/06/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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Midterm03 - Version 097/ABCAB midterm 03 Turner(58220 1 This print-out should have 19 questions Multiple-choice questions may continue on the next

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