gilbert (hlg574) – HW02 – tsoi – (55730) 1This print-out should have 18 questions.Multiple-choice questions may continue o nthe next column or page – find all choicesbefore answering.001 (part 1 of 2) 10.0 pointsAn elevator accelerates upward at 1.2 m/s2.The acceleration of gravity is 9.8 m/s2.What is the upward force exerted by thefloor of the elevator on a(n) 65 kg passenger?Correct answer: 715 N.Explanation:NamgWhen the elevator is accelerati ng upward,Fnet= m a = N − m gN = m a + m g.002 (part 2 of 2) 10.0 pointsIf the same elevator accelerates downwardswith an acceleration of 1.2 m/s2, what is theupward force exerted by the elevator floor onthe passenger?Correct answer: 559 N.Explanation:NamgWhen the elevator is accelerating down-ward,Fnet= m a = m g − N2N2= m g − m a.003 (part 1 of 2) 10.0 pointsConsider the 601 N weight held by two cablesshown below . The left-hand cable had tension590 N and makes an angle of θ with the wall.The right-hand cable had tension 730 N andmakes an angle of θ1with the ceiling.601 N590 N730 Nθ1θa) What is the angle θ1which the right-hand cable makes with respect to the ceiling?Round up to four significant digits. Expressunit as ”degrees”.Correct answer: 38.4808 degrees.Explanation:Observe the free-body diagram below.F2F1θ1θ2WgNote: The sum of the x- andy-components of F1, F2, andWgare equal to zero.Given : Wg= 601 N ,F1= 730 N ,θ1= 38.4808◦, andθ2= 90◦− θ .Basic Concepts:XFx= 0Fx1= Fx2F1cos θ1= F2cos θ2(1)F21cos2θ1= F22cos2θ2(2)andXFy= 0Fy1+ Fy2+ Fy3= 0F1sin θ1+ F2sin θ2− F3= 0F1sin θ1= −F2sin θ2+ F3F21sin2θ1= F22sin2θ2−2 F2F3sin θ2+ F23, since (3)F3sin θ3= F3sin 270◦= −F3, andF3cos θ3= F3cos 270◦= 0 .gilbert (hlg574) – HW02 – tsoi – (55730) 2Solution: Since sin2θ + cos2θ = +1 andadding Eqs. 2 and 3 , we haveF22= F21− 2 F1F3sin θ1+ F23sin θ1=F21+ F23− F222 F1F3θ1= arcsinF23+ F21− F222 F1F3= arcsin(601 N)2+ (730 N )22 (730 N) (601 N)−(590 N)22 (730 N) (601 N)= 38.4808◦.004 (part 2 of 2) 10.0 pointsb) What is the angle θ which the left-handcable makes with respect to the wall? Roundup to four significant digits. Express unit as”degrees”.Correct answer: 75.2391 degrees.Explanation:Using Eq. 1, we havecos θ2=F1F2cos θ1θ2= arccosF1F2cos θ1= arccos(730 N)(590 N)cos 38.4808◦= 14.7609◦θ = 90◦− θ2= 90◦− 14.7609◦= 75.2391◦.005 (part 1 of 2) 10.0 pointsHarry the painter swings year after yearfrom his bosun’s chair. His weight is 710 Nand the rope, unknown to him, has a breakingpoint of 405 N.Why doesn’t t he rope break when he issuppor ted as shown at the left above? Toanswer this, find the tension in the rope.Correct answer: 355 N.Explanation:In the left figure, Harry is supported bytwo strands of rope t hat share his weight, soeach strand suppo rts only 355 N, below thebreaking point. Total upward force suppliedby the ropes equals weight acting downward,giving a net force of zero and no acceleration.006 (part 2 of 2) 10.0 pointsOne day Harry is painting near a flagpole,and, for a change, he ties the free end of therope to the flagpole instead of to his chair asshown at t he right.Why did Harry end up taking his vacationearly? To answer this, find the tension in therope.Correct answer: 710 N.Explanation:In the right figure, Harry is now supportedby one strand, requiring the tension to be710 N. Since this is above the breaking pointof the rope, it breaks.007 (part 1 of 5) 10.0 pointsGiven: The pulley is massless and frictio nl essand attached to t he ceiling. A massless inex-tensible string is attached to the masses Mband Ma, where Mb> Ma. The tensions Ty,Tz, Tx, and the gravitational constant g ar emagnitudes.gilbert (hlg574) – HW02 – tsoi – (55730) 3ℓRωMbMaTyTxTzConsider the following possibilities and se-lect the correct choice.1. Ty= Mbg a nd Tx> Mag2. Ty= Mbg a nd Tx< Mag3. Ty> Mbg a nd Tx< Mag4. Ty< Mbg a nd Tx= Mag5. Ty< Mbg a nd Tx> Mag correct6. Ty> Mbg a nd Tx= Mag7. Ty< Mbg a nd Tx< Mag8. Ty= Mbg a nd Tx= Mag9. Ty> Mbg a nd Tx> MagExplanation:The mag ni tude of the tension in the stringis the same at all points along the stringT ≡ Ty= Tx. (1)Consider the free body diagramsMbMaTTMbgMagaaSince the larger mass will move down andthe smaller ma ss up, we can take motiondownward as positive for Mband motion up-ward as positive for Ma. Apply Newton’ssecond law to Maand Mbrespectively andthen combine the results, we have for massMa,XFx: T − Mag = Maa (2)T = Ma(g + a) or T > Mag ,and for mass Mb,XFy: Mbg − T = Mba (3)T = Mb(g − a) or T < Mbg .008 (part 2 of 5) 10.0 pointsConsider the following possibilities and selectthe correct choice.1. Ty> Tx2. Ty= Txcorrect3. Ty< TxExplanation:See Eq. 1 in Part 1.009 (part 3 of 5) 10.0 pointsConsider the following possibilities and selectthe correct choice.1. Tz< Mbg + Mag correct2. Tz> Mbg + Mag3. Tz= Mbg + Maggilbert (hlg574) – HW02 – tsoi – (55730) 4Explanation:Using Eq. 1 in Part 1, we haveTz= Ty+ Tx= Mb(g − a) + Ma(g + a)= Mbg + Mag − (Mb− Ma) a< Mbg + Mag010 (part 4 of 5) 10.0 pointsConsider the following possibilities and selectthe correct choice.1. Ty+ Tx< Tz2. Ty+ Tx> Tz3. Ty+ Tx= TzcorrectExplanation:Visualize a free-body diagram for the pul-ley. The forces are in equilibrium, soTz= Ty+ Tx≡ 2 T .011 (part 5 of 5) 10.0 pointsConsider the following possibilities and selectthe correct choice.1. k~axk < k~ayk2. k~axk = k~ayk correct3. k~axk > k~aykExplanation:The str ing is inextensible. All p oints on thestring accelerate equallya ≡ k~axk = k~ayk .We can add Eqs. 2 and 3 from Part 1 andobtainMbg − Mag = Maa + Mbaa =Mb− MaMa+ Mbg .012 (part 1 of 2) 10.0 pointsTwo masses m1and m2are connected in themanner shown.m2m1T1T2a =g2The system is accelerating downward withacceleration of magnitudeg2.Determine T2.1. T2=m2−12m1g2. T2=12(m2− m1) g3. T2=12m2− m1g4. T2=32m2g5. T2=12m2g correct6. T2= m2gExplanation:m2T2a =g2m2gFrom Newton’s Second Law,m2g − T2= m2g2.SoT2= m2g − m2g2=12m2g .013 (part 2 of 2) 10.0 pointsDetermine T1.1. T1= 2 (m1+ m2) g2. T1= (m1+ m2) g3. T1=52(m1+ m2) ggilbert (hlg574) – HW02 – tsoi – (55730) 54. T1=32(m1+ m2) g5. None of these6. T1=12(m1+ m2) g correctExplanation:m1+ m2T1a =g2(m1+ m2) gConsider m1and m2as a whole object withmass (m1+ m2), then(m1+ m2) g − T1= (m1+ m2)g2.SoT1=12(m1+ m2) g .014 10.0 pointsA 7.5 kg object hangs a t one end of a rope thatis attached to a support on a railroad