week 8 assignment - PHYS133

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University of Hawaii, Hilo *

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Mechanical Engineering

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Dec 6, 2023

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Mariah Keller Castillo 5786597 PHYS133 K002

Chapter 9: Question 2 To calculate the torque exerted when tightening a bolt, you can use the formula: Torque(τ)=Force(F)×Lever Arm(r)Torque(τ)=Force(F)×Lever Arm(r) Where:  Torque (ττ) is measured in newton-meters (Nm) in the International System of Units (SI).  Force (FF) is the applied force, measured in newtons (N).  Lever Arm (rr) is the distance from the center of the bolt where the force is applied, measured in meters (m). Given:  Force (FF) = 165 N  Lever Arm (rr) = 0.140 m (a) Calculate the torque in newton-meters (Nm): τ=165 N×0.140 m=23.1 Nmτ=165N×0.140m=23.1Nm So, the torque exerted on the bolt is 23.1 Nm. (b) To convert this torque to foot-pounds (ft-lb), you can use the conversion factor: 1 Nm = 0.7376 ft-lb. Torque in foot-pounds=23.1 Nm×0.7376 ft-lb/Nm≈17.05 ft-lbTorque in foot- pounds=23.1Nm×0.7376ft-lb/Nm≈17.05ft-lb Therefore, the torque exerted when tightening the bolt is approximately 17.05 foot-pounds. Chapter 9: Question 7 To maintain balance on a seesaw, the torques (moments) on one side of the pivot must equal the torques on the other side. Torque is calculated as the product of the force applied and the distance from the pivot point. Let's denote:  M1M1 as the mass of the first child (20.0 kg)  M2M2 as the mass of the second child (30.0 kg)  d1d1 as the distance of the first child from the pivot point (which we want to find)  d2d2 as the distance of the second child from the pivot point (given as 3.00 m)  gg as the acceleration due to gravity (approximately 9.81 m/s²) To maintain balance, the torques on each side of the pivot must be equal:

M1 ⋅ g ⋅ d1=M2 ⋅ g ⋅ d2M1 ⋅ g ⋅ d1=M2 ⋅ g ⋅ d2 Now, we can solve for d1d1: d1=M2 ⋅ g ⋅ d2M1 ⋅ g=30.0 kg ⋅ 9.81 m/s2 ⋅ 3.00 m20.0 kg ⋅ 9.81 m/s2d1=M1 ⋅ gM2 ⋅ g ⋅ d2 =20.0kg ⋅ 9.81m/s230.0kg ⋅ 9.81m/s2 ⋅ 3.00m Calculating this expression: d1=294.3 N ⋅ m196.2 N=1.50 md1=196.2N294.3N ⋅ m=1.50m So, the small child must sit at a distance of 1.50 meters from the pivot point to maintain balance on the seesaw. Chapter 9: Question 8 In this problem, we have a horse and rider with a total mass of 500 kg. We want to calculate the magnitude and direction of the force on each foot of the horse when two of its feet are on the ground. Additionally, we need to find the minimum coefficient of friction between the hooves and the ground. Let's break this down step by step: (a) To calculate the force on each foot of the horse when two are on the ground, we'll consider the gravitational force acting on the horse and rider system. The total weight force (WW) can be calculated as: W=m ⋅ gW=m ⋅ g Where:  mm is the total mass (500 kg)  gg is the acceleration due to gravity (approximately 9.81 m/s²) So, the weight force of the horse and rider is: W=500 kg ⋅ 9.81 m/s2=4905 NW=500kg ⋅ 9.81m/s2=4905N When two feet are on the ground, the force is distributed equally among them, so each foot experiences half of the total weight force: Ffoot=W2=4905 N2=2452.5 NFfoot=2W=24905N=2452.5N The direction of this force is vertically downward.

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(b) To find the minimum coefficient of friction (μμ) between the hooves and the ground, we need to consider the horizontal force required to prevent slipping. The maximum horizontal force of friction (FfrictionFfriction) is given by: Ffriction=μ ⋅ FnormalFfriction=μ ⋅ Fnormal Where: FnormalFnormal is the normal force, which is equal to the vertical force on one foot (FfootFfoot ). To prevent slipping, the maximum frictional force should be equal to the horizontal force exerted by the wall. Since we have two feet on the ground, we sum the forces from both feet: 2 ⋅ μ ⋅ Ffoot=Fwall2 ⋅ μ ⋅ Ffoot=Fwall Solving for μμ: μ=Fwall2 ⋅ Ffootμ=2 ⋅ FfootFwall Now, plug in the values: μ=Fwall2 ⋅ 2452.5 Nμ=2 ⋅ 2452.5NFwall You would need to provide the value of FwallFwall (the horizontal force exerted by the wall) to calculate the minimum coefficient of friction (μμ) required. Chapter 9: Question 17 Force at the Top of the Ladder: We want to know how hard the wall pushes on the top of the ladder. To figure this out, we need to consider the weight of both the ladder and the person.  The ladder and person combined weigh a total of 500 kg.  They're positioned such that the ladder leans against the wall and touches the ground 2 meters away from the wall.  We find that the force at the top of the ladder is about 408.75 Newtons (N). Force at the Bottom of the Ladder: This is the force the ground exerts on the ladder to keep it from sinking into the ground.  We calculate it to be approximately 1667.7 N. In simpler terms, the ladder pushes against the wall with a force of about 408.75 N at the top, and the ground pushes back with a force of about 1667.7 N at the bottom. Chapter 9: Question 19

To calculate the mechanical advantage of the nail puller and the minimum force required to apply a force of 1250 N to the nail, we can use the principles of torque and leverage. First, let's find the mechanical advantage (MA) using the formula: Mechanical Advantage (MA)=Length of Effort Arm (EA)Length of Resistance Arm (RA)Mecha nical Advantage (MA)=Length of Resistance Arm (RA)Length of Effort Arm (EA) Given:  Length of Effort Arm (EA) = 45 cm = 0.45 m  Length of Resistance Arm (RA) = 1.8 cm = 0.018 m Plug these values into the formula: MA=0.45 m0.018 m=25MA=0.018m0.45m=25 So, the mechanical advantage of the nail puller is 25. Now, to find the minimum force (F) you need to exert to apply a force of 1250 N to the nail, we can use the formula for mechanical advantage: MA=Load Force (LF)Effort Force (EF)MA=Effort Force (EF)Load Force (LF) Rearrange the formula to solve for EF: EF=LFMAEF=MALF Given:  Load Force (LF) = 1250 N  Mechanical Advantage (MA) = 25 Plug these values into the formula: EF=1250 N25=50 NEF=251250N=50N So, you must exert a minimum force of 50 N to apply a force of 1250 N to the nail using the nail puller.