Inclined Plane Annotated Scaled Drawing
Engagement:eight/27/14
Intro:In this activity nosotros tested a simple motorcar on our vex kit simple machine.We then had to calculate everything from that.We presented in class to anybody.Nosotros also had to accept notes when other presented.
Office 1 – Lever, Bicycle and Axle, and Pulley
Offset Class Lever |
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Create a scaled annotated drawing of the first grade lever.
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Calculate the platonic mechanical advantage of the lever organization.
Formula | Substitute / Solve | Terminal Answer |
IMA=De/Dr | IMA= | |
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Calculate the ideal attempt strength needed to overcome the known resistance force.
Formula | Substitute / Solve | Last Answer |
Fe=Fr x Dr/De | Fr=one*vi/6.25 | Fe=0.96lbs |
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Summate the bodily mechanical advantage of the lever organisation.
Formula | Substitute / Solve | Last Answer |
AMA=Fr/Fe | AMA=1/0.96 | AMA=ane.04 |
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Calculate the efficiency of the lever organisation.
Formula | Substitute / Solve | Last Respond |
eff=(AMA/IMA) * 100 | 1.04/1.04 | eff=100% |
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List and depict two examples of a outset form lever.
see saw or crowbar
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Calculate the platonic mechanical reward of the lever system.
Formula | Substitute / Solve | Final Answer |
IMA=De/Dr | IMA= 6/iii.5 | one.71 |
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Calculate the platonic endeavour forcefulness needed to overcome the known resistance force.
Formula | Substitute / Solve | Final Respond |
Fe*De=Fr*Dr | Fe*5.v=3*ane.07 | Fe=.58 lbs |
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Calculate the actual mechanical advantage of the lever arrangement.
Formula | Substitute / Solve | Final Answer |
AMA=Fr/Iron | 1.07/.5 | 2.fourteen |
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Calculate the efficiency of the lever organisation.
Formula | Substitute / Solve | Last Respond |
AMA/IMA | 2.14/one.71 | 125% |
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List and depict ii examples of a second form lever.
A wheelbarrow and a hammer
Third Form Lever |
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Create a scaled annotated drawing of the third class lever.
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Calculate the ideal mechanical advantage of the lever organization.
Formula | Substitute / Solve | Final Reply |
De/Dr | 6/12 | IMA=.5 |
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Calculate the ideal effort force needed to overcome the known resistance force.
Formula | Substitute / Solve | Final Reply |
Atomic number 26*De=Fr*Dr | 1.07*sixteen=half-dozen*Fe 12.84=6Fe 12.84/half-dozen | Fe=2.14 |
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Calculate the actual mechanical advantage of the lever system.
Formula | Substitute / Solve | Concluding Answer |
Fr/Fe | 1.07/2.14 | AMA=0.5 |
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Summate the efficiency of the lever system.
Formula | Substitute / Solve | Concluding Answer |
AMA/IMA | 0.v/0.5 *100 | eff=100% |
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Listing and describe two examples of a 3rd class lever.
Tweezers and scissors
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Is it possible for a beginning or second class lever to take a mechanical reward less than one, or for a 3rd class lever to have a mechanical advantage greater than one? Justify your answer.
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When y'all were solving for mechanical advantage, what units did the last answer require? Explain why.
Inches because
Bike and Axle |
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What is the diameter of the wheel ? 4.375 in
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What is the diameter of the axle ? 2.75 in
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Attach the resistance weight to the string attached to the axle. Use your fingers to turn the wheel. Based on where the applied effort and resistance are located, identify the altitude traveled by both forces during i full rotation.
D East =2x 3.14 x2 .1875=13.7 in
D R = 2x 3.14x 1.375=8.64
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Remove the resistance weight from the axle string and attach the weight to the wheel. Use your fingers to plough the axle. Based on where the applied try and resistance are located, identify the distance traveled past both forces during ane full rotation.
D E =8.64in
D R = 13.7 in
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Wrap the resistance weight effectually the axle using string. Use the force sensor fastened to the string wrapped around the wheel to create equilibrium. Based on where the applied attempt and resistance are located, identify the force required to concur the system in equilibrium.
F Due east =.726 lb
F R = 1.07 lb
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Wrap the weight around the wheel using string. Use the strength sensor attached to string on the axle to create equilibrium. Based on where the applied endeavour and resistance are located, place the forcefulness required to hold the arrangement in equilibrium.
F Due east =1.716 lb
F R = i.07 lb
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For the same resistance, is the effort forcefulness larger when the effort is applied to the wheel or when it is applied to the axle ? Explicate why.
The effort force is larger when the effort is practical to the axle. because the altitude is smaller on the axle.
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Create a scaled annotated drawing of the bike and axle system.
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Calculate the platonic mechanical reward of the wheel and beam system if the resistance force is applied to the axle .
Formula | Substitute / Solve | Final Respond |
IMA=De/Dr | 13.⅞.64 | one.59 |
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Calculate the platonic mechanical reward of the cycle and beam system if the resistance force is applied to the cycle .
Formula | Substitute / Solve | Final Answer |
IMA=De/Dr | 8.64/13.seven | .631 |
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Calculate the platonic try force needed to overcome the known resistance strength if the resistance forcefulness is applied to the wheel .
Formula | Substitute / Solve | Final Answer |
Fe x De=Fr x Dr | 13.7 ten i.07/eight.64 | 1.69 |
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Calculate the actual mechanical reward of your wheel and axle system if the resistance force is applied to the wheel .
Formula | Substitute / Solve | Final Respond |
AMA=Fr/Fe | 1.07/1.716 | .62 |
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Calculate the efficiency of the bike and axle system when the resistance strength is applied to the wheel .
Formula | Substitute / Solve | Final Respond |
%eff=AMA/IMA | .62/.631 | 98% |
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List and describe two examples of a bike and axle.
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Wheels n car and running cycle for hamster
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If you know the dimensions of a wheel and axle organisation used for an machine, how can you lot determine the distance covered for each axle revolution? Explicate whatsoever additional information and necessary formulas.
one time you get the bore of the wheel
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Why is the steering wheel on a school bus so large?
Because the distance is to large it is easier to turn the motorcoach
Stock-still Pulley |
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Create a scaled annotated drawing of the fixed pulley system.
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Summate the ideal mechanical advantage of the fixed pulley system.
Formula | Substitute / Solve | Concluding Answer |
IMA= # strands | IMA=1 | IMA=one |
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Summate the actual mechanical advantage of the fixed caster system.
Formula | Substitute / Solve | Final Reply |
AMA=F_r/F_e | AMA=(1lb)/(0.71)= | AMA=1.41 |
Calculate the efficiency of the fixed pulley system.
Formula | Substitute / Solve | Concluding Answer |
eff=(AMA/IMA) 100 | eff=one.28/1=1.28 | 128% |
Movable Pulley |
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Create a scaled annotated drawing of the pulley system.
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Summate the actual mechanical advantage of the pulley organisation.
Formula | Substitute / Solve | Concluding Reply |
AMA=Fr/Iron | .950 Lbs. / .660 Lbs. | one.44 |
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Calculate the ideal mechanical advantage of the pulley system.
Formula | Substitute / Solve | Final Answer |
IMA= De/Dr | nine ⅛ / 5 ¾ = | 1.59 or i.58695652 |
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Calculate the efficiency of the fixed caster system.
Formula | Substitute / Solve | Concluding Answer |
AMA/IMA | one.44/one.59 | 0.90566037735 91% |
Part ii – Inclined Aeroplane and Screw
Ramp and a Roller Coaster.
Calculate the efficiency of the fixed pulley organization.
IMA = iii Fe = .350 Fr = 1 lb
A. Elevators B. Exercise machines
Both strands are used to change direction.
The load is distributed over the two strands, one is combined to ceiling and the other is under the weight.
IMA is the number of supporting strands because it is the advantage of having that corporeality of cord.
Non parallel strings or the friction from the strings.
Screwjack and C clench
The screw is an inclined plane and is being slid up the entire manner.
Inclined airplane and wedge. The threads are slanted and the wedge pushed through the cloth. |
Role ii – Inclined Plane and Screw
Inclined Plane |
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Create a scaled annotated cartoon of the inclined plane system.
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Calculate the ideal mechanical advantage of the inclined plane system.
Formula | Substitute / Solve | Final Respond |
IMA = De/Dr | 15/eleven.75 | 1.27 |
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Calculate the platonic effort force needed to overcome the known resistance strength.
Formula | Substitute / Solve | Last Reply |
Fe*De=Fr*Dr | Fe*fifteen=1*eleven.75 | .78 |
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Calculate the actual mechanical advantage of the inclined plane system.
Formula | Substitute / Solve | Final Answer |
AMA = Fr/Fe | .85/1 | .85 |
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Calculate the efficiency of the inclined plane arrangement.
Formula | Substitute / Solve | Final Answer |
eff= AMA/IMA | .85/1.27 | .66=66% |
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List and describe two examples of an inclined airplane.
Ramp and a Roller Coaster.
Fixed Pulley |
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Create a scaled annotated cartoon of the stock-still pulley system.
|
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Calculate the ideal mechanical advantage of the stock-still pulley organisation.
Formula | Substitute / Solve | Concluding Answer |
IMA = # strands | IMA = 1 | IMA = 1 |
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Calculate the actual mechanical reward of the fixed caster arrangement.
Formula | Substitute / Solve | Final Reply |
AMA = F_R/F_E | AMA = (i lb)/(0.71 lb) | AMA = 1.41 |
Calculate the efficiency of the fixed caster system.
Formula | Substitute / Solve | Final Reply |
eff = AMA/IMA | eff = one.28 / 1 = 1.28 | 128% |
Block and Tackle |
IMA = 3
Fe = .350
Fr = 1 lb
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Calculate the bodily mechanical advantage of the caster arrangement.
Formula | Substitute / Solve | Concluding Respond |
AMA = Fr/Fe | AMA = 1 lb / .350 | AMA = 2.85 |
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Calculate the ideal mechanical advantage of the pulley system.
Formula | Substitute / Solve | Concluding Reply |
IMA = # of strands | IMA = # of strands = 3 | IMA = 3 |
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Calculate the efficiency of the fixed caster organisation.
Formula | Substitute / Solve | Terminal Reply |
Eff=AMA/IMA(100) | Eff = 2.85 / 3 | Eff = 95% |
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Describe two examples of a pulley organisation.
A. Elevators
B. Do machines
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The fixed pulley contained 2 strands. Explain the function of each strand.
Both strands are used to alter management.
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The movable pulley contained two strands. Explain the role of each strand.
The load is distributed over the two strands, one is combined to ceiling and the other is nether the weight.
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In the block and tackle system, explain how mechanical advantage relates to the number of strands.
IMA is the number of supporting strands because it is the advantage of having that amount of string.
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In a block and tackle organization with a mechanical reward of three, the effort is measured at 15 lbf. The resistance, when balanced, is measured at 42 lbf. What factors might account for the loss in energy?
Non parallel strings or the friction from the strings.
Spiral |
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Create a scaled annotated drawing of the screw organization.
|
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Calculate the ideal mechanical advantage of the spiral.
Formula | Substitute / Solve | Final Answer |
IMA = De/Dr | 13.seven / .123 | IMA = 111.38 |
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Calculate the ideal effort force needed to overcome the known resistance force.
Formula | Substitute / Solve | Concluding Answer |
Fe*De = Fr*Dr | 13.7 * Fe = .123*1.07 | Fe = .01 lbs |
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Calculate the actual mechanical advantage of the screw.
Formula | Substitute / Solve | Concluding Respond |
AMA = Fr/Fe | ane.07/.65 | AMA = one.646 |
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Calculate the efficiency of the screw.
Formula | Substitute / Solve | Final Answer |
Eff = AMA / IMA | 1.646 / 111.38 = .014 | Eff = 1.4 |
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Describe two examples of a screw.
Screwjack and C clamp
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Why do you think overcoming a resistance force using a screw is so easy?
The screw is an inclined plane and is beingness slid upwardly the entire way.
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The spiral is a combination of 2 simple machines. Place and defend what two simple machines you believe are combined to create a spiral.
Inclined aeroplane and wedge. The threads are slanted and the wedge pushed through the cloth.
Source: https://sites.google.com/a/student.leyden212.org/dominik-siemek-engineering-e-portfolio/poe/activity-1-1-1-a-vex-simple-machine-investigation
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