II. The Simple Machines (p )

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II. The Simple Machines (p )
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II. The Simple Machines (p.180-183) Machines - Ch. 7 II. The Simple Machines (p.180-183) Lever Pulley Wheel & Axle Inclined Plane Screw Wedge

A. Lever Lever a bar that is free to pivot about a fixed point, or fulcrum “Give me a place to stand and I will move the Earth.” – Archimedes Engraving from Mechanics Magazine, London, 1824 Resistance arm Effort arm Fulcrum

A. Lever Le must be greater than Lr in order to multiply the force. Ideal Mechanical Advantage (IMA) frictionless machine Effort arm length Resistance arm length Le must be greater than Lr in order to multiply the force.

A. Lever First Class Lever can increase force, distance, or neither changes direction of force

A. Lever Second Class Lever always increases force

A. Lever Third Class Levers always increases distance

B. Pulley Pulley grooved wheel with a rope or chain running along the groove a “flexible first-class lever” F Le Lr

B. Pulley Ideal Mechanical Advantage (IMA) equal to the number of supporting ropes IMA = 0 IMA = 1 IMA = 2

B. Pulley Fixed Pulley IMA = 1 does not increase force changes direction of force

B. Pulley Movable Pulley IMA = 2 increases force doesn’t change direction

B. Pulley Block & Tackle combination of fixed & movable pulleys increases force (IMA = 4) may or may not change direction

C. Wheel and Axle Wheel and Axle two wheels of different sizes that rotate together a pair of “rotating levers” Wheel Axle

C. Wheel and Axle Ideal Mechanical Advantage (IMA) effort force is usu. applied to wheel axle moves less distance but with greater force effort radius resistance radius

Problems Lr Le Lr = 20 cm IMA = Le ÷ Lr Le = 140 cm You use a 160 cm plank to lift a large rock. If the rock is 20 cm from the fulcrum, what is the plank’s IMA? GIVEN: Lr = 20 cm Le = 140 cm IMA = ? WORK: IMA = Le ÷ Lr IMA = (140 cm) ÷ (20 cm) IMA = 7 IMA Le Lr 20cm 160cm

Problems rr re re = 20 cm IMA = re ÷ rr rr = 5 cm A crank on a pasta maker has a radius of 20 cm. The turning shaft has a radius of 5 cm. What is the IMA of this wheel and axle? GIVEN: re = 20 cm rr = 5 cm IMA = ? WORK: IMA = re ÷ rr IMA = (20 cm) ÷ (5 cm) IMA = 4 IMA re rr 20 cm 5 cm

Problems rr re IMA = 6 re = IMA · rr re = ? re = (6)(4 cm) rr = 4 cm A steering wheel requires a mechanical advantage of 6. What radius does the wheel need to have if the steering column has a radius of 4 cm? GIVEN: IMA = 6 re = ? rr = 4 cm WORK: re = IMA · rr re = (6)(4 cm) re = 24 cm IMA re rr rr re

Problems Lr Le Fr = 150 N Le = IMA · Lr Fe = 15 N Le = (10)(0.3) You need to lift a 150 N box using only 15 N of force. How long does the lever need to be if the resistance arm is 0.3m? GIVEN: Fr = 150 N Fe = 15 N Lr = 0.3 m Le = ? MA = 10 WORK: Le = IMA · Lr Le = (10)(0.3) Le = 3 m Total length = Le + Lr Total length = 3.3 m 15N 0.3m ? 150N IMA Le Lr