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Machines - Ch. 7. Lever Pulley Wheel & Axle. Inclined Plane Screw Wedge. II. The Simple Machines (p.180-183). Resistance arm. Effort arm. Fulcrum. Engraving from Mechanics Magazine, London, 1824 . “Give me a place to stand and I will move the Earth.” – Archimedes. A. Lever.
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Machines - Ch. 7 • Lever • Pulley • Wheel & Axle • Inclined Plane • Screw • Wedge II. The Simple Machines (p.180-183)
Resistance arm Effort arm Fulcrum Engraving from Mechanics Magazine, London, 1824 “Give me a place to stand and I will move the Earth.” – Archimedes A. Lever • Lever • a bar that is free to pivot about a fixed point, or fulcrum
Effort arm length Resistance arm length A. Lever • Ideal Mechanical Advantage (IMA) • frictionless machine • 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
effort radius resistance radius C. Wheel and Axle • Ideal Mechanical Advantage (IMA) • effort force is usu. applied to wheel • axle moves less distance but with greater force
20cm Le 160cm Lr IMA Problems • 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
re 5 cm 20 cm rr IMA Problems • 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
re rr IMA Problems • 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 re rr
15N ? 0.3m 150N Le Lr IMA Problems • 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