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# Work in a Mechanical System - PowerPoint PPT Presentation

Unit 6. Work in a Mechanical System. Unit Vocabulary Pg 1 - 14. Force applied Mechanical work Work done Joule. Efficiency Ratio Radian. Unit Review Pg 15-16. # 1-18. Work involves forces that cause movement.

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## PowerPoint Slideshow about ' Work in a Mechanical System' - ivo

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Presentation Transcript

Force applied

Mechanical work

Work done

Joule

Efficiency

Ratio

# 1-18

### Work involves forces that cause movement

Work is getting something done. You can spend a lot of time and effort on something, but if nothing is accomplished, then no work has been done.”

• When a force or force like quantity causes something to move or change

• It can be as big as a crane lifting tons of steel or as small as a tiny charge moving in computer memory

• An applied force must act on an object

• The object must move when the force is applied

• Work is also done when a torque causes something to rotate

• A constant pressure causes a fluid to flow from one place to another

• Voltage causes charge to move from one place to another

• We can write ONE equation for work that serves as a pattern for the previous four equations

• When a force or torque moves an object

• Mechanical systems use force and torque to cause movement and, thus, to do useful work

• Using symbols the equation can be shortened to:

Units = Force x distance

= foot-pound or Newton-meters

• Given: A weight lifter applies a vertical force to raise a 200-lb barbell 5 ft.

• Find: The amount of mechanical work done in lifting the barbell

• Work = Force x Distance

• = (200lb)(5ft)

• = 1000 ft-lb

• Given: A football player pushes a pile of 4 lineman 6 yards. Each lineman weighs 200 pounds.

• Find: The amount of mechanical work done pushing the pile.

Work = Force x Distance

Force = (4 lineman x 200lbs)

Distance = 6 yards = 18 feet

Work = Force x Distance

Work = (800lb)(18ft)

= 14400 ft-lb

• Given: An electric truck is used to pull a loaded cart a distance of 100 meters. A force of 900 newtons is required to keep the cart moving at a constant speed.

• Find: The amount of mechanical work done on the cart.

• Work = Force x Distance

• = (900N)(100m)

• = 90,000 N-m

• A common SI unit for work or energy

• 1 N-m = 1 Joule

• 90,000 N-m = 90,000 J

• 55 N-m = 55 J

• Given: 75 Joules are acted on a tennis ball that is thrown

• If the tennis ball is thrown with a force of 15 Newtons

• Find: How far did the ball travel while the work was applied?

• 75 J = 75 N-m

• Work = (F) x (D)

• 75 N-m = (15 N) x (D)

• 5 m = D

• When work is done it can be rated by efficiency

• Comparison of the output work by the input work

• A pulley system = block and tackle

• Pull with force

• (F)

• Moves a distance

• (D)

• (w)

• Raised a distance

• (y)

• Input work = (F) x (D)

• Output work = (w) x (y)

• Efficiency is equal to a ratio of the output work to the input work

% Efficiency =

Output Work

Input Work

x 100%

• Given: The block and tackle shown is used to lift a 600 lb engine. It is raised 0.9 ft as the operator pulls with a force of 100lb over a distance of 6 ft.

• Find:

• Input work

• Output work

• Efficiency

• Input work = F x D

• F = 100 lb

• D = 6 ft

• Input work = 100 lb x 6 ft

• Input work = 600 ft-lb

• Output work = w x y

• w = 600 lb

• y = 0.9 ft

• Output work = 600 lb x 0.9 ft

• Output work = 540 ft-lb

• Efficiency = (Output/Input) x 100%

• Output = 540 ft-lb

• Input = 600 ft-lb

• Efficiency = (540 ft-lb)/(600ft-lb) x 100%

• Efficiency = 0.9 x 100%

• Efficiency = 90%

• Given: The block and tackle shown is used to lift a 1 kg mass. It is raised 15 cm as it is pulled with a force of 5.5 N over a distance of 34 cm.

• Find:

• Input work

• Output work

• Efficiency

1 kg = 9.8 N

100 cm = 1 m

100 cm = 1 m

Input work

• Input work = F x D

• F = 5.5 N

• D = 34 cm = 0.34 m

• Input work = F x D

• F = 5.5 N

• D = 34 cm

Input work = 5.5 N x 0.34 m

Input work = 1.87 N-m

Input work = 1.87 Joules

100 cm = 1 m

Output work

• Output work = (w) x (y)

• w = 1 kg = 9.8 N

• y = 15 cm = 0.15 m

• Output work = (w) x (y)

• w = 1 kg

• y = 15 cm

Output work = 9.8 N x 0.15 m

Output work = 1.47 N-m

Output work = 1.47 Joules

• Efficiency = (Output/Input) x 100%

• Output = 1.47 N-m = 1.47 J

• Input = 1.87 N-m = 1.87 J

• Efficiency = (1.47 J)/(1.87 J) x 100%

• Efficiency = 0.786 x 100%

• Efficiency = 78.6%

• Rotational work occurs when torque causes a mass to rotate through an angle

• The size of angles is commonly measured in degrees

• For this we must know π= 3.1416

• And Circumference = 2π r

Circumference = 360o = 2π r = 6.28 rad

Circumference = 360o = 2π r = 6.28 rad½ Circumference = 180o = π r = 3.14 rad

Circumference = 360o = 2π r = 6.28 rad¼ Circumference = 90o = ½ π r = 1.57 rad

Circumference = 360o = 2π r = 6.28 rad1/8 Circumference = 45o = ¼ π r = 0.785 rad

• This figure shows 1 radian is equal to 57.3 degrees

Circumference = 2 π r

Therefore a full circle must equal 2π r = (2)(3.14)(r) = 6.28 rad

• 7 revolutions

• 435 degrees

• ¾ revolutions

• 14 revolutions

• 26 degrees

• Note: The angle θmust be measured in radians

Given: An electric motor is used to drive a water pump. The motor provides a torque of 150 N-m.

Find: How much work is done while the motor turns 40 revolutions?

Torque = Force x Lever arm

Work = Torque x Angle

Torque (is given)= 150 N-m

Torque = Force x Lever arm

Work = Torque x Angle

Work = (150 N-m)(40 revolutions?)

Work = (150 N-m) (251.2 rad)

Work = 37,680 N-m or …

37,680 Joules

• Given: A simple mechanical winch has a crank handle 1 foot long. A force of 20 pounds is required to turn the crank.

• Find: How much work is done in turning the crank 5 revolutions.

• Hint: First we must find torque and the angle in radians in order to find work

Torque = Force x Lever arm

Work = Torque x Angle

Torque = Force x Lever arm

• Torque = (20 lb) (1 ft)

• Torque = 20 lb-ft

Work = Torque x Angle

Work = (20 lb-ft)(5 revolutions?)

Work = (20 lb-ft) (31.4 rad)

Work = 628 ft-lb

Θ = (10)(3.14)

?

Linear

Angular

Electrical

Thermal

### Work Done by a Winch Activity

Done together during lab period

### Work Done by a Winch Activity

Done individually during lab write up