Loading in 5 sec....

Work in a Mechanical SystemPowerPoint Presentation

Work in a Mechanical System

- 121 Views
- Uploaded on
- Presentation posted in: General

Work in a Mechanical System

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Unit 6

Work in a Mechanical System

Unit VocabularyPg 1 - 14

Force applied

Mechanical work

Work done

Joule

Efficiency

Ratio

Radian

Unit ReviewPg 15-16

# 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

- It is important to know that work is only done while the force is applied
- Examples: Throwing a ball or a vehicle coasting to a stop
- No work is done when the force is no longer applied (even if something is still moving)

- 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)

- (D)

- (w)

- (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

- Output work = w x y
- w = 600 lb
- y = 0.9 ft

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

- 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

1 kg = 9.8 N

100 cm = 1 m

- 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

1 kg = 9.8 N

100 cm = 1 m

- 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

- Rotational work occurs when torque causes a mass to rotate through an angle
- The size of angles is commonly measured in degrees
- It can also be measured in radians “rad”
- 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

1 Revolution = 6.28 radians1 Radian = 57.3 degrees

1 Revolution = 6.28 radians1 Radian = 57.3 degrees

- 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

1 revolution = 6.28 rad

Torque (is given)= 150 N-m

Torque = Force x Lever arm

1 revolution = 6.28rad

40 revolutions = 251.2 rad

Θ = 251.2 rad

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

1 revolution = 2π rad

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)

Θ = 31.4 rad

4 Types of Mechanical Work

?

Linear

Angular

Electrical

Thermal

Work Done by a Winch Activity

Work Done by a Winch Activity

Work Done by a Winch Activity

Work Done by a Winch Activity

Work Done by a Winch Activity

Done together during lab period

Work Done by a Winch Activity

Done individually during lab write up

Work Done by a Winch Laboratory