IOT POLY ENGINEERING 3-14 DRILL March 18, 2009 Solve the following problem in your notebook. It...

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IOT POLY ENGINEERING 3-14 DRILL March 18, 2009 Solve the following problem in your notebook. It takes a girl 1 minute to pull her 18 lb wagon a distance of 40 ft. The force required to pull the wagon is 12 lbs, which the girl applies to the wagon handle at 30 o with the horizontal. How much work does she do? How efficient is her force being applied in the direction of

Transcript of IOT POLY ENGINEERING 3-14 DRILL March 18, 2009 Solve the following problem in your notebook. It...

IOT

POLY ENGINEERING3-14

DRILL

March 18, 2009

Solve the following problem in your notebook.

It takes a girl 1 minute to pull her 18 lb wagon a distance of 40 ft. The force required to pull the wagon is 12 lbs, which the girl applies to the wagon handle at 30o with the horizontal. How much work does she do? How efficient is her force being applied in the direction of displacement? What power is used (units for power are ft-lb / s)?

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POLY ENGINEERING3-14

WORK

300

F = 12 lbs

Wt = 18 lbs

d = 40’

STEP #1: SKETCH and GIVENt = 1 min = 60 s

Wt = 18 lbs

d = 40 ft.

F = 12 lbs

W = ?

EFF = ?

P = ?

Fx = (12 lbs)cos 30o

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POLY ENGINEERING3-14

WORK

300

F = 12 lbs

Wt = 18 lbs

d = 40’

STEP #1: SKETCH and GIVENt = 1 min = 60 s

Wt = 18 lbs

d = 40 ft.

Fx = 10.4 lbs

W = ?

EFF = ?

P = ?

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POLY ENGINEERING3-14

STEP #2: WRITE EQUATIONFORMULASTEP #4: CHECK / BOX ANSWERSTEP #3: SUBSTITUTE / SOLVE

WORK

W = F x d

= Fx x d

= (10.4 lb)(40’)

W = 416 ft-lb

EFF = (Output/Input)x100%

= (F / Fx) x 100%

= [(10.4 lb)/(12 lb)] x 100%

EFF = 86.7%

P = W / t

= (416 ft-lb) / (60 s)

P = 6.93 ft-lb/s

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POLY ENGINEERING3-14

WORK1. Velocity, acceleration, force, etc. mean nearly the same

thing in everyday life as they do in physics.2. Work means something distinctly different.3. Consider the following:

1) Hold a book at arm’s length for three minutes.2) Your arm gets tired.3) Did you do work?4) No, you did no work whatsoever.

4. You exerted a force to support the book, but you did not move it.

5. A force does no work if the object doesn’t move

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POLY ENGINEERING3-14

WORK

• The man below is holding 1 ton above his head. Is he doing work?No, the object is not moving.

• Describe the work he did do:Lifting the 1 ton from the ground to above his head.

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POLY ENGINEERING3-14

WORK

WORK = FORCE x DISTANCE

The work W done on an object by an agent exerting a constant force on the object is the product of the component of the force in the direction of the displacement and the magnitude of the

displacement.

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POLY ENGINEERING3-14

WORK

WORK = FORCE x DISTANCE

W = F x d

Consider the 1.3-lb ball below, sitting at rest. How much work is gravity doing on the ball?

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POLY ENGINEERING3-14

WORKWORK = FORCE x DISTANCE

W = F x d

Now consider the 1.3-lb ball below, falling 1,450 ft from the top of Sears Tower. How much work will have gravity done on the

ball by the time it hits the ground?

F = 1.3 lbs W = F x dd = 1,450 ft. = (1.3 lb) x (1,450 ft.)W = ? W = 1,885 ft-lb

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POLY ENGINEERING3-14

A 3,000-lb car is sitting on a hill in neutral. The angle the hill makes with the horizontal is 30o. The distance from flat ground to the car is 200 ft. Begin with a free-body diagram. Then, calculate the weight component facing down the hill. Finally, calculate the work done on the car by gravity.

Wt = 3,000 lb

30o

Fw = ?d = 200’

WORKBack to our drill problem

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POLY ENGINEERING3-14

Wt = 3,000 lb

30o

Fw = ?d = 200’

WORK

60o

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POLY ENGINEERING3-14

WORK

60o

3000 lb.

x cos 60o = x / (3000 lb)

x = (3000 lb)(cos 600)

= (3000 lb)(1/2)

x = 1,500 lb.

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POLY ENGINEERING3-14

Wt = 3,000 lb

30o

F = 1,500 lb.d = 200’

WORK

F = 1,500 lb

d = 200 ft

W = ?

W = F x d

= (1500 lb) x (200 ft)

W = 300,000 ft-lb

EFFICIENCY

EFFICIENCY = x 100%OUTPUT INPUT

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POLY ENGINEERING3-14

Wt = 3,000 lb

F = 1,500 lb.

EFFICIENCY

FORCE APPLIED = 3,000 lb

EFFECTIVE FORCE = 1,500 lb

Back to our drill problem

INPUT

OUTPUT

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POLY ENGINEERING3-14

EFFICIENCY

FORCE APPLIED = 3,000 lb

EFFECTIVE FORCE = 1,500 lb

Back to our drill problem

INPUT

OUTPUT

EFFICIENCY = x 100%OUTPUT INPUT

EFF = x 100%1,500 lb 3,000 lb

EFF = 50%

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POLY ENGINEERING3-14

POWER

1. Three Buddhist monks walk up stairs to a temple.2. Each weighs 150 lbs and climbs height of 100’.3. One climbs faster than the other two.4. Who does more work?5. They all do the same work:

W = F x d (force for all three is 150 lb) = (150 lb)(100’)W = 15,000 ft-lb

6. Who has greater power?

POWER

Power is the rate of doing Work

P =

The less time it takes….The more power

Units:Watts, Horsepower,

Ft-lbs/s

W t

IOT

POLY ENGINEERING3-14

IOT

POLY ENGINEERING3-14

HOMEWORK

WORKSHEET

Problem Solving Steps:

1. Write given and sketch a diagram

2. Write equation/formula

3. Substitute values and solve

4. Check answer

5. Box answer