1. March 18, 2009
IOT POLY ENGINEERING
3-14
DRILL
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)?

2. STEP #1: SKETCH and GIVEN
IOT POLY ENGINEERING
3-14
WORK
STEP #1: SKETCH and GIVEN
t = 1 min = 60 s
Wt = 18 lbs
d = 40 ft.
F = 12 lbs
W = ?
EFF = ?
P = ?
F = 12 lbs
Fx = (12 lbs)cos 30o
300
d = 40’
Wt = 18 lbs

3. STEP #1: SKETCH and GIVEN
IOT POLY ENGINEERING
3-14
WORK
STEP #1: SKETCH and GIVEN
t = 1 min = 60 s
Wt = 18 lbs
d = 40 ft.
Fx = 10.4 lbs
W = ?
EFF = ?
P = ?
F = 12 lbs
300
d = 40’
Wt = 18 lbs

4. STEP #3: SUBSTITUTE / SOLVE STEP #4: CHECK / BOX ANSWER
WORK
IOT POLY ENGINEERING
3-14
STEP #3: SUBSTITUTE / SOLVE
STEP #4: CHECK / BOX ANSWER
STEP #2: WRITE EQUATIONFORMULA
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

5. IOT POLY ENGINEERING
3-14
WORK
Velocity, acceleration, force, etc. mean nearly the same thing in everyday life as they do in physics.
Work means something distinctly different.
Consider the following:
Hold a book at arm’s length for three minutes.
Your arm gets tired.
Did you do work?
No, you did no work whatsoever.
You exerted a force to support the book, but you did not move it.
A force does no work if the object doesn’t move

6. WORK 3-14 The man below is holding 1 ton above his head.
IOT POLY ENGINEERING
3-14
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.

7. WORK 3-14 WORK = FORCE x DISTANCE
IOT POLY ENGINEERING
3-14
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.

8. WORK 3-14 WORK = FORCE x DISTANCE W = F x d
IOT POLY ENGINEERING
3-14
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?

9. WORK 3-14 WORK = FORCE x DISTANCE W = F x d
IOT POLY ENGINEERING
3-14
WORK
WORK = 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 d
d = 1,450 ft = (1.3 lb) x (1,450 ft.)
W = ? W = 1,885 ft-lb

10. Back to our drill problem
WORK
IOT POLY ENGINEERING
3-14
Back to our drill problem
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.
d = 200’
Fw = ?
Wt = 3,000 lb
30o

11. IOT POLY ENGINEERING
3-14
WORK
d = 200’
Fw = ?
60o
Wt = 3,000 lb
30o

12. WORK 3-14 cos 60o = x / (3000 lb) x = (3000 lb)(cos 600)
IOT POLY ENGINEERING
3-14
WORK
cos 60o = x / (3000 lb)
x = (3000 lb)(cos 600)
= (3000 lb)(1/2)
x = 1,500 lb. x
60o
3000 lb.

13. WORK 3-14 F = 1,500 lb d = 200 ft W = ? W = F x d
IOT POLY ENGINEERING
3-14
WORK
d = 200’
F = 1,500 lb
d = 200 ft
W = ?
F = 1,500 lb.
Wt = 3,000 lb
30o
W = F x d
= (1500 lb) x (200 ft)
W = 300,000 ft-lb

14. EFFICIENCY

16. EFFICIENCY = x 100%
OUTPUT INPUT

17. Back to our drill problem
IOT POLY ENGINEERING
3-14
EFFICIENCY
Back to our drill problem
F = 1,500 lb.
Wt = 3,000 lb
FORCE APPLIED = 3,000 lb
EFFECTIVE FORCE = 1,500 lb
INPUT
OUTPUT

18. Back to our drill problem
EFFICIENCY
IOT POLY ENGINEERING
3-14
Back to our drill problem
FORCE APPLIED = 3,000 lb
EFFECTIVE FORCE = 1,500 lb
INPUT
OUTPUT
EFFICIENCY = x 100%
OUTPUT INPUT
EFF = x 100%
1,500 lb ,000 lb
EFF =
50%

19. POWER 3-14 Three Buddhist monks walk up stairs to a temple.
IOT POLY ENGINEERING
3-14
POWER
Three Buddhist monks walk up stairs to a temple.
Each weighs 150 lbs and climbs height of 100’.
One climbs faster than the other two.
Who does more work?
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
Who has greater power?

20. P = W t Units: POWER Watts, Horsepower, Ft-lbs/s 3-14
IOT POLY ENGINEERING
3-14
Power is the rate of doing Work
P =
The less time it takes….
The more power
Units:
Watts, Horsepower, Ft-lbs/s
W t

21. Problem Solving Steps:
IOT POLY ENGINEERING
3-14
HOMEWORK
WORKSHEET
Problem Solving Steps:
Write given and sketch a diagram
Write equation/formula
Substitute values and solve
Check answer
Box answer