1. Sub unit 6.1 ”Power in Mechanical Systems”
Introduction a 10 Minute video
Video shoes real life scenarios, equations, terms, and labels for both SI and English units.
After video we have the following power point lecture that discusses terms and their interpretations, formulas and their labels, and we finish with sample problems to identify the students abilities to use the new information.

2. Subunit 1 Power in Mechanical Systems
Power: Rate of doing work
Power rating: maximum rate at which each machine can do work

3. Linear Mechanical Systems
Work =
Force applied to object x
Distance object moves along the direction of the force acting on it
Power =
Work done by applied force
Time to do the work
W = f x d
P = W t
P =
f x d t
P = f x d t
P = f x V
Power = force x speed

4. Rotational Mechanical Systems
Work =
Torque applied x
Angle moved through
W = T x θ
P = W t
P =
T x θ t
Since ω = θ/t
P = T x ω
Power = torque x angular speed

5. Units for Mechanical Power
1 horsepower = 746 watts
1 horsepower = 550
ft•lbs
sec
English SI
ft•lbs
sec
Joule
sec
N•m
sec
watt
horsepower

6. Efficiency
Efficiency: ratio of work out to work in or ratio of power out (Pout) to power in (Pin)
Power In =
Work In
time
Power Out =
Work Out
time
% efficiency =
Work Out
Work In
X 100
Or…
% efficiency =
Power Out
Power In
X 100

7. Sample Problem Given: A steel casting weighs 200 pounds. It is
raised 3 ft in 4 sec at a constant speed.
Find : horsepower of cylinder used to lift the steel casting.
1st …
2nd …
P =
f x d t
P =
200 lb x 3 ft
4 sec =
150 ft•lb/sec
V = d/t =
3 ft
4 sec =
P = f x V
.75 ft/sec
150 ft•lb/sec
P = (200 lbs) (.75 ft/sec) =

8. Sample Problem Given: A crate weighs 980N and is lifted 2 meters
in 2 sec. The winch pulls the cable a dist.
Of 8 meters with a force of 272N in the
same 2 sec.
Find :
A.) Input power (in watts) supplied to block and tackle by the winch
B.) Output power (in watts) of block and tackle used in lifting crate
C.) Efficiency of block and tackle

9. Sample Problem Cont. A. B. C. Pin = Fin x Din Time = = 1088 N•m sec
1088 watts
Pout =
Fout x Dout
Time =
(980 N) (2m)
2sec =
980
N•m
sec =
980 watts
Efficiency =
Pout
Pin
X 100 =
980 watts
1088 watts =
90%
.90 x 100 =

10. Sample Problem Given: An electric motor has a shaft torque of
.73 lb•ft when rotating at 1800 rpm.
Find : horsepower of the shaft
P = T x ω
ω =
1800 ( )
rev
min
1 min
60 sec
6.28 rad. =
188.4 rad/sec
137.5
ft•lb
sec
P =
.73 lb•ft x rad/sec =
P =
137.5
ft•lb
sec x
1 hp .
550 ft•lb =
.25 hp

11. ( )( ) Sample Problem ουσ Given: a piston in a compressor has a
flywheel with a rate of 1100 rpm
(115rad/sec) Torque is 65 N•m
Find : hp of flywheel
P = T x ω
65 N•m x 115 rad/sec =
7475 watts ( )( )
7475 watts
1hp
746 watts =
10.0 hp
ουσ

12. Lab Demos