NAME______
Simple Machine Review
1) I am pushing a block up a ramp that is 3 meters high and 6 meters long. It is a 30-degree angle ramp. The block has a mass of 5 Kg. It is taking me 37 Newtons of force to push it.
Draw a picture.

What is the Effort Force? What is the Resistance Force?
Effort Force=Fin= 37 N, Resistance Force =Fout=
Weight=49 N
What is the Effort Distance? What is the Resistance Distance?
Effort dist=DistOut= 6 m, Resistance Dist= DistIn=
3 m
What
is the IDEAL mechanical advantage of this system?
IMA
= if WorkIn=WorkOut, so Fin*DistIn=Fout*DistOut
IMA=
DisIn/DistOut = Length/Height= 6/3=2:1
For
ramp Length/Height = 1/sinθ = IMA
For
lever IMA=Effort Arm/ResistanceArm
For
pulley IMA= # strings pulling up
For
wheel/axle= axle diameter/wheel diameter
What
is the ACTUAL mechanical advantage of this system?
AMA=
Fout/Fin= Weight/Effort= 49 N/37N= 1.32:1
What
is the Work OUT?
Work
= Fout*DistOut=Weight*Height=mgh=49N*3m=147 Joules
What
is the Work IN?
Work=Fin*DistIn=Effort*Length=37N*6m=222
Joules
What
is the Efficiency of this machine?
Efficiency
= WorkOut/WorkIn= 147J/222J= 66%
Or
AMA/IMA= 1.32/2 = 66%
Why?
33%
was lost, or 75 Joules, was lost to friction and changed from mechanical energy
to heat energy.
8-4 pg 301 #2
The efficiency of a squeaky pulley system is 73 percent. The
pulleys are used to raise a mass. What force is exerted on the machine if a
rope is pulled 18 m in order to raise a 58 kg mass a height of 3 m?
Work Out = Weight * Height
= mg h = 58*9.8 * 3 =
568.4 N * 3m = 1705 Joules
Efficiency = .73 =
WorkOut/WorkIn so .73 = 1705 J
/WorkIn
Work In =1705 J/ .73 =2335.89 Joules
Work In = Fin * Dist in
2335.9
J = Fin * 18m , Fin = 2335.9
Joule/ 18 m =129.77 N
Check: AMA = Fout/Fin=568.4/129.77=
4.38
IMA=Din/Dout=18/3=6
Efficiency=
AMA/IMA=4.38/6= 73%!
#3 a person pushes a 950N box up an incline. If the person
exerts a force of 250 N along the incline, what is the mechanical advantage of
the incline?
AMA= Fout/Fin=950 N/250 N
= 3.8:1
Pg 309
52 Why is it easier to loosen the lid from the top of a paint can with a long-handled screwdriver than with a short-handled screwdriver?
53. If a machine cannot multiply the amount of work, what is the advantage of using such a machine?
54. You are attempting to move a large rock using a long lever. Will the work you do on the lever be greater than, the same as, or less than the work done by the lever on the rock??? Explain.
53. You are attempting to move a large rock with a long lever. Is it more effective to place the lever’s axis of rotation nearer to your hands or nearer to the rock? Explain…
56. A perpetual motion machine is a machine that, when set in motion, will never come to a halt. Why is such a machine not possible?
57. If you were use a machine to increase the output force, what factor would have to be sacrificed? Give an example….
Winnie, a waitress, holds a 5 N tray stacked with twelve 3.5
N dishes in one hand. The length of her arm from her hand to her elbow is 30 cm
and her biceps exert a force 5 cm from her elbow, which acts as a fulcrum? What
type of lever is this? How much force must her biceps exert to allow her to
hold the tray?
Resistance Force: 5N+
12*(3.5)= 47N
1st class lever
IMA = effort arm/rest arm = .05/.3= .1667 or 1:6
MA= ForceOut/ForceIn =
1/6= 47/ForceIn , so Force in =282 Newtons
WorkIn=WorkOut Force*Dist=Force*Dist
ForceIn*5=47*30,
ForceIn=282 Newtons
A windmill uses sails blown by the wind to turn an axle that allows a grindstone to grind corn into meal with a force of 90 N. The windmill has sails of radius 6 m blown by a wind that exerts a force of 15 N on the sails, and the axle has a radius of 50 cm.
What is the IDEAL mechanical advantage of this system?
IMA=
DistIN/DistOut, 6 m/ .5 m = 12:1
What
is the ACTUAL mechanical advantage of this system?
AMA=ForceOut/ForceIn=90N/15N=6:1
What
is the Work OUT? =Force*Dist= 90N*.5m= 45
Joules
What
is the Work IN?= Force*Dist= 15 N*6m = 90
Joules
What
is the Efficiency of this machine?
Efficiency=AMA/IMA=
6/12 = 50% or WorkOut/WorkIn=45J/90J=50%
*** When building the pyramids, the ancient Egyptians were
able to raise large stones to very great heights by using inclines/ If an
incline has an ideal mechanical advantage of 4 and the pyramid is 15 m tall, how
much of an angle would the incline need in order for the Egyptian builder to
reach the top?
IMA= DistIn
(Length)/DistOut(Height)
4 = Length/15 m
Length= 60 m
Sinθ = Opp/Hyp =
15/60 = .25,
θ= sin-1 (.25) = 14.5°