Physics/Adva nced Physics
The Ramp simulation
Name K'24=­
Hr.
Open up the following website: http://phet.colorado.edu/simulations/sims.php?sim=The Ramp and click (/run
now." The simulation allows us to explore forces and motion on a horizontal surface and on a ramp. This
simulation allows us to change different features: object and its mass, the angle of the ramp, the frictional
characteristic of the ramp surface, etc.
Part A) Learning a bit about the simulation.
First off, when you open the simulation you get a screen like this:
There is a graph below and some controls on the right. This simulation includes lots of details we aren't really
ready to study yet: work and energy. However, it can show us forces and we are ready to study the properties of
how forces can be diagrammed on incline planes.
Choose File Cabinet
100 kg, . , 0.3
=
~PimlO
' " 225
IJ =0.4
There is a control on the right that allows us to change the object and the
coefficient of friction. Each object has a different mass. This mass is used to
find the weight of the object (W = mg). Remember that this coefficient of
friction is symbolized by mu (Jl). It represents the texture of the surface. A
bigger mu means more friction; a smaller mu means less friction.
Clicking under that and we can turn off the
friction. Often there are incline plane problems where
we neglect or ignore friction. We can turn this off and on by clicking this off
and on.
p=OJ'
Position
This next control is helpful. It moves your
object along the ramp. Move your object to
-6 and see where the object is moved to.
Look at what happens to your forces. Explain: htlO'\!\tU.I
1AtGt<,M:
t
'l
What happens to the direction of your normal force?
What happened to the frictional force?
til\s0<p ~
The Ramp simulation, p. 1, 12/23/2009
-6.0
V
Which way does the ~eight ~Iways point? L
OV\fu k,Of\=tt:rv\W Sv~"(.€.../
~<6- fflr ~
.
Now move the position to the reference point (0.0). Where is the object located now?
Now
move the object onto the ramp itself. Look at the forces. Make sure you understand why the normal force is now
angled. Why is friction now directed up the incline?
~CJ\IIA./
Etl.vhuv\ l< ptUt0hL1j
'00i\ect-
VVvO·U·,nq ~ V'~
J
'
Draw and label the forces o
(a) horizontal surface (b) on the incline
~
1\ /'\_
r:; t &tv~
rr
\[J Part B) Now choose the refrigerator:
Place it at -6.0 meters.
There are several ways to start the simulation. Under the ramp angle, you will see the
following command: Click on it and watch it change to a pause.
Under the object you just moved are the following commands. Do you see you can start
and stop the object under the applied force command?
Go ahead and click, Go!". And nothing will happen. So, we want to start the object
moving. We need to activate the applied force. Remember that applied forces are forces applied by people.
If
Use your arrows to add 100 N of applied
hot eJ'\6y\,( ~
right, nothing! Why not?
Forct-
~f\t~A
WO~hl~~
What happens?
v . That's
Bring it up to 200 N. Anything now? -=---lU...!..U..!.~_.
We need to put in the correct amount to get the object to move. We have to counter the frictional force with our
applied force. But can we calculate the amount of applied force we need? Well, we learned that:
friction force (Ff ) =IJ.*N (Normal force). But the normal force = ~et:)
can calculate this or maybe I should say that you can calculate this:
. So, the frictional force lJ.*m*g.
F} ~ 0,5· \"1 S' lL5; q B~/s "­
=
We
I
So, you should get Ff 0.5*175 kg*9.8 rn/s2 = 857.5 N. Plug this number into the Applied Force box and see what
happens. It doesn't look like the refrigerator is moving. Let's up our force to an even 860 N. Make sure the frig is
at a -6.0 position, make sure to hit, ({Clear" to clear the graph and then hit ({Go". Watch what happens. You may
need to watch it for Cl, while (Cjlround 20 sec of so). Explain what you see:
k~OK4- w-& ~ flf
1"
y-wr
We see the inclusion of a {fTotal" Force now. The total force is the net horizontal force. The Applied force of 860 N is now largerthan the frictional force of 857.5 *
N.
fCrc:s ~ unb~
t:l.~ \.:t
iY\0'VtA
---_/
The applied force is larger than the frictional force and we see a positive net
force. Consequently, the frig starts to accelerate. W~ ~ vt~.., ~.1. 4-."
The Ramp simulation, p. 2, 12/23/2009
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Part C: But what happens when the frig gets to the ramp? _ _~..!..-...I...--
Explain why the frig stops?
\ S i\-ct
_ _ _ _ _ _ _ _ _ _ _ __
Ol'\~ r-~ I -tl-e Pff LuJ ~4'<t.,
~~ eA'''M) ~
The applied force is now not large enough to counteract the frictional
force. Why would the force be adequate to move it on a horizontal
surface but not large enough to accelerate it up the incline?
MfJ-
wvbw- IT); ~ h
('CiA:
~~~ JeW-~'
q,'lm~
You see that the i1total" force has disappeared. Consequently, your
applied force is not great enough to create a net force.
Also, look at the direction of the forces on the incline plane. Weight is
still down but the normal force is now perpendicular to the incline
plane and the applied force and the friction force are parallel to the incline plan.
Appl fed wrce
Norma.i
But can we determine how much force we need to
counteract friction?
To understand inclines, you need to understand how forces
are diagrammed on incline planes for this scenario.
Frf{:tioJ'!
or
W (perpend icu lar)
."
First off, we rotate our coordinate axes to match the incline
plane. The x axis is parallel to the incline plane and the y
axis is perpendicular to the incline plane.
mgsin[9}
ormg or W/I (parallel)
+x
We can take the weight vector and find its two components.
The force into the incline plane (-y) is mgcos(9) or also called
W..1(perpendicular) or the perpendicular component of the
weight. This is balanced out by the Normal force (+y) as
there is no acceleration in this direction. So we can say that
N = mgcos(9).
-y
The x direction is along the incline plane. Look at the forces
and components. The applied force is up the plane (+X). We have friction and mgsin(9). mgsin(9) is also called
;Ie
the parallel component of the weight (WII). B~~~.aXis. Ttreapptte.d force is aloRg tt:lc
iil~ fJl.L)
1)(
a*is..
So to accelerate the frig, our applied force> friction + mgsin(9). Again, the friction force (h) = Il*N but the normal
force = mgcos(9). So, our applied force> Ilmgcos(9) + mgsin(9). Let's put in some numbers:
applied force> 0.5*175 kg*9.8 m/s2 *cos (10°) + 175 kg*9.8 m/s2 sin(100). applied force> 844.5 N + 298 N applied force> 1142.5 N The Ramp simulation, p. 3, 12/23/2009
~()\P
~
tlo '\.tv'\Il 7
Set the applied force 1143 N. Click, IIGo" and watch to see what happens. Again, it may take half a m nute to see
what happens. Does the refrigerator eventually accelerate up the ramp?
{ .
~v"
D~t,...
Pause the simulation and move the refrig back to the end of the ramp. S t the applied for~e for a bit larger,
"
ur-p
maybe 1200 N. Click, IIGd' and watch to see what happens:
.f1n...tv\
0-1:,tJ.v~
.
Part 0: Redo Part B with a different object. Follow the steps below:
I
Object choose:
F, l~
C.6tbl~t
D0
Mass of object =
Coefficient of Friction
~r L'f-
kg
(~) =
0 IS
...A Piano rN
Applied force> Frictional Force?
Applied force>
2fll t1
FileCabi net
100k~'J;.JJ= 0.3
Refrigtrator
175 kg, lJ = 0.5
Find the applied force needed to push the object along the horizontal surface.
Verify this with the simulation.
\ -==f N\tt
'\
11"225 kg,.~
=0.4 ~ o)ltl/b )
0 I~ l'\O
N. Ve.f ;LlltNiW 1\
I ~f\~vJ
Part'!: Redo Part C with the object you chose in part D. applied force> friction + WI/ (parallel) OR
o3 (fO())("lI~} Ci)S \0° applied force> + ~*(Normal Force) mgsin(8) OR [bO' l1(~ ~(I\(ol! \-b applied force > ~mgcos(8) + mgsin(8)
applied force >
~.)3
.L-'t0)\
\TO,\
N+
1-
0'
N
\1-0 l applied force>
4 s-q, f
N
Verify this with the simulation. Part F: What happens to the applied force needed if the angle increases? Redo part E with a medium angle. Don't change the object or the coefficient of friction. angle you chose?
~\)\)
applied force> friction +
W// (paral/el)
OR
applied force> + ~*(Normal Force) mgsin(8) OR
applied force > ~mgcos(8) + mgsin(8)
applied force>
L S q.0\
N+
t..t q 0
applied force>
7 t..t~." N Verify this with the simulation. \
The Ramp simulation, p. 4, 12/23/2009
N
/
~
Part G: What happens to the applied force needed if the angle increases? Redo part F with a larger angle than
in part F. Don't change the object or the coefficient of friction.
angle you chose?
bOO
applied force> friction +
WI; (paral/el)
OR
applied force> + IJ*(Normal Force) mgsin(8) OR
applied force> Ilmgcos(8) + mgsin(9)
applied force>
P-(t
N+<6'-1 g /1-
N
applied force>
Answering several questions: 111 parts E, F, and G, you used the same object with the same frictional coefficient (11). You varied the angle of the incline. Look at the applied force as your angle increased. Did the applied force increase or decrease or stay the same?
Look at the frictional force (Ilmgcos(9) ) part of your equatiqns on these parts. Did the frictional force increase or
. -}
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decrease or stay the same as the angle increased?...v
I
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Can you offer a mathematical explanation using the ideas of trig?
G:
Ol ~ ~'
>0
WIl./twY\. ~
Look at the WI; (parallel) or mgsin(8) part of your equations for these angles. Did this component increase or
decrease or stay the same as the angle increased? _ _ _
'1'_ _____
(J./j
._.._~-.".,--o.~f--f_e,_r a m athematica I exp Ian~~~!2.._
&- t I ,~IV\ &- J
Go back to the simulation and set the angle back to
10 degrees. Make sure the object is on the ramp
and that your guy is attempting to move it.
Now watch the forces and the lengths of the arrows
as you increase the angle. Discuss what happen~ to each force~
Normal:
vv-.''j G S.
Weight:
d,{te~~' 4G~~
Friction:
~ iV'-,-:?/"O) 8"
e
t
t
I
Take the angle to 90 degrees and explain what happens to the nor~al and frictional forces:
I
C(J)CjO ,,-0
*: The Ramp simulation, p. 5, 12/23/2009
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KampAngle
Part G: Allowing the object to slide down a friction-free surface. Go back and choose the refrig. Go back to a 10° angled ramp.
0.0
30.0
60.0
90.0
10.0 degrees
Place the object at the top of the ramp.
~ Frictionless
Make sure to Pause the object. Now remove the friction by clicking in the box: Notice that a skateboard is now under the frig. This indicates that we removed the friction. Remove the applied force by zeroing it out the applied force:
j,=m_ _ _ _'''''"';''';;''''"
Look at the force diagram now on the refrig. There is no change to the Normal and the Weight vectors. There is the "total" present on the diagram. That means that there is a net force. ~ cU\Af\
Now activate the "Go" and watch what happens to the frig: _ _ _ _ _ _ _ __
----
Why did the refrig accelerate down the ramp on its own? _ _...L.!~VY~-
~
~'-----
-
_ _ _ _ _ __
::::::::;;.
Let's look at this force diagram
detail:
Norma! force
We have two basic forces now: the normal force and the weight
vector.
Again, we can break the weight vector up into its two
components.
mg,sin{&j
,cfmg +y
or WI / (parallel)
+x
The force into the incline plane (-y) is mgcos(8). This is balanced
out by the Normal force (+y) as there is no acceleration in this
direction. So we can say that N = mgcos(8). This is the same as
when we were pushing the refrig up the incline.
/
/"
\\
.,.
X
-i
/~/
~-y
Why would the refrigerator accelerate down the ramp? The
sin(theta) component of the weight is unbalanced!
Atop the graph, we see this number:
%$~17.11
N'
We have been calling this WII. We can calculate this number: WII = mgsin(8) = 175 kg*9.8 m/s 2 *sin(100) = 297.81 N. We add the negative sign to indicate it is in the -x. This is exactly the same number that the simulation is showing us. Part H: Choose another object; choose a larger angle. Mathematically determine WII.
Object:
Mass:
Gr-vtk "3 e(J
kg ~
The Ramp simulation, p. 6, 12/23/2009
3 (}O CVi , If')
WII = mgsin(8) =
"'~
CO! N.
2--0 ~1 Verify this with the simulation.
Part I:
0
Mathematically make the angle approach 90
What happens to WII if the angle gets close to 90 degrees?
Try it on the simulation.
Do you see if your angle is 90 then WII = mgsin(8) =
mgsin(90 =mg(l). WII =mg =the weight of the object.
0
0
)
The object is in free fall (Weight
= Fgravity)
,nI.IIW
·W\t ..
The value for the refrig is -1715 N. But this is just its
weight: 175 kg* 9.8 mis/so
How much applied force would you need now to push the object up the
frictionless ramp?
\ 1l<)
Why the weight of the object, of course! You would be balancing the
weight and you would be able to lift the object
,'#$1J4I
iii at a constant velocity.
Do you understand why its state of motion
would be constant velocity?
,
o~w--cf
V·\I\.
h t'J Ivl~
-­
What would be-e-tltll'nie~~=':@dt~'s~s;ttaa-1t:ee-EoH=-f-Rm=Ho:~tion if your applied force> its we~~t ~Fgr!"itY}? Well, I put my applied force
to 1716 N and guess what happened to its state of motion? _ _ _ _......,f'A!U~_""'~_=___=___,_ _ _ _ _ _ _ __
~"pplied cree
Ii
I lifted the object and accelerated it as forces are now unbalanced. My velocity is
increasing.
716,00
t4
Wi,..1nl'o.N
Part J: Time to
play around a bit with the simulation. Try different objects, different coefficients of friction,
different applied forces, and different angles to come up with your own scenarios.
The Ramp simulation, p. 7, 12/23/2009