Class
Student
Date
_
_
Inclined Planes
y
~
\
\
x
..
.)
""
,,
,
\
\,
,
,
Fgy
,/
Wedges or slides in are called inclined planes in physics-speak
,/
At the top of the incline, you need to give a box some push to overcome the force of
static friction
,/
As you descend a sticky slide, the force of kinetic friction
,/
If the angle of the incline is 30°, the components of the gravity force are:
Fgx
,/
= mgcos600 = mgsin300
,/
= mgcos300
& Pyy = mgsin600
.
= Fny = mgcos300
The magnitude of the force of friction is: Ff = J.lFN = Jl.mgcos300
If no extra forces act on the x-direction (Le. a tension force along the incline) and the
motion of the block is with constant speed, then: Ff = Pyx
1.
opposes Y0':Jrm.otion
If no extra forces act on the y-direction (Le. a tension force along the incline) then the
normal force is: FN
,/
Ffk
== mgsin300
A crate weighing 562 N is resting on a plane inclined 30.0° above
the horizontal.
a)
Draw the free body diagram for the crate.
S{'I\c.e.
-t~
-r~e 0chon
Ro"c~,
c("O\-\e
foY'"ce
~eV"~to,.e
t,S
is
~t '("'est~hh.~~
F.f,S •
'\
'f...
If t~e
Ct, ~e.
Wl+4 +"'e
l~ "30°
Ll.\.cLc:"'te
o.",,~le
x- a~\'s
~o~~'joo
(>:
"'''''-d''Ce/ ~
I ~
bo"*.,.
b)
Find the components of the weight that are parallel and
perpendicular to the plane.
~
~()V-"""~
F~I'
F;
=:
~ca><
_ f:
ca..L -
c)
~'(-
--;::
-
f:~
COS
F~
f\'V\
=
600
bO°
=
562
Co~
~
60° := Z.cgl
56'2.. ~ l V\ {,oo =.
Y
N
g"1 rJ
Find the normal force acting on the block.
d) Find the force of ~tatic friction needed to make the crate
stay at rest.
_7
F~.s -::::. F~ 1\
-=
C
F~x = 2.~ IN.
,-V'£.t
((A-t
2.
0
-:::=
-(Q
S
+ ')
A 62-kg person on skis is going down a hill that is sloped at
3~. The coefficient of kinetic friction between the skis and
snow is 0.15.
F~CoS 5:/'
~x=
=: 6'2.I4>(tt.~'
~ )(,,5S3°
~
,s'~ '36(,N
~'1
'1-0\)(,~
F\I\~t~=
Q
W\o+~o"')
a) Draw the free body diagram for the skier.
(V\o
-= ~
51"
-::. b1.Ii('l.ln)
,=-'1fi~N
r "-F '"
F.f lC-:
:=.
b) What is the acceleration of the skier?
:=
'M
(/\x
'= 62 tl)c
2q~ ~ b2. ~x
Cl)( =.
~{,{,-'3
.
"
Sl""S')
FN'~ F~1 =-~
(o.'oolle)
r=~><- F~ tc-
>~\o\
2.q3
.
61..
N
0.\5 (~gbN)
~13N
)LO
--~--
(
Glx ''::.It.,.'l w".s1.] ;dow~
tl-,~ l~C((-.,('J p/~~
c) How fast is the skier going 5.0 s after starting from rest?
~
\Jf
=: Vi. + ttt
=-0 + 4.1 ~
(5,01)
:: F24
MiS)
3.
A block is being pulled with a rope along a frictionLess incline
as shown below. The magnit~de of the tension force is Fr = lSe N.
=
Ff
+~~,..e
fa, r lil
(fy~{ho~I~~r)
Q
Ff
\t\.,Q
-H•..-e
<) '"
F e,J)
a) Draw the free body diagram for the block on the incline.
b) What are the components of the force of gravity acting on the
block?
5 ~ I~d
e.
\IV
ra7~t~ve
:x - ~ )C l > ~ a l
7V~.ce
F~x -=. ,F~
~oho""
~~
coS 550
f ~'i:: F"~ Sl~550
e +~e
L'" cl ,eJ ,f'I ClI1 e
is
+~€
l.I
1>'\ ~
'lJ\l~
-
= ,5 ~
~f
(Q.81
( " . g I ~)
\."111
;1.)
c. .•••c~'1.pd
coS5~::-,
S l V\ 5~
-7
F~~t
=
-7
M.~
'
bb
N
15 \l~
FT
-
F~)(
'50 - 24
~b
-=
= Max
-= 150.x
'5~x
..a)(
=
G..~ ""'I
Sf ~.
j '\ )
l",c\,'ec.l
plt{l'1e,
&~
':: t 2., N
c) What is the acceleration of the block?
pi""" '\
~>
N
4.
A block weighing 100. Newtons is positioned on an incline that
makes an angle of 30. with the horizontal. The magnitude of the
friction force between the block and the incline is 10. Newtons.
A force of 120. Newtons is applied by pulling on a rope that
makes an angle of 30. with the incline, as shown.
0
0
a) Draw a free body diagram, and provide appropriate labels for
each of the forces.
(above)
b)" Calculate the component of the block's weight parallel to the
incline.
~
F~)c= F~ Co~
60°
- 100
Co$
60
c
=~
c) Calculate the magnitude and direction of the component of the
tension that is useful in moving the block up the incline.
d) Calculate the magn~tude and direction of the block's
acceleration ..
f"etx
-= W\~X
Fix - F-f''-- - F~"X -= ~ax
5 0 -= ~lao qx
\04 - 10 -
5.
A block of mass m slides at constant speed down a uniform plane
inclined at an angle e to the horizontal, as shown.
F'~
a) On the diagram, sketch and label arrows to represent the
frictional force, the normal force, and the force of gravity
acting on the block.
F",~tx
=0
F",~+'1:= 0
b) Express the component of the block's weight to the incline F11
and the component of the block's weight perpendicular to the
incline Fl in terms of the weight F of the block.
c)
F" -= ~)(
-==
F .1- --
-
Show that
c::
I
~1 -
F~ coS
f6t
S llll
(
=
(QO-8
)
q 0 -:8 )
=-
0
Ffl'-
J.t= tan(} •
FI\J
F" -
F-fIL
=
0
F'.f
p",
@=
I'
= FI, -= F~ $ t ~ e
-=
F..1. = F~ CO~ e
6.
(
You slide a 325-N trunk up a 20.0° inclined plane with a constant
speed by exerting a force.of 211 N parallel to the inclined
plane.
a.
Draw the free body diagram for the trunk.
~N
Q
,--
Q...,
'0
'10 - 20 -=-
component
~)( -=
of tl"'unk.Jsweight
=-
0
F~ Co> 70
~GJ)I'
== '" tJ
,0
along the plane.
325 N eo~ roo
7
Calculate the net force on the trunk.
C.
-)
---
FV\~t- = 0
d.
(
Calculate the magnitude and direction of the friction force.
l Ff ~j:= htfp
-
e.
Calculate the coefficient of friction.
Ffk-
f''- -;: F", ': F~'f':=
lOON
--305'"
~'1.S Sl"\ 10"
~ 305
f.
Ft\/
-
tV
Calculate the acceleration of the trunk.
2,(1 -
- F~x
ttl
~
[too.'"
J