I'm THE MAN! · Step 3: Solve for F1... F 1 = 20N cos33 o F 1 = 23.9N Step 4: Substitute value for...
Transcript of I'm THE MAN! · Step 3: Solve for F1... F 1 = 20N cos33 o F 1 = 23.9N Step 4: Substitute value for...
The stuff we have done...Worksheets: Labs:
• Newton's three Laws Research• Feeling the Force for the First Time• Newton: Gravity of Genius Video• Free the Body...Be the Body!• Forces in Equilibrium• Camelcorn Says...Keep Your Equilibrium• Yoda Feel the Force WarmUp• Keep Your Equinimity...• Friction is f = μn!• Springs Arrrrr f = μn too!• PreQuiz Review
• Friction• Hooke's Law
#1.
8N
20 N
Fy
33o
F1
Starting with the "Free Body" diagram below... First find mystery force F1, and then find Fy.
F1cos33o
Horizontal: Vertical:
ΣFx = 0 ΣFy = 0
8N
20 N
Fy
33o
F1F1sin33o
"Componentize" F1
Step 1
Step 2: Apply Newton's Second Law to both dimensions...
F1cos33o 20N = 0 F1sin33o 8N Fy = 0
F1cos33o = 20N Step 3: Solve for F1...
F1 = 20N cos33o
F1 = 23.9N Step 4: Substitute value for F1 into vertical equation and solve for Fy...
(23.9N)sin33o 8N = Fy
Fy = 5 N
EXAMPLE SOLUTION
The Camelcorn says...Keep Your Equilibrium!
#2.
3.2 lb
32 N
Fx
52o
F1
Find Mystery Forces: Fx and F1
14.24 N
32 N
Fx
52o
F1
F1sin52o
F1cos52o3.2 lb
Horizontal: Vertical:
ΣFx = 0 ΣFy = 0
Componentize
= =
F1cos52o Fx=32 N =F1sin52o 14.24 N
#3.
20 N
33o
F
Fx
Fx
20 N
Fsin33o
Fcos33o33o
F
Horizontal: Vertical:
ΣFx = 0 ΣFy = 0
Find Mystery Forces: Fx and F
= =Fcos33oFx = 20 NFsin33o =
(40N)cos30oF1cos48o
F1sin48o (40N)sin30o
Fy16N
#4. Find Mystery Forces: Fy and F1
Horizontal:
Vertical:
ΣFx = 0
ΣFy = 016N
F1 40N
=
=
(40N)cos30o F1cos48o
F1sin48o (40N)sin30o Fy 16N
18N
BONUS: Find Fx and θ
Horizontal:
Vertical:
ΣFx = 0
ΣFy = 0
5 lb
8 N
63o
θ
Fx
8 N
Fx
18Nsinθ
18Ncosθ
22.3N
22.3Ncos63o
22.3Nsin63o22.3Ncos63o 18NcosθFx
22.3Nsin63o 8 N 18Nsinθ
FEEL THE FORCE
F2F1
40o52o
46N
F2 cos40o
46N
F2 sin40oF1 sin52o
F1 cos52o
Componentize it!
Horizontal: Vertical:ΣFx = 0 ΣFy = 0= =
F2 cos40o F1 cos52o F1 sin52o F2 sin40o 46NSub
ReSub
T53N
θ24o
15N22o
20N
Ty
20N
Tx
53 sin24o
53 cos24o
Keep your equinimity during this equivical equilibrium problem of equipoise.
Find T and θ
15 sin22o
15 cos22o
Horizontal Statement of Equilibrium:
Vertical Statement of Equilibrium:
"Componentized" Free body Diagram
53 cos24oTx 15 sin22o
53 sin24o Ty 20N 15 cos22o
Trig Steps...
Hint: When unknown force and angle are together...find components of T then do trig steps at the end!
Freaky Friction f=μn1.
.5 NF=?
μ=.22
A chunk of ear wax collected from an old man with very hairy ears is pushed along the floor. What force F would keep the wax moving at a constant speed? Free Body Diagram
ƩFx = 0
ƩFy = 0
n
W
f
FF
2.13 lb
μ=?
A slug (mass = 1 slug) is dragged across the floor at a constant speed. What is the coefficient of friction μ for the surface combo of slug mucous on cement?
Free Body Diagram
ƩFx = 0
ƩFy = 0
Ff
n
wF f
n w
580 N
μ=?
40o
3. A stubborn 700 kg Camelcorn is dragged to the right by his snout at a constant speed what is μ?Free Body Diagram
ƩFx = 0
ƩFy = 0
f
580Nsin40o
580Ncos40o W
n
f
W n 580Nsin40o
580Ncos40o
7.8 N7.8 N30o
F
μ=.56
4.Free Body Diagram
ƩFx = 0
ƩFy = 0
A 7.8N jar of toe nail clippings is pushed along at a constant speed. Find the applied force F.
Oops they spilled!
Fcos30o
Fsin30o W
n
fFcos30o f
Fsin30o W n
That's OK I'll clean them up.
Springs Arrrr f=μn too!
1.
μ=.6
A 400 N booty chest is dragged along the poop deck with a spring. How much will the spring be stretched to keep it moving at a constant speed? Free Booty Diagram
ƩFx = 0 ƩFy = 0
k = 1500 N/m
x=?Fsp
f
n
W
Fsp f n W
2.
μ=?
A baby pirate (mass = .5 slug) is dragged across the poop deck at a constant speed. What is the coefficient of friction μ for the surface combo of baby booty on poop deck?
Free Booty Diagram
ƩFx = 0 ƩFy = 0
k = 25 lb/ft
x= 0.2 ft
n
W
Fsp
f
Fsp f n W
3.
μ= answer from #2
The same baby pirate (mass = .5 slug) is dragged across the poop deck at a constant speed again. What is the How much is the spring stretched this time?
Free Booty Diagram
ƩFx = 0
ƩFy = 0
k = 25 lb/ftx= ? ft
40o
n
W
Fspcosθ
Fspsinθ
fFspcosθf
Fspsinθn W
4.
μ=?40o
Free Booty Diagram for knot
k = 50 lb/ft
x= ? ft
The same baby pirate (mass =.5 slug) is suspended as shown. The weight of the booty chest is 400 N. Find x and μ.
*Hint: Draw a Free Bodyfor the place where all the ropesconnect.
ƩFx = 0 ƩFy = 0
Fsp
Tsinθ
Tcosθ
W
Fsp WTcosθ Tsinθ
T= f = μn = μWchest
T1= 45.2N, T3 = 75.1N
Componentized Free body
60cos53oT1cos37o
60sin53oT1sin37o
T3
53o37o
T1
T2
T3
Answer
53o37o 53o37o
T1
T2
T3
Free Body Diagram
445 N
#3
#4
I Feel so
naked.
*Note that I could have converted the rope tension into Newtons instead.
Step #1Free Body Diagram
Step #2ComponentizedFree Body Diagram
Step #4Convert and CalculateWJV and Wyoda in Newtonsthen add them to find Wtotal
Step #3Set up Newton's 2nd Law for Equilibrium in 2dimensions,sub in "special force" formulas for friction (Ffriction=μn) and springs(Fspring=kx) and solve for x.
* Note that I could have just as easily solved the horizontal equation for n and substituted into the vertical equation. Either way works.
Wtotal (Jesse and Yoda)
Fspring
friction
Fspringcosθ
Fspring
WJesse = 271 lb (4.45N/lb) = 1206 N
WYoda = mg = (15 kg)(9.8 m/s2) = 147 N Wtotal =WJesse+WYoda = 1353 N
ΣFx = 0 ΣFy = 0friction = Wtotal =
Step #5Sub in Wtotal and other values to calculate x.
μn =
μ
+
=
40ox = ?
k = 600 N/m
#5 If Jesse weighs 271 lbs and Yoda has a mass of 15 kg. If the coefficient of friction between the ground and Jesse's body is 0.4, how much would the spring be stretched in order to drag them along the ground at a constant speed?
μ = .4
25o
ANSWER: x = .84 m
Wtotal nnormal
friction
Fspringsinθ
Fspringsinθ
nnormal
nnormal + 0
kxcosθ Wtotal= kxsinθnnormal
Fspringcosθ
Wtotal ( kxsinθ ) kxcosθ
μWtotal
(
kxsinθμ = kxcosθ
μWtotal kxsinθμ= kxcosθ
)x= + ksinθμkcosθμWtotal
( )+ ksinθμkcosθ x=μWtotal
Solve for n and sub into friction
Distribute μ
Collect terms and factor out x
Solve for x
( )+ ksinθμkcosθ x=μWtotal
+ (.4)(600N/m)sin25o))((600N/m)cos25o x=(.4) (1353N)
x = .84m
θ