Transcript Chapter 5 Examples

Chapter 5
The Laws of Motion
EXAMPLES
Example 1: Action-Reaction


The force F12exerted BY
object 1 ON object 2 is equal
in magnitude and opposite in
direction to
F21 exerted BY object 2 ON
object 1

F12  F21
Example 2: Action-Reaction

The force Fhn exerted BY the hammer ON the nail is
equal in magnitude and opposite in direction to Fnh
exerted BY the nail ON the hammer
Fhn = – Fnh
Example 3: Action-Reaction

We can walk forward because
when one foot pushes
backward against the ground,
the ground pushes forward
on the foot.
– FPG ≡ FGP
Example 4: Normal Force
(a) m = 10 kg 
Weight: Fg = mg = 98.0N
The normal force is equal to the
weight!! Only this case
FN = mg = 98.0N
(b) Pushing force = 40N
FN = mg +40N = 138.0N
(c) Pulling force = 40N
FN = mg – 40N = 58.0N
Example 5: Accelerating the box
From Newton’s 2nd Law:
 F  ma
FP – mg = 100N – 98N = ma 
ma = 2N
The box accelerates upwards
because
FP > m g
Example 6: A Traffic Light at Rest
(Example 5.4 Text Book)

This is an equilibrium problem
 No movement, so a = 0

Upper cables are not strong as
the lower cable. They will break
if the tension exceeds 100N.

Will the light remain or will one
of the cables break?
Example 6: cont.

Find T3 from applying
equilibrium in the y-direction
to the light
F
y

 0  T3  Fg  0  T3  Fg  122N
Find x and y components for T1
and T2 :
T1x  T1 cos 37
T1 y  T1 sin 37
T2 x  T2 cos 53
T2 y  T2 sin 53
T1y
-T1x
T2y
T2x
Example 6: final.

Applying Newton’s 2nd Law to find the net force for each axis
particle in equilibrium:
for a
 F  T cos 37  T cos 53  0
2  F  T sin 37  T sin 53  (122 N )  0
(1)
x
y

1
1
2
In equation (1) solve for T2 in terms of T1 :
(3)

2
 cos 37 
T2  T1 
  1.33T1
 cos 53 
Substituting (3) into (2) and solving for T1 :
T1 sin 37  1.33T1 sin 53  (122 N )  0 
T1  73.4 N & T2  97.4 N

As we can see Both values are less than 100N, so the cables will not
break!!!
Example 7: Weight Loss
(Example 5.2 Text Book)
Apparent weight loss. The lady weights
65kg = 640N, the elevator descends with
a = 0.2m/s2.
What does the scale read (FN)?


From Newton’s 2nd law: ∑F = ma
FN – mg = – m a  FN = mg – m a
FN = 640N – 13N = 627N = 52kg Upwards!
FN is the force the scale exerts on the person, and
is equal and opposite to the force she exerts on
the scale.
Example 7: cont.


What does the scale read when the elevator
descends at a constant speed of 2.0m/s?
From Newton’s 2nd law: ∑F = 0
FN – mg = 0  FN = mg = 640N = 65kg
The scale reads her true mass!
NOTE: In the first case the scale reads an
“apparent mass” but her mass does not
change as a result of the acceleration: it stays
at 65 kg
Example 8: Normal Force
m = 10.0 kg  mg = 98.0N
Find: ax ≠ 0? FN?
if ay = 0
FPy = FPsin(30) = 20.0N
FPx = FPcos(30) = 34.6N
ΣFx = FPx = m ax 
ax = 34.6N/10.0kg 
ax = 3.46m/s2
ΣFy = FN + FPy – mg = m ay 
FN + FPy – mg = 0 
FN = mg – FPy= 98.0N – 20.0N
FN = 78.0N
m=10kg
FPy
FPx
Example 9: The Hockey Puck
Moving at constant velocity, with NO friction.
Which free-body diagram is correct?
(b)
Example 10: The Runaway Car
(Example 5.6 Text Book)



Replace the force of gravity with
its components:
Fgx = mgsin Fgy = mgcos
With: ay = 0 & ax ≠ 0
(A). Find ax
Using Newton’s 2nd Law:
y-Direction
ΣFy = n – mgcos = may = 0 
n = mgcos
Example 10: Final.

x-direction
ΣFx = mgsin = max  ax = gsin
Independent of m!!
(B) How long does it take the front of the car to reach the bottom?
x f  xi  vxi t  12 ax t 2 
d  12 axt 2  t 
2d
t 
ax
2d
g sin 
(C). What is the car’s speed at the bottom?
vxf 2  vxi 2  2ax d  vxf 2  2( g sin  )d 
vxf  2 gd sin 
Example 11: Two Boxes Connected by a Cord
Boxes A & B are connected by a cord (mass
neglected). Boxes are resting on a frictionless
table.
 FP = 40.0 N
 Find:
 Acceleration (a) of each box
 Tension (FT) in the cord connecting the
boxes
 There is only horizontal motion
With: aA = aB = a
 Apply Newton’s Laws for box A:
ΣFx = FP – FT = mAa (1)
 Apply Newton’s Laws for box B:
ΣFx = FT = mBa
(2)
Substituting (2) into (1):
FP – mBa = mAa  FP = (mA + mB)a 
a = FP/(mA + mB) = 1.82m/s2
Substituting a into (2) 
FT = mBa = (12.0kg)(1.82m/s2) = 21.8N

Example 12: Atwood’s Machine
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Forces acting on the objects:


Tension (same for both
objects, one string)
Gravitational force

Each object has the same
acceleration since they are
connected

Draw the free-body diagrams
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Apply Newton’s Laws

Solve for the unknown(s)
Example 12: cont.

Vary the masses and observe the values of the
tension and acceleration
 Note the acceleration is the same for both
objects
 The tension is the same on both sides of
the pulley as long as you assume a massless, frictionless pulley

Apply Newton’s 2nd Law to each
Mass.
ΣFy = T – m1g = m1 a
(1)
ΣFy = T – m2g = – m2 a (2)
Then:
T = m1g + m1 a
(3)
T = m2g – m2 a
(4)

Example 12: final.
Equating: (3) = (4) and Solving for a
m1 g + m1 a = m 2 g – m 2 a  m1 a + m 2 a = m 2 g – m1 g 
a (m1 + m2) = (m2 – m1)g 

m2  m1 
a
g (5)
m1  m2 

Substituting (5) into (3) or (4): T = m1g + m1 a (3) 
  m  m1  
T  m1 g  m1  2
g 
  m1  m2  
T  m1 g 
m1m1 g  m1m2 g  m1m2 g  m1m1 g
T

m1  m2
m1m2 g  m1m1 g

m1  m2
 m1m2
T  2
 m1  m2

g

Example 13: Two Objects and Incline Plane


Draw the free-body diagram for each object
 One cord, so tension is the same for both
objects
 Connected, so acceleration is the same for
both objects
Apply Newton’s Laws
xy plane:
ΣFx = 0 & ΣFy = m1 a 
T – m1g = m1 a 
T = m1g + m1 a (1)
 x’y’ plane:
ΣFx = m2 a & ΣFy = 0 
m2gsinθ – T = m2 a (2)
n – m2gcosθ = 0
(3)

Example 13: final.
Substituting (1) in (2) gives:
m2gsinθ – (m1g + m1 a) = m2 a  m2gsinθ – m1g – m1 a = m2 a

a (m1 + m2) = m2gsinθ – m1g 

m2 sin   m1 

a
g
 m1  m2 
Substituting a in (1): T = m1g + m1 a 
  m sin   m1  
T  m1 g  m1  2
g 
  m1  m2 

m1m1 g  m2 m1 g  m1m2 g sin   m1m1 g
T

m1  m2
T

m2 m1 g  m2 m1 g sin  m2 m1 g 1  sin  

m1  m2
m1  m2
Example 14: Pulling Against Friction

Assume: mg = 98.0N  n = 98.0 N,
s = 0.40, k = 0.30 
ƒs,max = sn = 0.40(98N) = 39N
Find Force of Friction if the force applied FA is:
A.
FA = 0  ƒs = FA = 0  ƒs = 0 Box does not move!!
B.
FA = 10N  FA < ƒs,max or (10N < 39N)
ƒs – FA = 0  ƒs = FA = 10N
The box still does not move!!

FA = 38N < ƒs,max  ƒs – FA = 0 
ƒs = FA = 38N
Force is still not quite large enough to move the box!!!
C.
FA = 40N > ƒs,max  kinetic friction.
This one will start moving the box!!!
ƒk  k n = 0.30(98N) = 29N.
The net force on the box is:
∑F = max  40N – 29N = max  11N = max 
ax = 11 kg.m/s2/10kg = 1.10 m/s2
D.
n
ƒs,k
ƒs,k
Example 14: Final.
ƒs,max = 39N
ƒk= 29N
Example 15: To Push or Pull a Sled
Similar to Quiz 5.7
Will you exerts less force if you push or pull the
girl? (θ is the same in both cases).
 Newton’s 2nd Law: ∑F = ma
x direction: ∑Fx = max  Fx – ƒs,max = max
 Pushing
y direction: ∑Fy = 0  n – mg – Fy = 0 
n = mg + Fy 
ƒs,max = μsn  ƒs,max = μs (mg + Fy )

FBD
n
Fx
ƒs,max
Pushing
Fy
Example 15: Final.
Pulling
y direction: ∑Fy = 0  n + Fy – mg = 0 
n = mg – Fy 
ƒs,max = μsn  ƒs,max = μs (mg – Fy )

FBD
FBD
NOTE: ƒs,max (Pushing) > ƒs,max (Pulling)
Friction Force would be less
if you pull than push!!!
Pulling
ƒs,max
n
Fy
Fx
Example 16: Why Does the Sled Move?
 To determine if the horse (sled) moves: consider only the
horizontal forces exerted ON the horse (sled) , then apply
2nd Newton’s Law: ΣF = m a.
 Horse:
T : tension exerted by the sled.
fhorse : reaction exerted by the Earth.
 Sled:
T : tension exerted by the horse.
fsled : friction between sled and snow.
Example 16: Final.
 Horse: If fhorse > T , the horse accelerates to the right.
 Sled: If T > fsled , the sled accelerates to the right.
 The forces that accelerates the system (horse-sled) is the
net force fhorse  fsled
 If fhorse = fsled the system will move with constant
velocity.
Example 17:
Experimental Determination of µs and µk

Tilted coordinate system:
K-Trigonometry:
Fgx= Fgsinθ = mgsinθ
Fgy= Fgcosθ = – mgcosθ
∑F = m a , ƒk,s  k,s n
ax ≠ 0 ay = 0


y direction: ∑Fy = 0 
n – mgcosθ = 0 
n = mgcosθ
(1)
x direction: ∑Fx = max 
mgsinθ – ƒk,s = max (2)
ax
Example 17: Final.


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The block is sliding down the plane,
so friction acts up the plane
This setup can be used to
experimentally determine the
coefficient of friction
From Equations (1) and (2) and the
fact that: k,s  ƒk,s /n
µ = tan θ
 For µs, use the angle where the
block just slips
 For µk, use the angle where the
block slides down at a constant
speed
Example 18: Sliding Hockey Puck
Example 5.12 (Text Book)

Draw the FBD, including the force of
kinetic friction
Given:
vxi = 20.0 m/s vxf = 0, xi = 0,
xf = 115 m
Find μk?
 y direction: (ay = 0)
∑Fy = 0 
n – mg = 0  n = mg (1)
 x direction: ∑Fx = max 
– μkn = max (2)
Example 18: Final.



Substituting (1) in (2) :
– μk(mg) = max  ax = – μk g
To the left (slowing down) & independent of the
mass!!
Replacing ax in the Equation:
vf2 = vi2 + 2ax(xf – xi) 
0 = (20.0m/s)2 + 2(– μk g)(115m) 
μk 2(9.80m/s2)(115m) = 400(m2/s2) 
μk = 400(m2/s2) / (2254m2/s2) 
μk = 0.177
Example 19: Acceleration of Two Objects
Connected with Friction
Example 5.13 (Text Book)
If: ƒk  kn, find: a
 Friction acts only on the object in contact with
another surface. Draw the FBD
Mass 1: (Block)
 y direction: ∑Fy = 0, ay = 0 
n + Fsinθ – m1g = 0 
n = m1g – Fsinθ (1)
 x direction: ∑Fx = m1a 
Fcosθ – T – ƒk = m1a 
Fcosθ – T – kn = m1a 
T = Fcosθ – kn – m1a (2)

a
FBD
Example 19:Final.
Mass 2: (Ball)




y direction: ∑Fy = m2a 
T – m2g = m2a 
T = m2g + m2a (3)
x direction: ∑Fx = 0, ax = 0
Substitute n = m1g – Fsinθ (1) into
T = Fcosθ – kn – m1a (2)
T = Fcosθ – k(m1g – Fsinθ ) – m1a (4)
Equate: (3) = (4) and solve for a:
m2 g  m2 a  F cos    k (m1 g  F sin  )  m1a 
m1a  m2 a  F cos    k m1 g   k F sin   m2 g 
m1  m2 a  F cos  k sin    k m1  m2 g 
F cosθ  μ k sinθ   μ k m1  m2 g
a
m1  m2 
FBD
a
Example 20: The Skier
If: FG = mg , ay = 0, Ffr  kFN, &
k = 0.10, then Find: ax
 FG components:
FGx= mgsin30o & FGy= – mgcos30o
 Newton’s 2nd Law
y direction: ∑Fy = 0  FN – mgcos30o = 0 
FN = mgcos30o (1)
x direction: ∑Fx = max 
mgsin30o – kFN = max (2)
Replacing (1) in (2)
mgsin30o – kFN = max (3)
 Substituting (1) into (3) and solving for ax
mgsin30o – kmgcos30o = max
ax = g(0.5) – 0.10g(0.87)  ax = 0.41g 
ax = 4.00m/s2

ax
Material for the Midterm

Material from the book to Study!!!



Objective Questions: 5-6-10
Conceptual Questions: 2-12-16
Problems: 10-12-19-20-25-28-34-46-47-48