dynamics problems. warm up the system below is in equilibrium. if the scale is calibrated in n, what...

Dynamics Problems

Warm UpThe system below is in equilibrium. If the scale is calibrated in N, what does it read?

5kg 5kg

Since the tension is distributed over the entire (mass-less) rope, the scale will read (9.8N/kg)(5kg) = 49N

Warm UpDetermine the Normal force in each of the following:

N w1 2N W W

0N

cosN w F

sinN F mg

m1 2

1 2sin sinN F F mg

F F

F

cosN w sinN F

F

2 1cos sinN mg F

F

W

F

m

2

1

Warm UpWrite F=ma for each scenario:

1 2 1 2cos cos sin sin

F ma

F F F F mg ma

m1 2F F

a

sin cos

F ma

ww w F a

g

2 1 2 1sin cos cos sin

F ma

mg F mg F ma

F

W

F

m

2

1

Example A pumpkin of unknown mass is suspended by a cord attached to the ceiling and

pushed away from vertical. When a 24.0 N force is applied to the pumpkin at an angle of 18.00 to horizontal, the pumpkin will remain in equilibrium when the cord makes an angle of 32.00 with the vertical.

(A) What is the tension in the cord when the pumpkin is in equilibrium ? (B) What is the mass of the pumpkin ?

Diagram Free body Diagram

1858

gF��

apF��

TF��

24.0N

(A) What is the tension in the cord when the pumpkin is in equilibrium ?

0

0

cos 58 24.0 cos 18 0

24.0 cos 18

cos 58

43.073

43.1

x x

x

T ap

F

F F

T N

NT

N

N

��

��

1858

gF��

apF��

TF��

24.0N

Since the pumpkin is in static equilibrium, we need only look at the horizontal components

(B) What is the mass of the pumpkin ?

1858

gF��

apF��

TF��

24.0N

Since the pumpkin is in static equilibrium, we need only look at the vertical components

2

0

0

sin 58 sin 18 0

sin 58 sin 18

43.073 sin 58 24.0 sin 18

9.81

4.4795

4.48

y y

y

T ap g

app

app

F

F F F

T F mg

T Fm

g

N Nms

kg

kg

��

��

Example The tension in the horizontal rope is 30N A) Determine the weight of the object

30N400

500


500

2TF

3 30TF N

1T wF F

2 cos 50TF

2 sin 50TF

A) Determine the weight of the object

500

2TF

3 30TF N

1T wF F

2 cos 50TF

2 sin 50TF The weight is in static equilibrium, so the appropriate net component forces must be zero.

Horizontal Component

2 3

2 3

2 3

32

0

0

cos 50 0

cos 50

cos 50

30

cos 50

46.672

x

T x T

T T

T T

TT

F

F F

F F

F F

FF

N

N

��

��

Vertical Component

2 1

2

2

0

0

sin 50 0

sin 50

46.672 sin 50

36

y

T y T

T w

w T

F

F F

F F

F F

N

N

��

��

The weight of the mass is 36N

Weight on a Wire A rope extends between two poles. A 80N weight hangs from it as per the A rope extends between two poles. A 80N weight hangs from it as per the

diagram.diagram. A) Determine the tension in both parts of the rope.A) Determine the tension in both parts of the rope.

100150

80N

T1T2


1015

1T 2T

1 cos 15T 2 cos 10T

1 sin 15T 2 sin 10T

GF

A) Determine the tension in both parts of the rope.

1015

1T 2T

1 cos 15T 2 cos 10T

1 sin 15T 2 sin 10T

GF

The weight is in static equilibrium, so the appropriate net component forces must be zero.

Horizontal Component

1 2

1 2

21

0

0

cos 15 cos 10 0

cos 10

cos 15

x

T x T x

F

F F

T T

TT

��

��

Vertical Component

1 2

1 2

22

2

0

0

sin 15 sin 10 0

cos 10sin 15 sin 10 80 0

cos 15

80

0.437527182.8

183

y

T y T y G

G

F

F F F

T T F

TT N

NT

N

��

��

1

182.8 cos 10

cos 15

186.4

186

NT

N

N

Example The system in the diagram is just on the verge of slipping. If a 7.0N weight is hanging from a

ring, what is the coefficient of static friction between the block and the tabletop?.

Free body Diagram

TcF��

GF��

TbF�� 30

cos 30TcF

sin 30TcF NF

��

GbF��

TbF��

fF��

Block Ring

35N

30

7.0N

The system in the diagram is just on the verge of slipping. If a 7.0N weight is hanging from a ring, what is the coefficient of static friction between the block and the tabletop?.

The weight is just in static equilibrium, so the appropriate net component forces must be zero. TcF

��

GF��

TbF�� 30

NF��

GbF��

TbF��

fF��

Block Ring

From Block:

35

Tb f

f N

f

Tbf

Tb

F F

F

mg

Fu

mg

F

N

From ring:

cos 30

sin 30

7

sin 30

Tb Tc

G Tc

Tc

F F

F F

NF

Combining:

35cos 30

357

cos 30sin 30

35

Tbf

Tc

Fu

NF

NN

N

cos 307

35 sin 30

0.35

f

N

N

Atwood’s Machine Example:

a)a) What are the tensions in the What are the tensions in the stringstring TT11 and and TT22 ??

b)b) Find the accelerationsFind the accelerations, , aa11 andand aa22, ,

of the massesof the masses..

Masses m1 = 10 kg and m2 = 20kg are attached to an ideal massless string and hung as shown around an ideal massless pulley.

Fixed Pulley

m1

m2

a1

a2

T1T2

Draw free body diagrams for each objectDraw free body diagrams for each object Applying Newton’s Second Law: Applying Newton’s Second Law:

TT11 - m - m11g = mg = m11aa1 1 (a) (a) TT22 - m - m22g = -mg = -m22aa22

=> => -T-T22 + m + m22g = +mg = +m22aa2 2 (b)(b)

ButBut TT11 = T = T22 = T = T

since pulley is idealsince pulley is ideal

andand aa11 = -a = -a22 =a =a

since the masses aresince the masses are connected by the stringconnected by the string

m2gm1g

Free Body Diagrams

T1 T2

a1 a2

--mm11g + T = mg + T = m1 1 aa (a)(a) -T + m-T + m22g = mg = m2 2 aa (b)(b)

Two equations and two unknownsTwo equations and two unknowns we can solve for both unknowns (we can solve for both unknowns (TT and and aa).).

Add (b) + (a): Add (b) + (a): g(mg(m22 – m– m1 1 ) = a(m) = a(m11+ m+ m2 2 ))

2 1

1 2

( )

( )

m ma g

m m

2

2

(20 10 )9.8

(10 20 )

3.27

kg kg ma

kg kg s

m

s

m2gm1g

Free Body Diagrams

T1 T2

a1 a2

Solve for Acceleration

--mm11g + T = mg + T = m1 1 aa (a)(a) -T + m-T + m22g = mg = m2 2 aa (b)(b)

Plug Plug aa into (b) and Solve for into (b) and Solve for TT

2 12 2

1 2

2 12 2

1 2

2 12

1 2

12

1 1\2

1 2

1 2

2

( )

( )

( )

( )

( )1( )

2

2 10 209.8

10 20

2

130.7

m mT m g m g

m m

m mT m g m g

m m

m mm g

m m

mmg

m m

mm g

m m

kg kg m

kg kg s

N

m2gm1g

Free Body Diagrams

T1 T2

a1 a2

Solve for T

m1

m2

a

a

TT

So we find:So we find:

2 1

1 2

( )

( )

m ma g

m m

g)mm(

mm2T

21

21

+=

Atwood Machine Review

Example A box with mass 18.0 kg is pulled along the floor. If the horizontal force is 95.0 N [E], what is the acceleration of the box?

Pulling a Box (Part 1)A box with mass 18.0 kg is pulled along the floor. If the horizontal force is 95.0 N [E], what is the acceleration of the box?

Free body Diagram

NF��

GF��

AF��

+y

+x

a

Pulling a Box (Part 1)A box with mass 18.0 kg is pulled along the floor. If the horizontal force is 95.0 N [E], what is the acceleration of the box?

NF��

GF��

AF��

Vertical Forces Horizontal Forces

0

0

0

y

N G

F

F F

N mg

N mg

��

��x x

A x

A

A

A

F ma

F ma

F ma

Fa

mF

m

��

��

Solving

2

95.0

18.0

5.28

AFam

N

kg

m

s

+y

+x

a

Example Two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg are pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T in the rope connecting both boxes?

Pulling a Box (Part 2)

Free body Diagram

NF��

GF��

AF��

TF�� +y

+x

a

A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T in the rope connecting both boxes?

NF��

GF��

TF��


NF��

GF��

AF��

TF��+y

+x

a


NF��

GF��

TF��

4.00 kg Box

1

1

1

x x

T

F m a

F m a

T m a

��

��

6.00 kg Box

2

2

2

x x

T A

A

F m a

F F m a

T F m a

��

Adding to eliminate T and find a

1

2

1 2

1 2

1 2

A

A

A

A

T m a

T F m a

F m a m a

F a m m

Fa

m m

Forces

+


NF��

GF��

AF��

TF��+y

+x

a

A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T at either end of the rope connecting both boxes?

NF��

GF��

TF��

1 2

2

20.0

4.00 6.00

2.00

AFam m

N

kg kg

m

s

Now for Tension

1

4.00 2.00

8.00

T m a

Nkg

kg

N

We could have used the other tension formula from Box 2 and obtained the same answer

Solve for Acceleration

Example A two boxes, one with a mass of 4.00 kg and the other with a mass of 6.00 kg is connected by a 1kg rope and they are pulled along the floor. If a horizontal force is 20.0 N [E] is applied to the 6.00 kg box, what is the acceleration of each box, and what is the tension T at either end of the rope connecting both boxes?

1.00kg


Free body Diagram

NF��

GF��

AF��

2TF�� +y

+x

a


NF��

GF��

1TF��

1.00 kg

Because the rope has mass, the two ends will experience different tensions


NF��

GF��

AF��

TF��+y

+x

a


NF��

GF��

TF��

4.00 kg Box

1

1

1 1

x x

T

F m a

F m a

T m a

��

��

6.00 kg Box

2

2

2 2

2 2

x x

T A

A

A

F m a

F F m a

T F m a

T F m a

��

Using F=ma for the system to find a

Forces

1 2

1 2

A rope

A

rope

F ma

F m m m a

Fa

m m m


NF��

GF��

AF��

TF��+y

+x

a


NF��

GF��

TF��

Now for T1Solve for Acceleration

1 2

2

20.0

4.00 1.00 6.00

1.82

A

rope

Fa

m m m

N

kg kg kg

m

s

1 1

4.00 1.82

7.27

T m a

Nkg

kg

N

Now for T2

2 2

20.0 6.00 1.82

9.09

AT F m a

NN kg

kg

N

Example A worker drags a 38.0 kg box along the floor by pulling on a rope attached to the box. The coefficient of friction between the floor and the box is us=0.450 and uk=0.410.

a) What are the force of friction and acceleration when the worker applies a horizontal force of 150N?b) What are the force of friction and acceleration when the worker applies a horizontal force of 190N?

38kg

Solution (Free Body Diagram)

a) What are the force of friction and acceleration of the worker applies a horizontal force of 150N?

38kg

GF��

The force of gravity down

NF�� The Normal force up

TF��

The applied force of tension to the right

fF��

Friction to the

left

+y

+x

Solution (Vector Components)

GF��

NF��

TF��

fF��

38kg+y

+x

0

0

0

38 9.80

372.4

y y

N G

N

N

F ma

F F

F mg

F mg

Nkg

kg

N

��

�� 0.450 372.4

168

f S NF F

N

N

Since the applied

force by the worker is only 150N, the box will not move

What are the force of friction and acceleration of the worker applies a horizontal force of 150N?

To determine if the box will move, we must find the maximum static friction and compare it to the applied force.

Solution (Vector Components)

GF��

NF��

TF��

fF��

38kg+y

+x

b) What are the force of friction and acceleration of the worker applies a horizontal force of 190N?

Since the applied force is greater than 168N from part a), we will have an acceleration in the x direction. So we will apply Newton’s 2nd Law in the horizontal direction.

2

190

190 0.410 372.4

38

0.982

x x

f T x

f T x

T f

K N

F ma

F F ma

F F ma

F Fa

mN F

mN N

kg

m

s

��

��

The acceleration of the box is 0.982 m/s2 [E]

FK=(0.410)(372.4N)=153N

Example A train of three masses is pulled along a frictionless surface. Calculate the tensions in the ropes.

8 kg 5 kg 13 kg30 N

We can find the acceleration of the train by treating the three masses as one unit.

2

30 8 5 13

1.15 [ ]

F ma

N kg kg kg a

ma left

s

Tension in rope T1

T1

T1F

1 1

1 1

230 8 1.15

20.8

A

A

F ma

F T m a

T F m a

mN kg

s

N

Tension in rope T2

T2T1

1 2 2

2 1 2

220.8 5 1.15

15.1

F ma

T T m a

T T m a

mN kg

s

N

T2

2

213 1.15

15.1

F ma

T ma

mkg

s

N

or

Example

A farmer uses his tractor to pull a sled loaded with firewood. Suppose the chain pulls the sled at and angle of 300 above the horizontal. How hard does the tractor have to pull to keep the sled moving with a constant velocity if the sled and firewood has a weight of 15,000 N and uk=0.40?

300

Solution (Free Body Diagram)Solution (Free Body Diagram)

300

GF��

The force of gravity down

NF�� The Normal force up

The applied force of tension at 300

fF��

Friction to the

left TF��

30 cos 30TF

sin 30TF

Tension broken down into components


Solution (Force Components))

300

GF��

NF��

fF��

TF��

30 cos 30TF

sin 30TF


Vertical Components

0

0

sin 30 0

sin 30

y

N G Ty

N T

N T

F

F F F

F mg F

F mg F

��

��

Horizontal Components

0

cos 30 0

cossin 30 30 0

x x

f Tx

k N

T

T

k T

F ma

F F

g

F

Fm Fu

F

��

��

We will need FN, so solve for FN

Since we have a constant velocity, acceleration is 0

+y

+x

Solution (Force Components)

300


Solve for FT

sin 30 cos 30 0

sin 30 cos 30 0

sin 30 cos 30

sin 30 cos 30

sin 30 cos 30

k T T

k k T T

k T T k

T k k

kT

k

u mg F F

u mg u F F

u F F u mg

F u u mg

u mgF

u

0.40 15000

0.40 sin 30 cos 30

5628.384

5600

T

NF

N

N

+y

+x

GF��

NF��

fF��

TF��

30 cos 30TF

sin 30TF

Example A group of children toboggan down a hill with a 300 slope. Given that the coefficient of kinetic friction is 0.10, calculate their acceleration and the speed they will obtain after 6.0s.


A group of children toboggan down a hill with a 300 slope. Given that the coefficient of kinetic friction is 0.10, calculate their acceleration and the speed they will obtain after 6.0s.

+x

+y

Choose axis orientation to match the direction of motion and the normal to the surface

Object

GF��

GF��

Force of gravity is straight down

NF��

NF��

Normal is perpendicular to

the surface

fF��

fF��

Force of friction opposes direction

of motion

30 cos 30GF

sin 30GF

30 cos 30GF

sin 30GF

Decompose gravity into axis

components

Solution (Force Vectors)


+x

+y

GF��

NF��

fF��

30 cos 30GF

sin 30GF

y direction

0

0

cos 30 0

cos 30

cos 30

y

N Gy

N G

N G

N

F

F F

F F

F F

F mg

��

��Remember to solve for FN because we

will need it later

x direction

0

0

sin 30

cos 30 sin 30

cos 30 sin 30

x x

f Gx x

k N G x

k x

x k

F ma

F F ma

u F F ma

u mg mg ma

a u g g

+x

+y

GF��

NF��

fF��

30 cos 30GF

sin 30GF

Acceleration

0

2 2

2

2

cos 30 sin 30

0.10 9.80 cos 30 9.80 sin 30

4.05

4.1

x ka u g g

m m

s s

m

sm

s

Example 9: Solution (Force Vectors)


Speed

20 4.05 6

24.3

24

f iv v at

m ms

s sm

sm

s

Example Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient of kinetic friction between all surfaces is 0.20. Find the magnitude of the horizontal force, F, necessary to drag block B to the left at a constant speed if A and B are connected by a light, flexible cord passing around a fixed, frictionless pulley.


Block A in the diagram below weighs 1.50 N and block B weighs 4.30 N. The coefficient of kinetic friction between all surfaces is 0.20. Find the magnitude of the horizontal force, F, necessary to drag block B to the left at a constant speed if A and B are connected by a light, flexible cord passing around a fixed, frictionless pulley.

Box B

Object B

Force of Gravity from A and B

GF�� GF

��

NF�� NF

��

Normal Force

AF��

Applied Force

AF��

TF�� TF

�� ( )f tableF��

( )f tableF��

Friction from Table

Tension

( )f box AF�� ( )f box AF

��

Friction from A

+y

+xa



Box A

GF��

NF��

AF��

TF�� ( )f tableF

�� ( )f box AF��

Object

AGF��

AGF��

Force of Gravity from A only

NF�� NF

��Normal Force

TF��

TF��

Tension

fF��

fF��Friction

+y

+x

a

+y

+x

a



GF��

NF��

AF��


�� ( )f box AF��

AGF��

NF��

TF��

fF��

+y

+x

a

A B

x-direction for Box A

0

0

0

0

0

x

f T

f T

k N T

k A T

T k A

F

F F

F F

F F

m g F

F m g

��

( ) ( )

( ) ( )

0

0

0

0

x

f box A f table TA

A k N A k N table T

A k A k A B T

A k A k A B T

F

F F F F

F F F F

F m g m m g F

F m g m m g F

��

x-direction for Box B

3

A k A k A B k A

k A k B

F m g m m g m g

m g u m g

This was determined using Box A

0

0

0

0

0

x

f T

f T

k N T

k A T

T k A

F

F F

F F

F

m

g

F

g F

F m

��

( ) ( )

( ) ( )

0

0

0

0

x

f box A f table TA

A k N A k N table T

A k A k A B T

A k A k A B T

F

F F F F

F F F F

F m g m m g F

F m g m Fm g

��



GF��

NF��

AF��


�� ( )f box AF��

AGF��

NF��

TF��

fF��

+y

+x

a

A B

3

3 0.20 1.50 0.20 4.30

1.76

A k A k BF m g u m g

N N

N

ExampleHow much force is needed to give the blocks an acceleration of 3.0 m/s2 if the coefficient of kinetic friction between the blocks and the floor is 0.20 ? How much force does the 1.50 kg block exert on the 2.0 kg block?

1.5kg

2.0kg

1.0kg

Example (free body diagram)How much force is needed to give the blocks an acceleration of 3.0 m/s2 if the coefficient of kinetic friction between the blocks and the floor is 0.20 ? How much force does the 1.50 kg block exert on the 2.0 kg block?

1.5kg

2.0kg

1.0kg

3 Blocks taken as a Single Unit

Object

GF��

Force of Gravity

NF��

Normal Force

AF��

Applied

fF��

Friction

a

+y

+x

1.5kg

2.0kg

1.0kg

Example (force vectors)How much force is needed to give the blocks an acceleration of 3.0 m/s2 if the coefficient of kinetic friction between the blocks and the floor is 0.20 ? How much force does the 1.50 kg block exert on the 2.0 kg block?

1.5kg

2.0kg

1.0kg

GF��

NF��

AF��

fF�� a

+y

+x

21.5 2.0 1.0 3.0 0.20 1.5 2.0 1.0 9.8

22.32

22

x x

f A

f A

k N A

A

A

F ma

F F ma

F F ma

F F ma

mg F ma

F ma mg

m Nkg kg kg kg kg kg

s kg

N

N

Example (free body diagram)How much force is needed to give the blocks an acceleration of 3.0 m/s2 if the coefficient of kinetic friction between the blocks and the floor is 0.20 ? How much force does the 1.50 kg block exert on the 2.0 kg block?

1.5kg

2.0kg

1.0kg

2.0 kg Block and 1.0 kg taken as a Single Unit

Object

2.0 1.0GF

��Force of Gravity

NF��

Normal Force

AF��

Applied

fF��

Friction

a

+y

+x

Since we are considering 2.0kg and 1.0kg block as a unit, then the Force is the push

of 1.5 kg block on the combined block

Example (force vectors)How much force is needed to give the blocks an acceleration of 3.0 m/s2 if the coefficient of kinetic friction between the blocks and the floor is 0.20 ? How much force does the 1.50 kg block exert on the 2.0 kg block?

2.0kg

1.0kga

+y

+x

2.0 1.0GF

��

NF��

1blockF��

fF��

1

1

1

1

22.0 1.0 3.0 0.20 2.0 1.0 9.8

14.88

15

x x

f block

N block

block

block

F ma

F F ma

uF F ma

umg F ma

F ma mg

m Nkg kg kg kg

s kg

N

N

��

��

A 5.00 kg object placed on a frictionless, horizontal table is connected to a string that passes over a pulley and then is fastened to a hanging 9.00 kg object.

a)Find the acceleration of the two objectsb)Find the tension in the string.

Step 1: Free body Diagram

FG=9 kg

FTFT

The easiest way to choose the signs for the forces is to logical choose what you believe will be correct direction and follow that direction from one object to the other. If your final answer is negative, it just means your initial direction choice was wrong.

+

+

Example

m1


a)Find the acceleration of the two objects

FG=9 kg

FTFT

+

+

Example (Solution)

HorizontaHorizontal

1

1T

F m a

F m a

Vertical

2

2T G

F m a

F F m a

Adding these equations will remove the force of tension, thus giving allowing us to solve for acceleration.

2

2

2

1

2

1

12

6

9

.3

9.8

88.2 14

T

T G

G

F m a

F F m a

F m a m a

mkg m m a

s

N

am

s

kg a

Since the answer is positive our initial direction choice was correct.

m1

Note: don’t just say that a force of (9.8x9) is pulling on the 5 kg mass. This will give an acceleration of 17.6, which is greater than the acceleration due to gravity of the falling 9 kg mass, and you would not have tension in the rope. Ugly!


b)Find the tension in the string.

FG

FTFT

+

+

Example (Solution)

HorizontaHorizontal

1

1T

F m a

F m a

We need only substitute the acceleration value into either the horizontal or vertical equation.

1

25

31

6.

.5

3

TF m a

mkg

s

N

Vertical

2

2T G

F m a

F F m a

m1

Example Blocks A, B, and C are placed as in the figure and connected by ropes of negligible mass. Both A and B weigh 25 N each, and the coefficient of kinetic friction between the blocks and the surface is 0.35. Block C descends with constant velocity.a)Determine the tension in the rope connecting Block A and B.b)What is the weight of Block C?c)If the rope connecting A and B were cut, what would be the acceleration of C?

ExampleBlocks A, B, and C are placed as in the figure and connected by ropes of negligible e mass. Both A and B weigh 25 N each, and the coefficient of kinetic friction between the blocks and the surface is 0.35. Block c descends with constant velocity.a)Determine the tension in the rope connecting Block A and B.b)What is the weight of Block C?c)If the rope connecting A and B were cut, what would be the acceleration of C?

Block A

Object

GF��

GF��

Gravity

NF��

NF��

Normal

TF��

TF��

Tension

Block B

GF��

GF��

Gravity

NF��

NF��

Normal

( )T CF��

( )T CF��

( )T AF��

Tension from C

36.9

Object

( )T AF��

Tension from A

+x+yfF��

fF��

FrictionfF

��

fF��

Friction

Example (Force Vectors)Blocks A, B, and C are placed as in the figure and connected by ropes of negligible e mass. Both A and B weigh 25 N each, and the coefficient of kinetic friction between the blocks and the surface is 0.35. Block c descends with constant velocity.a)Determine the tension in the rope connecting Block A and B.b)What is the weight of Block C?c)If the rope connecting A and B were cut, what would be the acceleration of C?

GF��

NF��

TF��

fF��

GF��

NF��

( )T CF��

( )T AF��

36.9fF

��

( ) ( )

0

0

0

0

x

f A T A

k N T A

k A T A

F

F F

F F

m g F

��

0

0

sin 36.9 0

cos 36.9 sin 36.9 0

x

f B T A Gx T C

k BN B T A T C

k B BT A T C

F

F F F F

F F m g F

m g F m g F

��

�� CT CF m g

Block A Block B Block C

cos 36.9Bm g

sin 36.9Bm g


GF��

NF��

TF��

fF��

GF��

NF��

( )T CF��

( )T AF��

36.9fF

��

0k A T Am g F cos 36.9 sin 36.9 0k B BT A T Cm g F m g F CT CF m g

Block A Block B Block C

cos 36.9Bm g

sin 36.9Bm g

cos 36.9 sin 36.9 0k B BT CAm g F m mg g

Applying Block C’s equation to

Block B

0.35 25

8.75

8.8

k AT AF m g

N

N

N

Solving for the tension between block A and B

cos 36.9 sin 36.9

0.35 25 cos 36.9 8.75 25 sin 36.9

30.757

31

C k B BT Am g m g F m g

N N N

N

N

The weight of Block C


GF��

NF��

TF��

fF��

GF��

NF��

( )T CF��

( )T AF��

36.9fF

�� cos 36.9Bm g

sin 36.9Bm g

2

) 8.8

) 31

) 1.5

T A

C

a F N

b m g N

mc a

s

2

sin 36.9

sin 36.9

30.7 0.35 25 cos 36.9 25 sin 36.9

2.55 3.14

1.53

C B B Cf B

C Bf B

B C

F ma

m g F m g m m a

m g F m ga

m m

N N N

kg kg

m

s

When the rope is cut:

Example A small rubber stopper is hung from the rear view mirror in a truck. The cord makes an angle of 9.20 with the vertical as the truck is accelerating forward. Determine the magnitude of the acceleration.

9.2

Example (Free Body Diagram)A small rubber stopper is hung from the rear view mirror in a truck. The cord makes an angle of 9.20 with the vertical as the truck is accelerating forward. Determine the magnitude of the acceleration.

9.2

Object

GF��

GF��

Gravity

TF��

TF��

Tension

cos 9.2TF

sin 9.2TF

9.2

Tension broken into components

We will look at this from outside the truck (ie the ground) because we would prefer an inertial frame of reference.

Example (Force Vectors)A small rubber stopper is hung from the rear view mirror in a truck. The cord makes an angle of 9.20 with the vertical as the truck is accelerating forward. Determine the magnitude of the acceleration.

GF��

TF��

cos 9.2TF

sin 9.2TF

9.2

0

0

cos 9.2 0

cos 9.2

y

G Ty

T

T

F

F F

mg F

mgF

��

��

Vertical Forces Horizontal Forces

sin 9.2

sin 9.2

cos 9.2

sin 9.2

sin 9.2

cos 9.2

x x

T

T

x

x

F ma

F ma

am

m

m

g

F

g

2

2

tan 9.2

9.8 tan 9.2

1.6

xa g

m

s

m

s

Example Calculate the acceleration of a box that experiences a magnetic force of repulsion of 12N from a wall and a coefficient of friction of 0.40 between the wall and the box. The box is also being pushed against the wall with a force of 25N at 300 to the horizontal.

5.0 kg

30

25N

Example (Free Body Diagram)

Calculate the acceleration of an box that experiences a magnetic force of repulsion of 12N from a wall and a coefficient of friction of 0.40 between the wall and the box. The box is also being pushed against the wall with a force of 25N at 300 to the horizontal.

5.0 kg

30

25N

Object

Gravity

Magnetic

Applied

Normal

Since the gravity force down (5x9.8) is greater

than force up (25sin(30), the box slides down, so

friction is up.

Friction

25sin 30

25cos 30

Example (Vector Forces)Calculate the acceleration of an box that experiences a magnetic force of repulsion of 12N from a wall and a coefficient of friction of 0.40 between the wall and the box. The box is also being pushed against the wall with a force of 25N at 300 to the horizontal.

5.0 kg

30

25N

25sin 30

25cos 30 a

Horizontal

0

0

0

25 cos 30

x

Ax M N

Ax M N

N Ax M

N M

F

F F F

F F F

F F F

F N F

��

��

Vertical

25 sin 30

25 sin 3 25 cos 300

y y

Ay f G y

Ay f G y

A

N

y f Gy

k

Mk

F ma

F F F ma

F F F ma

F F Fa

mN mg

m

N mg

F

N

m

F

��

��

+y

+x

Example (Insert Numbers)Calculate the acceleration of an box that experiences a magnetic force of repulsion of 12N from a wall and a coefficient of friction of 0.40 between the wall and the box. The box is also being pushed against the wall with a force of 25N at 300 to the horizontal.

5.0 kg

30

25N

25sin 30

25cos 30 a

2

2

25 sin 30 25 cos 30

25 sin 30 0.4 25 cos 30 12 5 9.8

5.0

6.5

k My

N N F mga

mm

N N N kgs

kg

m

s

+y

+x

ExampleThree hundred identical objects are connected in series on a ramp. Calculate the force between objects 120 and 121 if each object has a mass of 0.50 kg and the applied force that is pulling them has a magnitude of 300N. The objects are accelerating at 0.88 m/s2, and the coefficient of kinetic friction is 0.15.

Example (Free Body Diagram)

Three hundred identical objects are connected in series on a ramp. Calculate the force between objects 120 and 121 if each object has a mass of 0.50 kg and the applied force that is pulling them has a magnitude of 300N. The objects are accelerating at 0.88 m/s2, and the coefficient of kinetic friction is 0.15.

For Masses 121 to 300

Big Mass

Gravity

GF��

NF��Normal

( )T aboveF��

Tension from above

fF��

Friction sin 10mg

cos 10mg 10

Treat objects 300 thru 121 as a single mass

of 180*0.5kg

Example (Vector Forces)Three hundred identical objects are connected in series on a ramp. Calculate the force between objects 120 and 121 if each object has a mass of 0.50 kg and the applied force that is pulling them has a magnitude of 300N. The objects are accelerating at 0.88 m/s2, and the coefficient of kinetic friction is 0.15.

For Combined Masses 121 to 300

GF��

NF��

( )T aboveF��

fF��

sin 10mg

cos 10mg 10

+y +xa

y-axis

0

0

0

cos 10 0

cos 10

y

N Gy

N Gy

N

N

F

F F

F F

F mg

F mg

��

��

x-axis

sin 10

sin 10

cos 10 sin 10

x x

f Gx T x

f Gx T

N T

T N

T

F ma

F F F ma

F F F ma

F mg F ma

F ma F mg

F ma mg mg

��

��

Example (Insert Values)Three hundred identical objects are connected in series on a ramp. Calculate the force between objects 120 and 121 if each object has a mass of 0.50 kg and the applied force that is pulling them has a magnitude of 300N. The objects are accelerating at 0.88 m/s2, and the coefficient of kinetic friction is 0.15.

For Combined Masses 1 to 120

GF��

NF��

( )T aboveF��

fF��

sin 10mg

cos 10mg 10

+y +xa

2 2 2

cos 10 sin 10

180 0.5 0.88 0.15 180 0.5 9.8 cos 10 180 0.5 9.8 sin 10

362.6477

360

TF ma mg mg

m m mkg kg kg

s s s

N

N

ExampleA person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left across a level floor when u=0.400,

a)Find the normal force on the crateb)Find the force of friction on the cratec)Find the acceleration of the crate.

300

a

Example (Free Body Diagram)A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left across a level floor when u=0.400,


300

a

Fg

FN

Ff

Fa

300

+y

+x

Example (Vector Forces)A person exerts a force of 175N at W300S on a 20.0 kg crate which slides to the left across a level floor when u=0.400,


Fg

FN

Ff

Fa

300

a

y-axis

0yF ��

0N Gy AyF F F ��

0N Gy AyF F F

sin 30 0N AF mg F

sin 30N AF mg F

220.0 9.8 175 sin 30

196 87.5

283.5

284

N

mF kg N

s

N N

N

N

Insert Values

+y

+x



Fg

FN

Ff

Fa

300

a

f k NF F��

0.400 283.5fF N��

113fF N

Friction

113.4fF N��

+y

+x



Fg

FN

Ff

Fa

300

a

x xF ma��

f Ax xF F ma ��

cos 30f A xF F ma

113.4 175 cos 30 xN N ma

38.1544xma N

38.1554

20.0x

Na

kg

21.9077x

ma

s

21.91x

ma W

s

Acceleration

+y

+x

ExampleCalculate the unknowns for each accelerated block.

18 kg

23.7m

s

0.2

a)

F 6 kg 40N

0.2

b) a

m41N

0.3

c)2

5.6m

s

F maA fF F ma

A fF ma F

218 3.7 0.2 18 9.8A

m NF kg kg

s kg

102AF N

A NF ma F

Example (Solution)Calculate the unknowns for each accelerated block.

18 kg

23.7m

s

0.2

a)

F

F maA fF F ma

A fF ma F

218 3.7 0.2 18 9.8A

m NF kg kg

s kg

102AF N

A NF ma F


F maA fF F ma

A fF Fa

m

40 0.2 6 9.8

6

NN kg

kga

kg

24.7m

as

A NF Fa

m

6 kg 40N

0.2

b) a


F maAF mg ma

Am a g F

2 2

41

5.6 0.3 9.8

Nm

m ms s

4.8m kg

AFma g

m41N

0.3

c)2

5.6m

s

ExampleDetermine the acceleration of the cart that is required to prevent Sir Isaac Newtant from falling. The coefficient of static friction between the block and Newtant is μs

ExampleDetermine the acceleration of the cart that is required to prevent Sir Isaac Newtant from falling. The coefficient of static friction between the block and Newtant is μs

If I do not fall, then the friction force, Ff, must balance my weight mg, that is Ff = mg

The only horizontal force is the Normal Force .Therefore

F=N=ma

Putting this together we obtain: f s

s

s

s

s

F N

ma

mg

g

ga

ma

a

Attached bodies on two Attached bodies on two inclined planesinclined planes

all surfaces frictionless

peg is frictionless

m1

m2

smooth peg

1 2

How will the bodies move?How will the bodies move?

From the free body diagrams for each body, and the chosen coordinate system for each block, we can apply Newton’s Second Law:

Taking “x” components:

1) T1 - m1g sin 1 = m1 a1X

2) T2 - m2g sin 2 = m2 a2X

But T1 = T2 = T

and -a1X = a2X = a

(constraints)

m1

yx

m2

x y

T1

N

m1g

1

m2g

T2

N

2

Solving the equationsSolving the equationsUsing the constraints, we get 2 eqn and 2 unks,

solve the equations.

T - m1gsin 1 = -m1 a (a)

T - m2gsin 2 = m2 a (b)

Subtracting (a) from (b) gives:

m1gsin 1 - m2gsin 2 = (m1+m2 )a

So:

am m

m mg

1 1 2 2

1 2

sin sin

Two-body dynamicsTwo-body dynamics

In which case does block In which case does block mm experience a larger acceleration? In experience a larger acceleration? In (1) (1) there is a there is a 10 kg10 kg mass hanging from a mass hanging from a rope. In (2) a hand is providing a constant downward force of rope. In (2) a hand is providing a constant downward force of 98.1 N98.1 N. In both cases the ropes and pulleys . In both cases the ropes and pulleys are massless.are massless.

(a)(a) Case (1) (b)(b) Case (2) (c)(c) same

m

10kga a

m

F = 98.1 N

Case (1) Case (2)

SolutionSolution

m

10kga

Add (a) and (b):

mWg = (m + mW)a

W

W

mm

gma

Note:

WW mm

mgmT

(a)

(b)

T = ma (a)

mWg -T = mWa (b)

For case (1) draw FBD and write FNET = ma for each block:

mW=10kg

SolutionSolution

The answer is (b) Case (2) In this case the block experiences a larger acceleratioin

m

N198a

.T = 98.1 N = ma For case (2)

m

10kga

Case (1)

kg10m

N198a

.

m

a

F = 98.1 N

Case (2)

m

N198a

.

Problem: AccelerometerProblem: Accelerometer A weight of mass A weight of mass mm is hung from the ceiling of a car with a massless is hung from the ceiling of a car with a massless

string. The car travels on a horizontal road, and has an acceleration string. The car travels on a horizontal road, and has an acceleration aa in the in the xx direction. The string makes an angle direction. The string makes an angle with respect to the with respect to the vertical (vertical (yy) axis. Solve for ) axis. Solve for in terms of in terms of aa and and gg..

a

i i

Accelerometer...Accelerometer... Draw a Draw a free body diagramfree body diagram for the mass: for the mass:

What are all of the forces acting?What are all of the forces acting?

m

TT (string tension)

mgg (gravitational force)

i i

Accelerometer...Accelerometer...

Using components Using components (recommended):(recommended):

ii: : FFX X = T= TXX = T sin = T sin = ma= ma

jj:: F FY Y = T= TY Y mg mg

= T cos = T cos mg = 0 mg = 0

TT

mgg

m

maa

jj

ii

TX

TY

Accelerometer...Accelerometer...

Using components Using components ::

ii: : T sin T sin = = mama

jj:: T cos T cos - mg = 0- mg = 0

Eliminate Eliminate T T ::mgg

m

maa

T sin = ma

T cos = mgtan

a

g

TX

TY

jj

ii

TT

Accelerometer...Accelerometer... Alternative solution using vectorsAlternative solution using vectors (elegant but not as systematic): (elegant but not as systematic):

Find the total vector force Find the total vector force FFNETNET::

TT

mgg

FFTOT

m

TT (string tension)


Accelerometer...Accelerometer... Alternative solution using vectorsAlternative solution using vectors (elegant but not as systematic): (elegant but not as systematic): Find the total vector force Find the total vector force FFNETNET::

Recall that Recall that FFNET NET = m= maa::

So So

maa

tanma a

mg g

TT

mgg

tana

g

m

TT (string tension)


Accelerometer...Accelerometer... Let’s put in some numbers:Let’s put in some numbers:

Say the car goes from Say the car goes from 00 to to 60 mph60 mph in in 10 seconds10 seconds:: 60 mph = 60 x 0.45 m/s = 27 m/s60 mph = 60 x 0.45 m/s = 27 m/s.. Acceleration Acceleration a = Δv/Δt = 2.7 m/sa = Δv/Δt = 2.7 m/s22.. So So a/g = 2.7 / 9.8 = 0.28 a/g = 2.7 / 9.8 = 0.28 ..

= arctan= arctan (a/g) = 15.6 deg(a/g) = 15.6 deg

tana

g

a

UnderstandingUnderstanding

A person standing on a horizontal floor feels two forces: the downward pull of gravity and the upward supporting force from the floor. These two forces

(A) Have equal magnitudes and form an action/reaction pair(B) Have equal magnitudes but do not form an action/reaction pair(C) Have unequal magnitudes and form an action/reaction pair(D) Have unequal magnitudes and do not form an action/reaction

pair(E) None of the aboveBecause the person is not accelerating, the net force they feel is zero. Therefore the magnitudes must be the same (opposite directions. These are not action/reaction forces because they act of the same object (the person). Action/Reaction pairs always act on different objects.

dynamics problems. warm up the system below is in equilibrium. if the scale is calibrated in n, what...

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