Home work

Thesis1. Hair tension and it’s applications

2. Frictions and their applications

3. Frictional reduction

4. The moon movements

5. Water moving by moon gravitation

6. Solar system movements

1. Net force and Newton's first law2. Newton's second law3. Newton's third law4. Frictional forces5. Gravitation6. Circular motion7. Centripetal force8. Static equilibrium and reference frames

Teflon is actually a substance called polytetrafluoroethylene (PTFE), which is considered to be the most slippery substance that exists

Review about rotation

Learning check

Part 7

Centripetal Force

Centripetal Force: Appear with CMFc= m.a=m.v2/RPerpendiculer to v

Exercise: Calculating Centripetal

accelaration for the Moon?

Giving: Moon takes 27.3 days (2.36.106s) to orbit the Earth with average distance of 3.84.108m radius of the moon = 1 737.4 kilometers

Determine the centripetal force acts to the moon?Determine the gravity from the moon’s surface?

mass of the Moon = 7.36 × 1022 kilograms

Part 8

Static equilibrium and reference frames

Static Equilibrium

0F

Static Equilibrium :The object is either at rest

Dynamic Equilibrium: The object has its center of mass moving with a constant velocity.

1-Translational Equilibrium: Equilibrium in linear motion

However for object with size this is not sufficient. One more condition is needed. What is it?

For rotational equilibrium: the net torque acting on the object about any axis must be zero.

0

For an object to be in static equilibrium, it should not have linear or angular speed.

0CMv 0

CMd

d

F

-F

Conditions for Equilibrium

What happens if there are many forces exerted on the object?

If an object is in translational and rotational static equilibrium, the net torque must be 0 about any arbitrary axis.

Because the object is not movingnot moving, no matter what the rotational axis is, there should not be a motion.

Suppose that forces acting on x-y plane, giving torque only along z-axis. What is the conditions for equilibrium in this case?

Two vector equations turn to three equations.

0F 0xF 0yF

0 0z

O

F1

F4

F3

F 2

F5

r5 O’r’

How did we solve equilibrium problems?

1. Identify all the forces and their directions and locations

2. Draw a free-body diagram with forces indicated on it

3. Write down vector force equation for each x and y component with proper signs

4. Select a rotational axis for torque calculations Selecting the axis such that the torque of one of the unknown forces become 0.

5. Write down torque equation with proper signs6. Solve the equations for unknown quantities

Learning checkA uniform 40.0 N board supports a father and daughter weighing 800 N and 350 N, respectively. If the support (or fulcrum) is under the center of gravity of the board and the father is 1.00 m from CoG, what is the magnitude of normal force n exerted on the board by the support?

F D

n

1m x

MBgMFgMFg

Determine where the child should sit to balance the system.

Help

Since there is no linear motion, this system is in its translational equilibrium

xF 0

yF gM B 0gM F gM D n

The net torque about the fulcrum by the three forces is:

Therefore to balance the system, the daughter must sit

x mgM

gM

D

F 00.1 mm 29.200.1350

800

0 gM B 00.1 gM F xgM D 0

Therefore the magnitude of the normal force

N11903508000.40n

F Dn

1m x

MBgMFgMFg

Equilibrium Test

Determine the position of the child to balance the system for different position of axis of rotation (half of X at the right side).

What is the conclusion of this exercise?

F D

n

MBgMFg MFg

1m x

x/2

Rotational axis

Solution

No matter where the rotation axis is, net effect of the torque is identical.

Since the normal force is

The net torque about the axis of rotation by all the forces is

n gMgMgM DFB 2/xgM B 0 2/00.1 xgM F 2/xn 2/xgM D

Therefore

The net torque can be rewritten 2/xgM B 2/00.1 xgM F

2/xgMgMgM DFB 2/xgM D

xgMgM DF 00.1 0

x mgM

gM

D

F 00.1 mm 29.200.1350

800

F Dn

MBgMFg MFg

1m x

x/2

Rotational axis

Mechanical Equilibrium

A uniform ladder of length l and weight mg = 50 N rests against a smooth, vertical wall. If the coefficient of static friction between the ladder and the ground is s=0.40, find the minimum angle min at which the ladder does not slip.

FBDmg

P

f

n

O

l

Result

First the translational equilibrium, using components xF Pf 0

yF nmg 0

From the rotational equilibrium O minmin sincos

2 Pl

lmg 0

min

P

mg

2tan 1 51

40

50tan 1

N

N

The maximum static friction force just before slipping is, therefore, max

sf ns NN 20504.0 P

Thus, the normal force is n mg N50

mg

P

f

n

O

Inertial Reference Frames:

• A Reference FrameReference Frame is the place you measure from.– It’s where you nail down your (x,y,z) axes!

• An Inertial Reference Frame (IRF) is one that is not accelerating or having constant velocity.– We will consider only IRFs in this course.

• Valid IRFs can have fixed velocities with respect to each other. – Remember that we can measure from different vantage

points.Inertial Reference Frame (IRF) is different from Non-

Inertial Reference Frame (NIRF) that is accelerating a0

Learning check

Choosing an IRF from these

a) a free fall

b) an UCM

c) a butterfly with 5m/s of velocity

d) a fly with 5m/s2 of acceleration

Home work

1-

2-

3-

4-A person holds a 50.0N sphere in his hand. The forearm is horizontal. The biceps muscle is attached 3.00 cm from the joint, and the sphere is 35.0cm from the joint. Find the upward force exerted by the biceps on the forearm and the downward force exerted by the upper arm on the forearm and acting at the joint. Neglect the weight of forearm.

O

FB

FU mg

d

l

5-A uniform horizontal beam with a length of 8.00m and a weight of 200N is attached to a wall by a pin connection. Its far end is supported by a cable that makes an angle of 53.0o with the horizontal. If 600N person stands 2.00m from the wall, find the tension in the cable, as well as the magnitude and direction of the force exerted by the wall on the beam.

8m

53.0o

2m

FBD

R T

53.0o

Tsin53

Tcos53

Rsin

Rcos

Result

From the rotational equilibrium

Using the translational equilibrium

First the translational equilibrium, using components

cosR

sinR

0.53cosT

NNT 2006000.53sin

7.710.53cos313

0.53sin313800tan 1

And the magnitude of R is

Rcos

0.53cos T N582

1.71cos

0.53cos313

xF cosR 0

yF sinR 0

0.53cosT

0.53sinT N600 N200

00.80.53sin T 0NT 313

00.2600 N mN 00.4200

8m

53.0o

2mFBD

R T

53.0o

Tsin53Tcos53

Rsin

Rcos

???

6-

7-

End of chapter 2Thank you very much for your attention

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