+ All Categories
Home > Documents > Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ......

Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ......

Date post: 01-Sep-2018
Category:
Upload: vudieu
View: 217 times
Download: 0 times
Share this document with a friend
84
1 Unit 4 Statics Static Equilibrium Translational Forces Torque
Transcript
Page 1: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

1

Unit 4Statics

Static Equilibrium

Translational Forces

Torque

Page 2: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

2

Dynamics vs Statics

Dynamics: is the study of forces and motion.

We study why objects move.

Statics: is the study of forces and NO motion.

We study why objects DO NOT move.

Page 3: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

3

Recall Newton’s First Law

All objects remain at rest, or continue to move at a constant velocity unless acted upon by an external unbalanced force.

Thus, an object will stay at rest if all of the forces acting on the object balance each other.

We say that the object is in a state of STATIC EQUILIBRIUM if it is at rest.

Page 4: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

4

An object moving with a constant velocity is in a state of DYNAMIC EQUILIBRIUM.

Page 5: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

5

In Static Equilibrium

We must consider:

Translation Forces

Forces that make objects move from one place to another.

Rotational Forces

Forces that make an object rotate about a point.

Page 6: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

6

Translation Forces

Activity

Draw a free body diagram.

What is the mass of the object?

Page 7: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

7

Page 8: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

8

Find the tensions in the ropes below.

0xF

Horizontal Forces Vertical Forces

60.0 kg

T2 T1

T2

T1

1 2 0x xT T

1 2sin42 sin26o oT T

1 2x xT T

0yF

1 2 0y y gT T F

1 2y y gT T F

1 2cos42 cos26o oT T mg

21

sin26

sin42

o

o

TT

22 cos42sin

cos226

sin6

42o

o

oo m

TT g

2 cos42 csin26

os26sin42

o oo

o gT m

1 sin42oT

2 sin26oT

2 cos26oT

1 cos42oT

Page 9: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

9

2

cos4sin26

sin422 cos26

oo

oo

mgT

2 sin26

sin

60 9.8

cos42 cos242

6

Nkg

o oo

o

kgT

2 424T N

1 278T N

21

sin26

sin42

o

o

TT

sin26

sin42

424 o

o

N

Page 10: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

10

OR

60.0 kg

T2 T1

T1

T2

mg

Why do the vectors connect together?

The net force is zero

We can find the tensions using the Law of Sines

sin sin sinA B Ca b c

Page 11: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

11

Find T1

60.0 kg

T2 T1

T1

T2

mg

180 42 26o o o

We need to find the missing angle first.

1

sin26 sin112

T mg

112o

1 sin112 sin26T mg

1

60.0 9.8 sin26

sin112

Nkgkg

T

1 278T N

Page 12: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

12

Find T2

60.0 kg

T2 T1

T1

T2

mg

2

sin42 sin112

T mg

2 sin112 sin42T mg

2

60.0 9.8 sin42

sin112

Nkgkg

T

2 424T N

Page 13: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

13

Find the mass of the snowflake

45.0

o

10.0 N

30.0

o

T1

T1

10.0 N

mg

We need to find the missing angle first.

180 45 30o o o 105o

sin45 sin105

10.0N mg

sin45 10.0 sin105mg N

10.0 sin105

9.8 sin45Nkg

Nm

1.39m kg

Page 14: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

14

Tension Worksheet

Page 15: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

15

Torques(Rotational Forces)

Archimedes said "Give me a lever long enough and a fulcrum on which to place it, and I shall move the world."

Page 16: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

16

Watch what happens when you don’t study Torque

Page 17: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

17

Page 18: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

18

Page 19: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

19

Page 20: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

20

Page 21: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

21

Page 22: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

22

Page 23: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

23

Page 24: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

24

Page 25: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

25

Boss man, you ain't gonna believe this!I left my hook hanging in the water while we were at lunch and a BIG fish almost pulled my rig into the water.

Page 26: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

26

Page 27: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

27

Page 28: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

28

Page 29: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

29

Page 30: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

30

Centre of Mass(Gravity Spot)

The centre of mass is a single point in a body at which its entire mass is considered to be concentrated.

An object can balance on a point only if its center of mass is directly above the point.

Alternatively, if you hang an object from a string, the object's center of mass will be directly below the string.

Page 31: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

31

It is usually located near the more massive part of an object.

The centre of mass is also called the centre of gravity, which is the point at which the force of gravity acts. The force of gravity is equal on both sides of the object’s centre of gravity.

Page 32: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

32

The center of mass is an important point on an aircraft, as it defines the amount of mass forward or behind the center of gravity that needs to be moved in order to pitch the plane up or down without applying any external forces

Page 33: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

33

Page 34: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

34

Page 35: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

35

Page 36: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

36

Location of Centre of Mass

For uniformly, regular shaped objects the centre of mass is at the geometric centre. For example the centre of mass of a ball

(sphere) is located at the centre of the ball.

For non-uniform shaped objects, the centre of mass is located at its balancepoint in a gravitational field.

Page 37: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

37

Example:Find the centre of mass of the following:

1. A uniform 10 m log?

2. A uniform 15 m extension ladder?

3. A broom?

Page 38: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

38

4. A Triangle

Page 39: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

39

6.A boomerang? Note: In the case of a boomerang the centre of mass is not located within the actual boomerang!

Page 40: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

40

A male versus A female

http://hypertextbook.com/facts/2006/centerofmass.shtml

Page 41: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

41

A person's center of mass is slightly below his/her belly button, which is nearly the geometric center of a person.

Males and females have different centers of mass Females' centers of mass are lower than those of

males. The average ratio of center of mass to height in

females is approximately 0.543 and the average ratio of center of mass to height in males is approximately 0.560.

Page 42: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

42

Where is the centre of mass of a hammer?

When a force is not directed through an object’s centre of mass, the force will cause the object to rotate.

Example: An unbalanced hammer.

Page 43: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

43

Look in text pages 233

Page 44: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

44

Page 45: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

45

Torque

Torque: The rotational force caused by a force acting at a distance from a pivot point.

Examples: Opening a pop bottle

Tightening a screw

Using brakes in a car

Opening a door

Page 46: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

46

Page 47: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

47

Equation for Torque

where

is Torque

r is the distance a Force is applied

from a pivot point (aka point of rotation)

is the component of the force that acts perpendicular to the surface.

What are the units for Torque?

F r

F

Nm

Page 48: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

48

Is Torque a vector quantity?

Yes! The direction is:

clockwise or counterclockwise

Clockwise is takento be positivein the text.

Counterclockwise is takento be positive in mathematics and inengineering.

Page 49: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

49

Examples of Torque

1.Find the torque supplied to a 75 cm wide door by the following forces applied at the edge of the door:

A) 20N

B) 30 N at 30o to the perpendicular to the door

Page 50: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

50

2.Suppose a bolt requires 65 Nm to be properly tightened. What force is required if the wrench is 15 cm long?

Page 51: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

51

Not tested on public, but useful info in real life!!

3.In the old British system (and in the US) torque is measured in ft-lbs. How many Nm is one ft-lb?

One foot is 0.305 m.

One kilogram is 2.2 pounds

Page 52: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

52

Static EquilbriumConsists of 2 Parts

First Part

Translation Forces

where is the sum of ALL of the

forces acting on an object through its centre of mass

1 2 3 ...net nF F F F F 1

n

ii

F

0

netF external

Page 53: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

53

Static EquilbriumConsists of 2 Parts

Second Part

Torques (or Moments of Force)

where is the sum of ALL of the

(or moments of force) acting on an object.

NOTE: These forces are NOT acting through the object’s centre of mass. WHY NOT?

There is NO rotation force (torque) if a force acts through an object’s centre of mass

1 2 3 ...net n 1

n

ii

0

net torques

Page 54: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

54

Static Problems Torque(we did translational forces in unit 2)

Seesaw Problems

Cantilever Problems

Crane Problems

(or Strut and cable )

Ladder Problems

Page 55: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

55

Torque QuestionsSeesaw problems:

1. Pat and Tyler are playing on a 4 m long seesaw that is supported at the centre. If Pat has a mass of 30 kg and sits at one end of the seesaw, where should Tyler (mass = 35 kg)

sit so that the seesaw balances?

Page 56: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

56

2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted at its center. A 20.0 kg child sits on one end of the seesaw.

a) Where should a person push with a force of 220 N in order to hold the seesaw level?

b) Where should a 40.0 kg child sit to balance the seesaw?

Page 57: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

57

3. Consider the diagram below :A load weighing 60 N is to be supported by a force F applied at the end of a 5.0 kg, uniform, lever as shown. What Force is necessary if the fulcrum is placed at position A?

A

1.2m

40cm

60 N

F

Page 58: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

58

4. What force is needed to balance the 10.0 kg mass if:

2.0 m

1.0 m

10.0 kg

F

A) the seesaw is massless. B) the seesaw has a mass of 42 kg

Page 59: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

59

5. In order to hang a mass of M1=30.0kg from the horizontal flat roof of a building, a plank of length 2.4 m is placed on the roof. A rock of mass M2= 15.0 kg is placed on one end. How far can the end of the plank reach over the building without tipping over?

Page 60: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

60

Cantilever Problems

1. A 1.5 x 10 3 kg car is crossing a 120 m long flat bridge which is supported at both ends. When the car is 32 m from one end, what force must each end support be able to provide?

A B

Page 61: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

61

2. A 4.0 m diving board is supported by two blocks, one at the end and the second 1.0 m from the end. A 60.0 kg person stands at the end of the board. Find the forces given to the board by the two supports if:

A)The board is a Canadian Tire special - no mass.

A B

Page 62: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

62

Page 63: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

63

2. A 4.0 m diving board is supported by two blocks, one at the end and the second 1.0 m from the end. A 60.0 kg person stands at the end of the board. Find the forces given to the board by the two supports if:

B) The mass of the board is 40.0 kg.

A B

Page 64: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

64

3. A uniform rod of mass 13 kg and length 3.0m rests on two points, one at its left end and one at the centre point. What are the contact forces on the rod at these points? Comment on the stability of this situation.

Page 65: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

65

4. Two people of unequal strength must carry a uniform beam of length L while holding it horizontal. The weaker of the two holds the beam at one end.

A)How far from the other end must the stronger person hold the beam in order to support three quarters of the weight?

A B

Page 66: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

66

B) Is there a way in which the stronger person can carry the beam at one end and still support more than one half the weight of the beam?

Page 67: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

67

5.In the diagram below, a 10.0 m uniform horizontal beam, weighing 1.00 × 102 N is supported by a rope at each end. If a 4.00 × 102 N box is positioned 2.0 m from the left end of the beam, what is the tension in each of the support ropes (T1 andT2)? (June 2005 - Public)

Page 68: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

68

Crane Problems (or Suspended load from a strut and cable)

1. Determine the tension in the cable and the compression force in the boom to support the 1.0 x 10 2 kg object. The angle between the boom and the supporting cable is 37° .

Popular PublicQuestions

Page 69: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

69

2. A crane is used to lift a uniform 24 m long pipe with a mass of 730 kg as shown in the diagram below.

What minimum tension is required by the cable to lift the end of the pipe off the ground?

Page 70: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

70

3.A traffic light hangs from a structure as shown. The uniform aluminum pole AD is 4.0 m long and weighs 5.0kg.

The weight of the traffic light is 10.0 kg.

A)Determine the tension in the horizontal, massless cable CD.

Page 71: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

71

3.A traffic light hangs from a structure as shown. The uniform aluminum pole AD is 4.0 m long and weighs 5.0kg.

The weight of the traffic light is 10.0 kg.

B)Determine the vertical and horizontal components of the force exerted by the pivot A on the aluminum pole

Page 72: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

72

4. A uniform rod of length L and mass of 4.0 kg is hinged at the left end. A 25.0 kg sign is suspended from the right end. A guy

wire is connected to the end of the rod and fastened to a wall.

A) Draw a free body diagram for the rod .

B) Determine the tension in the guy wire.

Page 73: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

73

June 2004 Public

(i) Sketch the free body diagram for the rod in the diagram above. Label all forces. (2 marks)

(ii) If the mass of the block is 5.0 kg and the rod is uniform with a mass of 0.40 kg, what is the magnitude of the tension in the wire? (3 marks)

Answer

Page 74: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

74

Page 75: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

75

Page 76: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

76

Ladder Problems

For problems involving a ladder (or other object) leaning against a wall, the wall should be considered frictionless.

These problems should be limited to three forces acting in different directions

(i.e., three different sets of components).

Page 77: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

77

1. A 8.2 kg ladder is resting against a wall such

that the angle made with the ground is 75o.

Find the force friction required by the ground to keep the ladder from moving.

Page 78: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

78

2. A 4.0 m ladder is resting against a building such that the foot of the ladder is 0.75 m away from the building. The ladder weighs 99 N. Find the force of friction required at ground level to keep the ladder from sliding if a 75 kg person stands at following locations along the ladder:

A) 1.0 m from the bottom of the ladder.

Page 79: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

79

2. A 4.0 m ladder is resting against a building such that the foot of the ladder is 0.75 m away from the building. The ladder weighs 99 N. Find the force of friction required at ground level to keep the ladder from sliding if a 75 kg person stands at following locations along the ladder:

B) at the middle of the ladder.

Page 80: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

80

2. A 4.0 m ladder is resting against a building such that the foot of the ladder is 0.75 m away from the building. The ladder weighs 99 N. Find the force of friction required at ground level to keep the ladder from sliding if a 75 kg person stands at following locations along the ladder:

C) 3.0 m from the bottom.

Page 81: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

81

2. A 4.0 m ladder is resting against a building such that the foot of the ladder is 0.75 m away from the building. The ladder weighs 99 N. Find the force of friction required at ground level to keep the ladder from sliding if a 75 kg person stands at following locations along the ladder:

D) 3.7 m from the bottom

Page 82: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

82

3. A 5.0 m long ladder leans against a wall at a point 4.0 m above the ground as shown. The ladder is uniform and has a mass of 12.0 kg. A 55 kg painter is standing 3.0 m up the ladder. Assuming the wall is frictionless (but the ground is not) determine the forces exerted on the ladder by the ground and the wall.

Page 83: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

83

Page 84: Unit 4 Statics - Centre for Distance Learning and … 4 2015.pdf · Torques (Rotational Forces) ... 2.A playground seesaw with a total length of 5.0 m and a mass of 30.0 kg is pivoted

84

http://surendranath.tripod.com/Applets/Dynamics/CM/CMApplet.html


Recommended