Forces and Mass
Force
:
a
force
is an interaction that causes an affected object to
be pushed or pulled in a certain direction.
we can
not see
the force but we can
see the
effect
of force
It can also change the shape or size of a body on which it
acts
Units of force:
Newtons (N)
1000 N = 1kilonewtons (kN)
1000 000 N = 1 meganewton (MN)
Mass
– is the amount of matter contained in an object.
Which can be found by using an ordinary scales.
Measured in kilograms
(kg)
Weight
– mass of an object in which the force of the earth
exerts on its , this force is called the force of gravity.
[mass (kg) x 9.81 ].Measured in Newtons
(N)
Example:
A body has a mass of 100kg. What is its weight?
Solution:
Weight
= 100 x 9.81 =
981 N
Therefore we
need a force more than 981 N
in order to lift
this 100kg body mass
.
Types of force s
Basically forces are of two types:
Contact forces
Non contact forces or At a distance forces
Contact Forces
Those forces which represent the result of physical contact between two
objects, where one of the object exerts force on the other.
Examples of contact forces:
Stretching of the spring of a balance
pushing of a pram
kicking of a football
Pushing the door
Hitting the ball
Types of contact force
1.
Applied force:
Force due to the action of muscles is called applied
force. It is also called muscular force
2.
Normal force:
It is a contact interaction force between surfaces. It
always acts perpendicular to the surfaces and out of the
surface
3.
Frictional force
Frictional force comes from interactions with a surface
as an object moves or tries to move relative to the
surface
.
4.
Air resistance force
Force that acts in a direction opposite motion through a
gas is called air resistance force.
5.
Spring force
Spring force comes from displacement of molecules. It
is always opposite the displacement of spring
.
6.
Tensile forces
force applied which tends to increase the length of an object
7.
Compressive forces
force is applied which tends to shorten the length of an object
8.
Shearing forces
force is applied which causes one layer of the material tend
to slide over another layer
Non Contact Forces
Those forces which do not involve physical contact between the two
objects but act through the space between the two.
Non contact force examples
1.
Gravitational force
Attractive force exerting between two objects is called
gravitational force. It originates with the presence of
matter.
2. Electromagnetic force
The electromagnetic force which includes basic
electric and magnetic interactions and is responsible
for the binding of atoms and the structure of solids.
Weak nuclear force
Strong nuclear force
Effects of forces on Engineering materials:
Tension
- material will stretch
Amount of stretch depends upon :
the cross section area
type of metal
Size of force applied
Compression
– material shortens, caused buckling and breaking if material
is thin and long.
Bending
– bend under the action of forces applied to them. Amount
of bending depending on the size of force and the dimension
of the beam.
Notching
done to make bending easier on beams.
Twisting
– force that caused rotation.
Example:
A spanner used to rotate the nut.
Stress
– amount of force a material can withstand.
The thickness or the cross sectional area
of the material will mostly determined its
stress contend.
Ultimate stress
- maximum stress that a material can withstand
ULTIMATE STRESSES FOR METALS IN TENSION
Strain
– measured the amount of distortion an object
undergoes when force is applied.
(
have no units)
Strain = Alteration in length / Original length
Example:
A bar metal 2500 mm long has a tensile force applied to it. Under the
action of the force its length increases by 0.75 mm.
What is the strain?
Solution:
Strain = AL / OL
= 0.75
mm
/ 2500
mm
=
0.0003
Example:
A wire 3m long and having a cross-sectional area of 0.08mm
2
has a
force of 80N applied to it. Under the action of this force it stretches
to 3.09m. Calculate the
stress
in the wire, and also find the
strain.
Force as a Vector Quantity
A vector is a quantity that possessors both
size
(amount of force)
and
direction
(angle).
Force which acts on a body in a particular direction
Vertical Force
Resultant force
Horizontal force
Two forces acting on the object if the object is free to move thus
will move in the direction of the
resultant force
5N
6N
Resultant Force = 5 N + 6N =
11 N
10N
6N
Resultant Force = 10N – 6N =
4N
1.
2.
Representation of Forces
–
representing of forces graphically
Key Factors:
Magnitude
(size)
Direction
(angle)
Point of application
Sense (away /towards) the point of application
Procedures in representing forces:
Faintly draw a horizontal line and mark point of
application.
From the point of application draw a faint line with the
direction.
Chose a suitable scale.
Show its sense.
Therefore:
Magnitude will be represented by a length
Arrow represents the sense
Angle at which the line inclined represents the
direction
Example:
A force of 15 N pushes towards point X and it acts in a direction
which is 45° to the vertical. Represent this force graphically
.
Solution:
45°
Scale : 1cm = 5N
X
3cm
Moments of force
–
force that have turning effect (rotation)
Fulcrum
– point about which the object turns
Clockwise moments
– rotates clock wise direction distance from
the fulcrum
Units of moments
depend on the units of force and the units of
distance
Anti- clockwise moments
– rotates anticlockwise direction
distance from the fulcrum
Clockwise moments = Force (N) x Distance (mm,cm,m)
Anti-clockwise moments = Force (N) X Distance (mm, cm, m)
Example 1:
A tommy bar is used to operate a box spanner. What turning moment
do the forces produce?
100mm
100mm
Solution:
Turning Moment = Clockwise Moment
= ( F X D ) + ( F X D )
= (60N X 100mm) + (60N X 100mm)
= (600Nmm) + (600Nmm)
=
12000Nmm
60N
60N
Principle of Moments
– the sum of the clockwise moments is
equal to the sum of anti-clockwise direction
PRINCIPLE OF MOMENTS
Example 2:
1. Find the force F required to balance the lever shown:
F
60mm
30mm
Solution:
“For Balance the anti-clockwise moments must equal to the clockwise”
Clockwise Moment =
Anti-Clockwise Moment
F X D =
F X D
F X 30mm =
50N X 60mm
F30mm =
3 000 Nmm
F =
3 000 N
mm
/ 30
mm
F =
100N
50N
Example 3:
Find the Force of R?
R
60N
40mm
120mm
Solution:
Clockwise Moments =
Anti-Clockwise Moments
F X D =
F X D
60N X 120 =
R X 40mm
720Nmm =
R40mm
720N
mm
/
40
mm
=
R
180N
=
R
Example 1:
In this diagram the Fulcrum (Pin) prevents it to fall to the ground.
Therefore the fulcrum pin holds the total load.
Fulcrum Pin
1000N
2000N
4m
2m
Solution:
Total Upward Force
= Total Downward Force
Fulcrum Pin
= 1000N + 2000N
Fulcrum Pin
=
3000N
Example 2:
Calculate the Force on the Spring and the force on the pivot
when a load of 200N is applied as shown.
F (spring)
R
200N
Solution:
F = force on the spring
Clockwise moment about the Pivot
=
Anticlockwise moment about the Pivot
F X D
=
F X D
F X 2cm
=
200N X (20cm +2cm)
F2cm =
4400Ncm
F
= 4400N
cm
/2
cm
F
=
2200N
20cm
2cm
Pivot
Cont.
R =
Force on the Pivot
Total Upward Force
= Total Downward Force
2200N
= 200N + R
2200N –
200N
= R
2000N
= R
Centre of Gravity –
the
centre of gravity of a body is the point at
which all the mass of the body may be assumed
to concentrated.
Balance and Unbalance
– A body will only balanced if it is supported
at its centre.
Example 1:
A uniform bar of metal 4 meters long and weighing 180 newtons
is supported at 1 metre from one end. Find the force needed to balance
the
bar
if this force is applied at 0.5 metre from the same end as shown
.
Solution:
180N will be concentrated on the centre
2m
1m
0.5m
1m
180N
1m
0.5m
F
F
Cont.
Taking the moments at the point which it was supported.
Clockwise moment =
Anticlockwise moment
F X D =
F X D
F X 0.5m =
180N X 1m
F0.5m =
180Nm
F =
180N
m
/ 0.5
m
F =
360N
Forces, such as contact and non-contact forces, interact with objects to cause motion or deformation. Mass is the amount of matter in an object, measured in kilograms. Learn about applied force, normal force, frictional force, air resistance, spring force, tensile forces, compressive forces, and shearing forces. Discover how to calculate weight from mass and understand the units of force. Explore the effects of forces on objects and their surroundings.
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Presentation Transcript
Force Force
Force Force : a force forceis an interaction that causes an affected object to be pushed or pulled in a certain direction. we can not see not see the force but we can see the effect effect of force It can also change the shape or size of a body on which it acts see the Units of force: Units of force: Newtons (N) 1000 N = 1kilonewtons (kN) 1000 000 N = 1 meganewton (MN)
Mass Mass is the amount of matter contained in an object. Which can be found by using an ordinary scales. Measured in kilograms (kg) (kg) Weight Weight mass of an object in which the force of the earth exerts on its , this force is called the force of gravity. [mass (kg) x 9.81 ].Measured in Newtons (N) Example: Example: (N) A body has a mass of 100kg. What is its weight? Solution: Solution: Weight Weight = 100 x 9.81 = 981 N Therefore we need a force more than 981 N need a force more than 981 N in order to lift this 100kg body mass.
Types of force Types of force s s Basically Basically forces are of two types forces are of two types: : Contact forces Non contact forces or At a distance forces Contact Forces Contact Forces Those forces which represent the result of physical contact between two objects, where one of the object exerts force on the other. E Examples xamples of contact of contact forces: forces: Stretching of the spring of a balance pushing of a pram kicking of a football Pushing the door Hitting the ball
Types of contact Types of contact force force 1. 1. Applied force: Applied force: Force due to the action of muscles is called applied force. It is also called muscular force 2. Normal force: Normal force: It is a contact interaction force between surfaces. It always acts perpendicular to the surfaces and out of the surface 3. 3. Frictional force Frictional force Frictional force comes from interactions with a surface as an object moves or tries to move relative to the surface. 4. Air Air resistance resistance force force Force that acts in a direction opposite motion through a gas is called air resistance force. 5. Spring force Spring force Spring force comes from displacement of molecules. It is always opposite the displacement of spring. 2.
6. 6. Tensile forces Tensile forces force applied which tends to increase the length of an object 7. 7. Compressive forces Compressive forces force is applied which tends to shorten the length of an object 8. 8. Shearing forces Shearing forces force is applied which causes one layer of the material tend to slide over another layer
Non Non Contact Contact F Forces orces Those forces which do not involve physical contact between the two objects but act through the space between the two. Non Non contact force contact force examples examples 1. 1. Gravitational force Gravitational force Attractive force exerting between two objects is called gravitational force. It originates with the presence of matter. 2. Electromagnetic force 2. Electromagnetic force The electromagnetic force which includes basic electric and magnetic interactions and is responsible for the binding of atoms and the structure of solids. Weak nuclear force Strong nuclear force
Effects of forces on Engineering materials: Effects of forces on Engineering materials: Tension Tension - material will stretch Amount of stretch depends upon : Amount of stretch depends upon : the cross section area type of metal Size of force applied Compression Compression material shortens, caused buckling and breaking if material is thin and long. Bending Bending bend under the action of forces applied to them. Amount of bending depending on the size of force and the dimension of the beam. Notching Notching done to make bending easier on beams. Twisting Twisting force that caused rotation. Example: Example: A spanner used to rotate the nut.
Stress amount of force a material can withstand. The thickness or the cross sectional area of the material will mostly determined its stress contend. Stress (N/m ) = force carried by the bar (N) cross sectional area (m ) Ultimate stress - maximum stress that a material can withstand
ULTIMATE STRESSES FOR METALS IN TENSION ULTIMATE STRESSES FOR METALS IN TENSION METAL METAL ULTIMATE STRESS ULTIMATE STRESS (MN / m ) 110 400 (MN / m ) Cast Iron Mild Steel Aluminum Brass Copper 80 410 200
Example: Example: A short bar in compression has a cross-sectional area of 20 mm and it carries a force of 10 000 N Find the stress of the bar. Solution: Solution: Stress (N/m ) = force carried by the bar (N) cross sectional area (m ) Cross - Sectional Area = 20 mm = (20 mm / 1000 000 mm /m ) Stress Stress = 10 000 N (20 / 1000 000) m = (10 000 N X 1000 000 m ) 20 = 500 000 000 N/m = 500 000 000 N/m 1000 000 N/MN = 500 MN / m
Strain Strain measured the amount of distortion an object undergoes when force is applied.(h have no units) ave no units) Strain = Alteration in length / Original length Strain = Alteration in length / Original length Example: Example: A bar metal 2500 mm long has a tensile force applied to it. Under the A bar metal 2500 mm long has a tensile force applied to it. Under the action of the force its length increases by 0.75 mm. action of the force its length increases by 0.75 mm. What is the strain? What is the strain? Solution: Solution: Strain = AL / OL Strain = AL / OL = 0.75 = 0.75 mm mm / 2500 / 2500 mm mm = = 0.0003 0.0003
Example: Example: A wire 3m long and having a cross-sectional area of 0.08mm2 has a force of 80N applied to it. Under the action of this force it stretches to 3.09m. Calculate the stress stress in the wire, and also find the strain. strain. Solution: Solution: Stress Stress = fcb (N) = fcb (N) cross = 80 N (0.08 = (80N = (80N x x 1 000 000 m = 1 000 = 1 000 000 000 N = = 1000 MNm 1000 MNm2 2 Strain = Strain = AL AL / OL / OL = = 0.09 m /3m 0.09 m /3m = = 0.03 0.03 cross- -sectional (m sectional (m2 2) ) 0.08mm mm2 2/1 000 000 /1 000 000mm 1 000 000 m2 2) ) 0.08 000 000 N/m /m2 2 / / 1 1 000 000 mm2 2/m /m2 2) ) 0.08 000 000 N N/MN /MN
Force as a Vector Force as a Vector Q Quantity uantity A vector is a quantity that possessors both size size (amount of force) and direction direction (angle). Force which acts on a body in a particular direction Vertical Force Vertical Force Resultant force Resultant force Horizontal force Horizontal force Two forces acting on the object if the object is free to move thus will move in the direction of the resultant force resultant force
5N 6N 1. Resultant Force = 5 N + 6N = 11 N 10N 6N 2. Resultant Force = 10N 6N = 4N
Representation of Forces Representation of Forces representing of forces graphically Key Factors: Key Factors: Magnitude (size) Direction (angle) Point of application Sense (away /towards) the point of application Procedures in representing forces: Procedures in representing forces: Faintly draw a horizontal line and mark point of application. From the point of application draw a faint line with the direction. Chose a suitable scale. Show its sense.
Therefore: Therefore: Magnitude will be represented by a length Arrow represents the sense Angle at which the line inclined represents the direction Example: Example: A force of 15 N pushes towards point X and it acts in a direction which is 45 to the vertical. Represent this force graphically. . Solution: Solution: Scale : 1cm = 5N 45 X 3cm
Moments of force Moments of force force that have turning effect (rotation) Fulcrum Fulcrum point about which the object turns Clockwise moments Clockwise moments rotates clock wise direction distance from the fulcrum Clockwise moments = Force (N) x Distance ( Clockwise moments = Force (N) x Distance (mm,cm,m mm,cm,m) ) Units of moments Units of moments depend on the units of force and the units of distance Anti Anti- - clockwise moments clockwise moments rotates anticlockwise direction distance from the fulcrum Anti Anti- -clockwise moments = Force (N) X Distance (mm, cm, m) clockwise moments = Force (N) X Distance (mm, cm, m)
Example 1: A tommy bar is used to operate a box spanner. What turning moment do the forces produce? 60N 100mm 100mm 60N Solution: Turning Moment = Clockwise Moment = ( F X D ) + ( F X D ) = (60N X 100mm) + (60N X 100mm) = (600Nmm) + (600Nmm) = 12000Nmm
PRINCIPLE OF MOMENTS Principle of Moments the sum of the clockwise moments is equal to the sum of anti-clockwise direction Clockwise Direction = Anti-clockwise Direction (Force x Distance) = (Force X Distance)
Example 2: 1. Find the force F required to balance the lever shown: 30mm 60mm F 50N Solution: For Balance the anti-clockwise moments must equal to the clockwise Clockwise Moment = Anti-Clockwise Moment F X D = F X D F X 30mm = 50N X 60mm F30mm = 3 000 Nmm F = 3 000 Nmm / 30mm F = 100N
Example 3: Find the Force of R? R 60N 40mm 120mm Solution: Clockwise Moments = Anti-Clockwise Moments F X D = F X D 60N X 120 = R X 40mm 720Nmm = R40mm 720Nmm / 40mm = R 180N = R
Reaction at the Fulcrum Example 1: In this diagram the Fulcrum (Pin) prevents it to fall to the ground. Therefore the fulcrum pin holds the total load. Fulcrum Pin 4m 2m 1000N 2000N Solution: Total Upward Force = Total Downward Force Fulcrum Pin = 1000N + 2000N Fulcrum Pin = 3000N
Example 2: Calculate the Force on the Spring and the force on the pivot when a load of 200N is applied as shown. 200N F (spring) 20cm 2cm Pivot R Solution: F = force on the spring Clockwise moment about the Pivot = Anticlockwise moment about the Pivot F X D = F X D F X 2cm = 200N X (20cm +2cm) F2cm = 4400Ncm F = 4400N cm/2 cm F = 2200N
Cont. R =Force on the Pivot Total Upward Force = Total Downward Force 2200N = 200N + R 2200N 200N = R 2000N = R
Centre of Gravity Centre of Gravity thecentre of gravity of a body is the point at which all the mass of the body may be assumed to concentrated. Balance and Unbalance A body will only balanced if it is supported at its centre.
Example 1: A uniform bar of metal 4 meters long and weighing 180 newtons is supported at 1 metre from one end. Find the force needed to balance the bar if this force is applied at 0.5 metre from the same end as shown. Solution: F 180N will be concentrated on the centre 0.5m 1m 2m 1m 180N F 0.5m 1m
Cont. Taking the moments at the point which it was supported. Clockwise moment = Anticlockwise moment F X D = F X D F X 0.5m = 180N X 1m F0.5m = 180Nm F = 180Nm / 0.5m F = 360N
