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Copyright Sautter 2003
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Statics
• When objects are subjected to forces and the net force does
not equal zero translational (place to place) motion occurs.
• When a rigid body is subjected to a net force which is not
zero, translation occurs. In addition, rotational motion may
occur and even when the net force does equal zero!
• Rotational motion is the result of applied torques often times
called moments. Torques means the “twist” experienced by a
body. This twisting is the result of an applied force multiplied
by the perpendicular distance to pivot point or center of
rotation. This pivot point is often referred to as a fulcrum in
some applications.
• The following slide shows torques as applied to a seesaw
device. When the sum of the torques equals zero (clockwise
torques equals counterclockwise torques) the angular
acceleration equals zero. If the system is not initially rotating
then no rotation occurs when equal torques act.
Σ
Σ
Large mass
Small perpendicular distance
rcc
fulcrum
Fcc
Counterclockwise
Torque
Fcc x rcc
Small mass
Large perpendicular distance
rc
Fc
Clockwise
Torque
Fc x rc
Analyzing Force Systems
• Concurrent refers to the fact that all the forces are acting at a
point. In order to analyze a system of concurrent forces we
must first draw a “Free Body Diagram” for the system.
• A Free Body Diagram selects a point of force interaction and
then shows each of the individual forces acting at that point
using vector arrows.
• Each force is then resolved into its horizontal (x) and vertical
(y) components using cosine and sine functions.
• In order for the point to be in translational equilibrium the
sum of forces acting on the point must equal zero, that is all x
forces must add to zero and all y forces must add to zero.
• When a rigid body is contained in the system, torques must
also be considered. The sum of torques around any point on
the body must be zero to ensure rotational equilibrium.
Σ
Rope tension
pull
Point of
Force
Interaction
Θ1
Θ2
100 lbs
W =100 lbs
SYSTEM
FREE BODY DIAGRAM
Σ
Rope tension
pull
TY
Θ1
Θ2
TX
PX
PY
100 lbs
W =100 lbs
ΣFX = 0
TX = PX
ΣFY = 0
TY + PY = W
Σ
rope tension`
pull
TY
Θ1
TX
Θ2
PY
PX
W =100 lbs
By substitution
P( Cos Θ2 / Cos Θ1) Sin Θ1 + P Sin Θ2 = W
With given angles and weight pull P can now
be found. With the value of P found, tension T
can also be found
TX = T Cos Θ1
TY = T Sin Θ2
Px = P Cos Θ2
PY = P Sin Θ2
HORIZONTAL FORCES
ΣFX = 0
TX = PX
T Cos Θ1 = P Cos Θ2
T = P Cos Θ2 / Cos Θ1
VERTICAL FORCES
ΣFY = 0
TY + PY = W
T Sin Θ1 + P Sin Θ2 = W
Problems in Statics
A 100 lb weight is suspended from the center of a wire which make
an angle of 200 with the ceiling. Find the tension in the wire.
ceiling
200
200
Free Body Diagram
T1
T2
200
200
100 lbs
TX1
100 lbs
TX2
Σ FX = 0
By substitution
T Sin Θ + T Sin Θ = W
2 T Sin Θ 2 = W
T =100 / 2 Sin 200
T = 146 nt
TX1 = TX2
T1 Cos Θ1 = T2 Cos Θ2
Θ1 = Θ 2
T1 = T2
Σ FY = 0
TY + PY = W
T Sin Θ1 + P Sin Θ2 = W
Problems in Statics
A 100 lb weight hangs from the ceiling by two wires making angles of 30 and 45
degrees respectively with the ceiling. Find the tension in each of the wires.
Free Body Diagram
ceiling
300
450
T1
300
100 lbs
Σ FY = 0
TY1 + TY2 = W
T1 Sin Θ1 + T2 Sin Θ2 = W
By substitution
T2 (Cos 450 / Cos 300) Sin 300 + T2 Sin 450 = W
T2 (0.707 / 0.866) (0.5)+ T2 0.707 = 100
T2 = 89.7 lbs
T2
450
100 lbs
Σ FX = 0
TX1 = TX2
T1 Cos 300 = T2 Cos 450
T1 = T2 Cos 450 / Cos 300
T1 = T2 Cos 450 / Cos 300
T1 = 89.7(.0.707 / 0.866)
T1 = 73.2 lbs
Statics Problems with Rigid Supports
• A stick or rigid support is often called a strut or
sometimes a boom. These struts used in statics problems
are of two types, weighted and weightless.
• When weightless struts are used, any torques created by
the weight of the strut itself are neglected.
• When weighted struts are used, the torque created by the
strut’s weight must be considered. Since the strut can be
considered as a rectangular solid, its center of mass is
located at the geometric center (midpoint or at one half its
length)
• The following problem is solved on the next slide:
• A weightless strut of length l, supports a 100 pound
weight at its end and is held perpendicular to the wall by
a cable which makes an angle of 45 degrees with the strut.
The cable is attached to the wall above the strut. Find the
tension in the cable.
Problems in Statics - (weightless struts)
τcc = r x T
tension
Counter
Clockwise
r
torque
Θ=450
pivot
Analyzing
Torques in
Rigid Bodies
r = torque arm
of cable tension
Length of strut ( l )
Push of strut
Free Body Diagram
of Forces acting
on body
τ =τ
cc
c
rxT=lxw
Sin Θ = r / l
r = l Sin Θ
l Sin Θ T = l x w
canceling l gives
100 lbs
T = (l x w) / l Sin Θ
T = w / Sin Θ
weight
T = 100 lb / sin 450 = 141 lbs
Clockwise
torque
τc = l x w
Σ τ= 0
T
P
W
Σ FX = 0
Σ T cos Θ = P
141 cos 450 = P
P = 99.7 lbs
Statics Problems with Rigid Supports
• The following problem involving a weighted strut
is solved on the next slide:
• A strut of uniform density and of length l
weighing 50 pounds, supports a 100 pound
weight at its end and is held perpendicular to the
wall by a cable which makes an angle of 45
degrees with the strut. The cable is attached to
the wall above the strut. Find the tension in the
cable, the compression (push) of the strut and the
forces acting on the hinge (pivot point).
Problems in Statics - (struts of uniform mass)
τcc = r x T
Στ=0
weight of strut = 50 lbs
Center of Mass at
mid point
Counter
tension
Clockwise
torque
of cable r
cc
r = torque arm
of cable tension
Θ=450
Length of strut = l
pivot
Weight of strut
at Center of Mass
Clockwise
torque
of strut
τs = w
strut
xl/2
100 lbs
hanging
c
cc
r x T = (ws x l / 2) + (wh x l)
r = l sin Θ
l sin Θ x T = (ws x l / 2) + (wh x l)
canceling l gives
T = ((ws / 2) + (wh)) / sin Θ
T= ((50/2) + 100) / sin 450
T = 125 / .707 = 177 lbs
Clockwise
torque
of weight
τw = w
τ =τ
τ = sτ + τw
xl
Continued on next slide
FREE BODY DIAGRAM FROM THE PROBLEM
ON THE PREVIOUS SLIDE
Tension in cable = 177 lbs
lift
of
cable
Θ=450
Σ FY = 0
Push of strut
opposite pull
of cable
Weight = 100 lbs
Forces on hinge
Vector
sum
125 lbs
25 lbs
push of strut = pull of cable
pull of cable = T cos 450
Push of strut = 177 x 0.707
P = 125 lbs
The up pull is that of the cable
and the hinge (pivot)
They must balance the weight
of 100 lbs.
Pull of cable = T sin 450 = 177 (0.707)
Pull of cable = 125 lbs
Upward force of hinge = 25 lbs
Force net = (25 2 + 125 2 ) 1/2 = 127 lbs
Θ = tan-1 (25 / 125) = 11.3 0
Forces & Torques
( Translational & Rotational Equilibrium )
• When the sum of forces applied to a body equals zero the body
experiences no acceleration. If it is at rest initially, it will remain
at rest. If it is in motion, its motion will continue at a constant
rate.
• When the sum of forces on a body initially at rest equals zero,
although the body will not translate (move linearly) it still may
rotate (spin) !
• To insure rotational equilibrium, the sum of torques acting on the
body must also be zero. All torques tending to rotate the body
clockwise must be balanced by all torques tending to rotate the
body in the counterclockwise direction.
• Torque is the product of applied force times the perpendicular
distance between the line of action of the force (the straight
direction of the force) and the pivot point (point around which
rotation occurs). When applying torques to a body, not only the
force but the location of the the force must be considered.
τA
τB
upward
push of A
upward
push of B
center of mass
10 cm
Arbitrary
Reference
point
50 cm
A
Force Equilibrium
90 cm
B
τwt
ΣF=0
FA + FB = weight of plank
weight
of
plank
Torque Equilibrium
Στ = 0, cτc = τc
τA + τB = τwt
TORQUES & FORCES ON A RIGID BODY
A 1.0 meter plank of mass 10 kg is supported at each end. The supports
are placed 10 cm from one end and 10 cm from the other. A 60 kg man
stands 30 cm from one end, Find force on each support.
τA
τB
upward
push of B
upward
push of A
10 cm
Arbitrary
Reference
point
30 cm 50 cm
A
weight
Wt of man of
plank
τman
τplank
Continued on next slide
90 cm
B
Force Equilibrium
ΣF=0
FA + FB = wt of plank + wt man
FA + FB = (10 x 9.8) + (60 x 9.8)
FA + FB = 686 nt
TORQUES & FORCES ON A RIGID BODY
(cont’d)
τA
τB
upward
push of B Torque Equilibrium
upward
push of A
10 cm
Arbitrary
Reference
point
30 cm 50 cm
A
weight
Wt of man of
plank
τman
τplank
90 cm
B
Στ = 0, cτc = τc
τA + τB = τp+τman
Wt man = 588 nt, Wt plank = 98 nt
(10 x FA) + (90 x FB) = (50 x 98) + (30 x 588)
10 FA+ 90 FB= 22540
From previous slide FA + FB = 686
FA = 686 - FB
10 (686 - FB )+ 90 FB= 22540
FB = 15680 / 80 = 196 nt
FB = 686 – 196 = 490 nt
Center of Mass for Complex Bodies
• The center of mass of a body is the point where all the mass
could be concentrated and give the same effect as the actual
distribution of mass within the body.
• The center of mass of a regularly shaped, symmetrical body
(such as a sphere, a cylinder, a rectangular solid, etc.) of
uniform density, is located at the geometric center of the
object.
• For bodies consisting of several connected points of mass
the center of mass can be found by adding the torques
caused by each of the masses using a selected reference
point to measure the torque arms and then dividing that
sum the the sum of the masses.
• The reference used may be any point on or even outside the
body but all measurements of the torque arms (sometimes
called moment arms) must be made from this point.
The Center of Mass of a body is a point where all the mass of the
object could be concentrated and give the same results as the
actual distribution of mass within the body. Center of Mass is
a simplifying technique used in problem solving.
The Center of Mass of a symmetrically shaped object of uniform
density is located at its geometric center.
Finding the Center of Mass of a Body consisting of several
Individual connected points of mass.
m1
Arbitrary
reference
point
x1
m2
x2
m3
x3
Reference
point
Find the center of mass of the object shown.
It is of uniform density.
Center of mass
segment #1 ( 1,2 )
2 ft
6 ft
Center of mass
segment #2 ( 2,5 )
2 ft
xcm = ((m x 1 ) + (m x 2)) / (m + m)
Xcm = 3m / 2m = 1.5 ft
4 ft
Center of mass
Of object (1.5, 3.5)
ycm = ((m x 2 ) + (m x 5)) / (m + m)
ycm = 7m / 2m = 3.5 ft
Torques & Non Perpendicular Forces
• Since torque is defined in terms of a perpendicular
torque arm (distance from the line of action of the force
to the center of rotation), when forces are applied at
angles other than 900, only the perpendicular
component of the force must be used to calculate the
torque!
• To determine the components of a force we use sine and
cosine functions as described in previous programs on
forces and vectors.
• Forces applied to a body whose line of action pass
through the center of rotation have no perpendicular
distance between the pivot point and the direction of
the force and therefore cause no torque since any force
value times zero gives zero.
τ
0x1
=
2
cos
30
5
C
Torque = force x perpendicular distance from pivot
τ
F5 = 2 lbs
1
300
F2 = 5 lbs
F1 = 2 lbs
1 ft
= 0x2
1 ft
2 cos 300
2
450
1 ft
Perpendicular
component of
F5 is used to
calculate torque
τ
c
= 0x5
Line of action
passes through
pivot for F1 & F2
Torque arm = 0
c = center of rotation
( pivot point )
1 ft
F3 = 1 lb
τ
C4
=1 x 3
F4 = 3 lbs
τ
C3
=1x1
τ
Push
of
C wall
wall
½ Length of ladder
Moment arm
rwall
Wt of ladder
Wt of man
Angle = Θ
Moment arm
rladder
τ
CC ladder
Push
of
floor
Moment arm
rman
Force
of
friction
Selected pivot point
τ
CC man
LADDER PROBLEMS
A uniform ladder rests against a wall. It weighs 50 lbs and
makes a 300 angle with the floor. A 200 lb climbs ¼ of the way
up the ladder. What force of friction with the floor is required ?
Push
of
τ
C wall
L = length of ladder
rwall = L sin 300
rladder = ½ L cos 300
rman ¼ L cos 300
= Pwall x rwall
wall
½ Length of ladder
Σ τ= 0, τcc = τc
rwall
Wt of ladder
rladder
CC ladder=
τ
+
CC man
=
τ
C wall
Continued on next slide
floor pivot point
rman
wtladder x rladder
CC ladder
friction
Θ
τ
τ
Wt of man
τ
CC man
= wt man x rman
LADDER PROBLEMS (cont’d)
wtladderx ½ L cos 300 + wt man x ¼ L cos 300 = Pwall x L sin 300
The only horizontal forces acting on the system are the
push of the wall and the force of friction. They must be
equal and opposite ! Ffriction = Pwall
Pwall = (wtladderx ½ L cos 300 + wt man x ¼ L cos 300 ) / L sin 300
Pwall = ((50 x ½ x 0.866) + (200 x ¼ x 0.866)) / 0.500
Pwall = Ffriction = 130 nt
Up ward force on the floor = wt of man + wt of ladder
Ffloor = 200 + 50 = 250 lbs
130 nt
Θ
Force on
Foot of ladder
250 lbs
Fladder = (( 1302) + (2502))1/2 = 282 nt
Θ = tan-1 (130 / 250 ) = 27.50
A picture hangs from a wire which makes a 300 angle with the
Vertical. The picture weighs 8.0 nt. Find the tension in the wire ?
(A) 4.0 nt (B) 4.6 nt (C) 8.0 nt (D) 9.2 nt
Click
here for
answers
A 100 kg box is hung from two ropes which make angles of 20 and 40 degrees
respectively. Find the tension in the 200 rope.
(A) 57 nt (B) 727 nt (C) 74.2 nt (D) 39.5 nt
A 30 lb weight is hung from the end of a 6 foot beam which weighs 20 lbs.
Where is the balance point as measured from the 30 lb weight ?
(A) 2.4 ft (B) 2.6 ft (C) 0.8 ft (D) 3.5 ft
A ladder weighs 240 nt and rests against a wall at a 600 angle. A 600 nt
man stands ¾ of the distance from the top of the ladder. What is the force
of the ladder against the wall ?
(A) 240 nt (B) 840 nt (C) 440 nt (D) 606 nt
A pipe 2.0 meters long has another similar pipe 1.0 meter long connected to
its end to form a T. Find the center of gravity from the T intersection .
(A) 10 cm (B) 67 cm (C) 16 cm (D) 84 cm