Explains interaction of forces in non accelerating systems **More good stuff available at: www.wsautter.com and http://www.youtube.com/results?search_query=wnsautter&aq=f

1. Copyright Sautter 2015

2. 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. 2

3. Σ
Counterclockwise
Torque
Fcc x rcc
Clockwise
Torque
Fc x rc
rcc rc
Fcc
Fcfulcrum
Small mass
Large perpendicular distance
Large mass
Small perpendicular distance
Σ
3

4. 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.
4

5. Σ
100 lbs
W =100 lbs
Rope tension
pull
Θ1
Θ2
FREE BODY DIAGRAMSYSTEM
Point of
Force
Interaction
5

6. 100 lbs
W =100 lbs
Rope tension
pull
Θ1 Θ2
PXTX
PY
TY
ΣFX = 0
TX = PX
ΣFY = 0
TY + PY = W
Σ
6

7. Σ
W =100 lbs
pull
Θ1 Θ2
PXTX
PY
TY
ΣFX = 0
TX = PX
T Cos Θ1 = P Cos Θ2
T = P Cos Θ2 / Cos Θ1
ΣFY = 0
TY + PY = W
T Sin Θ1 + P Sin Θ2 = W
TX = T Cos Θ1
TY = T Sin Θ2
Px = P Cos Θ2
PY = P Sin Θ2
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
rope tension`
VERTICAL FORCES
HORIZONTAL FORCES
7

8. 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
100 lbs
200 200
Free Body Diagram
Σ FX = 0
TX1 = TX2
T1 Cos Θ1 = T2 Cos Θ2
Θ1 = Θ2
T1 = T2
200 200
100 lbs
TX1
TX2
T1 T2
Σ FY = 0
TY + PY = W
T Sin Θ1 + P Sin Θ2 = W
By substitution
T Sin Θ+ T Sin Θ = W
2 T Sin Θ 2 = W
T =100 / 2 Sin 200
T = 146 nt
8

9. 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.
ceiling
100 lbs
300 450
300 450
100 lbs
T1
T2
Free Body Diagram
Σ FX = 0
TX1 = TX2
T1 Cos 300 = T2 Cos 450
T1 = T2 Cos 450 / Cos 300
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
T1 = T2 Cos 450 / Cos 300
T1 = 89.7(.0.707 / 0.866)
T1 = 73.2 lbs 9

10. 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.
10

11. Problems in Statics - (weightless struts)
100 lbs
tension
Push of strut
weight
Θ=450
r
Clockwise
torque
Counter
Clockwise
torque
pivot
τc = l x w
Length of strut ( l )
τcc = r x T Σ τ= 0
τcc = τc
r x T = l x w
Sin Θ = r / l
r = l Sin Θ
l Sin Θ T = l x w
canceling l gives
T = (l x w) / l Sin Θ
T = w / Sin Θ
T = 100 lb / sin 450 = 141 lbs
T P
W
Σ FX = 0
Σ T cos Θ = P
141 cos 450 = P
P = 99.7 lbs
Free Body Diagram
of Forces acting
on body
Analyzing
Torques in
Rigid Bodies
r = torque arm
of cable tension
11

12. 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).
12

13. Problems in Statics - (struts of uniform mass)
100 lbs
tension
Θ=450
r
Clockwise
torque
of strut
τs = wstrut x l / 2
Length of strut = l
Weight of strut
at Center of Mass
Clockwise
torque
of weight
τw = whanging x l
Counter
Clockwise
torque
of cable
Σ τ = 0
τcc = τc
τcc = sτ+ τw
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
τcc = r x T
pivot
T= ((50/2) + 100) / sin 450
T = 125 / .707 = 177 lbs
weight of strut = 50 lbs
Center of Mass at
mid point
r = torque arm
of cable tension
Continued on next slide 13

14. FREE BODY DIAGRAM FROM THE PROBLEM
ON THE PREVIOUS SLIDE
Tension in cable = 177 lbs
Θ=450
Weight = 100 lbs
Push of strut
opposite pull
of cable
lift
of
cable
Σ FY = 0
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 lbsForces on hinge
125 lbs
25 lbs
Vector
sum Force net = (25 2 + 125 2 ) 1/2 = 127 lbs
Θ = tan-1 (25 / 125) = 11.3 0 14

15. 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. 15

16. 16
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