Tue Sept 29
• Assign 5 – Friday
• Exam Mon Oct 5
– Morton 235
– 7:15-9:15 PM
– Email by Wednesday
afternoon at the latest if
conflict
– Q/A Sunday 1-3,4-6,7-9
Today:
• Center of Gravity
• Static Equilibrium
Sign convention for torques:
(-) CW torque
(+) CCW torque
Equilibrium
ΣF = 0 and Στ = 0
In 2-d: ΣFX = 0 and ΣFY = 0
Goal: Write expression for Στ and ΣF
Choose axis where don’t know force and don’t care
At which point and at which direction would the least amount of force
be required to hold the lever stationary?
Longest lever arm - perpendicular force
Torque – More Formal
• Pick axis. Can calculate torque due to any force.
• Torque (τ) = (mag of Force) x (perpendicular lever arm)
• τ = F r⟂
τ = F r sinθ
Axis of Rotation
τ=Fr
τ=0
Perpendicular Lever arm – line drawn through hinge and
perpendicular to line of action.
Torque – Perpendicular Lever Arm
axis
• May have to calculate using trig
• Be careful to make sure you use
right trig function
May be given to you
Perpendicular Lever arm – line drawn through hinge and
perpendicular to line of action.
Pushups – Varying Force – How to make pushups easier
(On homework
presume horizontal)
θ
Tilted Pushups:
• Torque from weight smaller
because of angle
• Arms still pushing perpendicular
• Less force required
Two Different Approaches
Find lever arm through trig
Look at Components of force
F
r⟂
F
θ
θ
d
τ = F r⟂
τ = F (d cos θ)
θ
d, r⟂
τ=F
τ = F⟂ d
τ = (F cosθ) d
F⟂
A cook holds a 2.00 kg carton of milk at arm's length. The bicep
muscle exerts a force Fb on the forearm. What is the expression for the
torque of the bicep muscle on the forearm? Use the elbow as the axis
of rotation.
(1) (F cosθ) 8cm
b
(2) (Fbsinθ) 8cm
(3) Fb 8cm
(4) -(Fbcosθ) 8cm
(5) -(Fbsinθ) 8cm
(6) -Fb 8cm
(7) None of the above
A cook holds a 2.00kg carton of milk at arm's length. What force Fb
must be exerted by the bicep muscles? (ignore the weight of the
forearm for now).
A cook holds a 2.00 kg carton of milk at arm's length. The bicep
muscle exerts a force Fb on the forearm. What direction is the force of
the elbow on the forearm? (Choose 1,3,5,7 only if exactly on axis.
3
Choose zero for no force)
2
4
5
1
6
FE
8
7
To balance torque, need to push down.
To balance forces horizontally, need to push to left.
A cook holds a 2.00 kg carton of milk at arm's length. What force Fb
must be exerted by the bicep muscles? (ignore the weight of the
forearm for now).
"Write the expression for the sum of the torques."
Write out each torque in terms of variables or quantities given.
Choose axis at elbow since don't know those forces yet.
Στ = 0
Fg(25cm+8cm) – Fbcos(75º)(8cm) + FELBOW(0cm) = 0
Fg = 2.00kg*9.8 = 19.6N
Fb = 312N
What is the magnitude of the horizontal component of the force that
the elbow exerts on the forearm?
(1) 61.2N
(2) 292N
(3) 301N
(4) 307N
(5) 312N
(6) 332N
ΣFx = 0
Fbsin(75) – FELBOW,X = 0
FELBOW,X = 312N*sin(75) = 301N
What are the magnitude and direction of the force exerted on the
forearm by the elbow?
If we can find the
components of the force,
we can reconstruct the full
force vector.
301N
θ
FELBOW,X = 301N
ΣFy = 0
-Fg + Fbcos(75) – FELBOW,Y = 0
FELBOW,Y = 312N*cos(75) – 19.6N
FELBOW,Y = 61N
61N
Center of Gravity (Center of Mass)
• If extended object, can choose one place
to apply Force due to Gravity (weight)
• This is called the Center of Gravity (CG)
Finding CG:
• If symmetrical, along line of symmetry – "real center”
• If irregular, can balance or hang
Stability
If CG over or under point(s) of support then stable.
Stable Equilibrium: displace from
equilibrium and it returns
Unstable equilibrium: displaced but does not
return to equilibrium
Stability
Unstable equilibrium
Neutral equilibrium: displaced
stable in new state
Stability - Stance
Lower CG, more stable in this situation
Hit sideways at CG, easier to knock higher CG off balance
A child is sitting on an
adult’s leg. Given the
measurements shown to the
right, what is the torque
due to the child if we use
the point shown as the axis
of rotation?
(1) +0.388 Nm
(3) + 3.80 Nm
(5) + 3.72 Nm
(7) + 37.2 Nm
axis
(2) – 0.388 Nm
(4) – 3.80 Nm
(6) – 3.72 Nm
(8) – 37.2 Nm
τ = -F*r⟂ = -mg*r⟂ = -(10 kg) (9.8 m/s2) (0.38 m)
τ = -37.2 Nm
A child is sitting on an adult’s axis
leg. Given the measurements
shown to the right, what is
the torque due to weight of
the leg if we use the point
shown as the axis of rotation?
(1) +0.8 Nm
(3) + 3.20 Nm
(5) + 7.84 Nm
(7) + 15.8 Nm
(2) – 0.8 Nm
(4) – 3.20 Nm
(6) – 7.84Nm
(8) – 15.8 Nm
Use location of CG for point of application
τ = -F*r⟂ = -mg*r⟂ = -(4 kg) (9.8 m/s2) (0.20 m)
τ = -7.84 Nm
A child is sitting on an adult’s axis
leg. What is the tension in
upper leg muscle?
Στ(about knee) = 0
-τChild – τLeg + τMuscle = 0
- 37.2 Nm – 7.84 Nm + F * 0.02m= 0
F = 45.0Nm / 0.02m
F = 2252N
A 800 N painter stands on a 400 N scaffold (a uniform rectangular
board) with one rope supporting each end. The painter stands onequarter of the length of the scaffold from the left side. Compare the
tension in the two ropes.
(1) TL > TR
(2) TL < TR
(3) TL = TR
What is the tension in each rope?
L/2 from left or
right end
Στ(about left rope) = 0
-800*(0.25L) -400*(0.5L)+TR*L = 0
Στ(about right rope) = 0
-TL*L + 800*(0.75L)+ 400*(0.5L) = 0
Draw, Draw, Draw!!!!
Look for similarities
If don’t know L, take “leap of
faith” and stick it in as a
variable. See if it cancels.
Break forces into components perpendicular and parallel to the
board, bridge, arm, tray, fishing pole, wrench, ….
Στ = 0, ΣFX = 0, ΣFY = 0
Look for similarities
35°
Στ = 0, ΣFX = 0, ΣFY = 0
A 1500N beam is attached to a wall by a hinge. The
beam is 4.0m long, and the mass of the beam is
distributed evenly along the beam. The beam is
supported by a cable that is attached 1.0m to the beam
from the wall and makes an angle of 35° from the
vertical. The goal is to find the tension in the cable.
35°
1m
4m
At what point does it make the most sense to place the axis of rotation?
(1) Hinge
(2) Point Cable attaches (3) Center of beam
(4) Right End
A 1500N beam is attached to a wall by a hinge. The
beam is 4.0m long, and the mass of the beam is
distributed evenly along the beam. The beam is
supported by a cable that is attached 1.0m to the beam
from the wall and makes an angle of 35° from the
vertical. The goal is to find the tension in the cable.
What is the lever arm for the force due to gravity?
Use the hinge as the axis of rotation.
(1) 0m
(2) 1m
(3) 2m
35°
4m
1m
Fg
(4) 3m
Center of gravity applied at center of board.
Center of board is 2m from the hinge.
(5) 4m
A 1500N beam is attached to a wall by a hinge. The
beam is 4.00m long, and the mass of the beam is
distributed evenly along the beam. The beam is
supported by a cable that is attached 1.00m to the
beam from the wall and makes an angle of 35° from
the vertical. The goal is to find the tension in the
cable.
35°
1m
4m
What is the tension in the cable?
(1) 1500N (2) 2460N (3) 3000 N (4) 3660 N (5) 5230 N (6) 6000N (7) 7320N
Στ(about hinge) = 0
T(cos35°)*(1m) -1500*(2m)+FHINGE* 0 = 0
T = (3000N*m)/(cos35°*1m)
T = 3660N
A 1500N beam is attached to a wall by a hinge. The
beam is 4.00m long, and the mass of the beam is
distributed evenly along the beam. The beam is
supported by a cable that is attached 1.00m to the
beam from the wall and makes an angle of 35° from
the vertical.
What direction does the hinge push on the beam?
(1) up and to the right
(2) directly to the right
ΣFX = 0
-3660*sin35° + FxHinge = 0
FxHinge = +2099 N
35°
(3) down and to the right
ΣFY = 0
3660cos35° - 1500N + FyHinge = 0
FyHinge = -2998N + 1500 N
FyHinge = -1498N
Or put axis at 1m:
CG applies CW torque
Hinge must apply CCW torque.
A cable is attached to an elliptical rim and attached via a pulley to a
weight. The black dot represents the axis. In which configuration is
the torque on the ellipse the greatest?
(1)
(2)
Longer lever arm for 2
Exercise machines or bikes – may find
elliptical gears or cams – adjusts force at
different points in motion
You have fixed lever arm (black handle)
but lever arm for ‘load’ is adjusting
throughout cycle
A picnic basket weighing 100N is hanging from the end of a 3.0mlong beam. The beam weighs 250 N and is supported by a cable
which attaches to the middle of the beam and makes an angle of 30°
relative to the perpendicular to the beam. The beam is attached to the
wall by a hinge.
Suppose a bear weighing 1250N walks out onto the beam. The cable
can only withstand a tension of 1800N. How close can the bear get to
the basket before the cable snaps?