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

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

Sign convention for torques:

(-) CW torque

(+) CCW torque

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⟂

Perpendicular Lever arm – line drawn through hinge and perpendicular to line of action.

τ = F r τ = 0

τ = F r sinθAxis of Rotation

Torque – Perpendicular Lever Arm

Perpendicular Lever arm – line drawn through hinge and perpendicular to line of action.

• May have to calculate using trig • Be careful to make sure you use

right trig function May be given to you

axis

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

τ = F r⟂

τ = F (d cos θ)

τ = F

τ = F⟂ d

τ = (F cosθ) d

θθ

Fr⟂θ

F F⟂

d d, r⟂

Look at Components of force

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) (Fbcosθ) 8cm

(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. Choose zero for no force)

1

23

4

5

6

7

8

To balance torque, need to push down.

To balance forces horizontally, need to push to left.

FE

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?

FELBOW,X = 301N

If we can find the components of the force, we can reconstruct the full force vector.

ΣFy = 0

-Fg + Fbcos(75) – FELBOW,Y = 0

FELBOW,Y = 312N*cos(75) – 19.6N

FELBOW,Y = 61N

301N

61Nθ

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 equilibriumNeutral 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?

axis

τ = -F*r⟂ = -mg*r⟂ = -(10 kg) (9.8 m/s2) (0.38 m)

τ = -37.2 Nm

(1) +0.388 Nm (2) – 0.388 Nm (3) + 3.80 Nm (4) – 3.80 Nm (5) + 3.72 Nm (6) – 3.72 Nm (7) + 37.2 Nm (8) – 37.2 Nm

A child is sitting on an adult’s 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?

axis

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

(1) +0.8 Nm (2) – 0.8 Nm (3) + 3.20 Nm (4) – 3.20 Nm (5) + 7.84 Nm (6) – 7.84Nm (7) + 15.8 Nm (8) – 15.8 Nm

A child is sitting on an adult’s leg. What is the tension in upper leg muscle?

axis

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 one-quarter 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?

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

L/2 from left or right end

Draw, Draw, Draw!!!!

Look for similarities

Break forces into components perpendicular and parallel to the board, bridge, arm, tray, fishing pole, wrench, ….

Στ = 0, ΣFX = 0, ΣFY = 0

If don’t know L, take “leap of faith” and stick it in as a variable. See if it cancels.

Look for similarities

Στ = 0, ΣFX = 0, ΣFY = 0

35°

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.

4m

35°

1mAt 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.

4m

35°

1mWhat is the lever arm for the force due to gravity? Use the hinge as the axis of rotation. (1) 0m (2) 1m (3) 2m (4) 3m (5) 4m

Center of gravity applied at center of board. Center of board is 2m from the hinge.

Fg

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.

Στ(about hinge) = 0

T(cos35°)*(1m) -1500*(2m)+FHINGE* 0 = 0

T = (3000N*m)/(cos35°*1m)

T = 3660N

4m

35°

1mWhat is the tension in the cable? (1) 1500N (2) 2460N (3) 3000 N (4) 3660 N (5) 5230 N (6) 6000N (7) 7320N

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.

35°

What direction does the hinge push on the beam? (1) up and to the right (2) directly to the right (3) down and to the right

ΣFX = 0 -3660*sin35° + FxHinge = 0

FxHinge = +2099 N

Or put axis at 1m: CG applies CW torque Hinge must apply CCW torque.

ΣFY = 0 3660cos35° - 1500N + FyHinge = 0

FyHinge = -2998N + 1500 N FyHinge = -1498N

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 2Exercise 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 3.0m-long 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?