Torque τ = F·ℓ F and ℓ must be perpendicular. Units: N*m (enter them this way into computer) F...

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Torque • τ = F·ℓ •F and must be perpendicular . • Units: N*m (enter them this way into computer) F Axis

Transcript of Torque τ = F·ℓ F and ℓ must be perpendicular. Units: N*m (enter them this way into computer) F...

Page 1: Torque τ = F·ℓ F and ℓ must be perpendicular. Units: N*m (enter them this way into computer) F Axis ℓ.

Torque

• τ = F·ℓ• F and ℓ must be perpendicular.

• Units: N*m (enter them this way into computer)

F

Axis

Page 2: Torque τ = F·ℓ F and ℓ must be perpendicular. Units: N*m (enter them this way into computer) F Axis ℓ.

A board is placed in a support. F1 and F2 push down on the board. If the board stays stationary, how do the forces compare?

Support (fulcrum)

(1) F1 > F2

(2) F1 = F2

(3) F1 < F2

F1 F2

Page 3: Torque τ = F·ℓ F and ℓ must be perpendicular. Units: N*m (enter them this way into computer) F Axis ℓ.

Which of the following forces provides the greatest torque about the given axis? (the black dot represents the axis of rotation)

2F

F

2.5ℓ

1.5ℓ

2F

A B C

(1) A (2) B (3) C (4) All the same.|τA| = 2F*1.5ℓ = 3F*ℓ

|τB| = 2F*ℓ = 2F*ℓ

|τC| = F*2.5ℓ = 2.5F*ℓ

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Page 4: Torque τ = F·ℓ F and ℓ must be perpendicular. Units: N*m (enter them this way into computer) F Axis ℓ.

Which of the following signs (+ or -) are true?

2F

F

2.5ℓ

1.5ℓ

2F

A B C(1) - A , - B , + C

(2) + A , - B , - C

(3) - A , + B , + C

Sign Convention: (+) for CCW rotation; (-) for CW

τA : CW (-)

τA : CCW (+)

τA : CCW (+)

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Page 5: Torque τ = F·ℓ F and ℓ must be perpendicular. Units: N*m (enter them this way into computer) F Axis ℓ.

Levers

Small force with long lever arm

can exert large force with short

level arm

Class 1 Lever – Fulcrum between input effort and output load

• Pliers

• Catapult

• Oars

• Crowbar

• Seesaw

Page 6: Torque τ = F·ℓ F and ℓ must be perpendicular. Units: N*m (enter them this way into computer) F Axis ℓ.

Equilibrium

• Equilibrium – Accel = 0

• Static Equilibrium – Accel = 0 AND velocity=0

– Stationary

What happens? What happens?

Page 7: Torque τ = F·ℓ F and ℓ must be perpendicular. Units: N*m (enter them this way into computer) F Axis ℓ.

Example.A uniform board of 10 m is supported at the center. If a person (80 kg mass) is standing 2 m away from the center, how much force is required to balance the board at the other end?

Support (fulcrum)

F1 F2

Page 8: Torque τ = F·ℓ F and ℓ must be perpendicular. Units: N*m (enter them this way into computer) F Axis ℓ.

A force of 5N is applied downward 0.3m to the left of the fulcrum. Where would one exert a force of 1 N downward to balance the the stick?

(1) 0.2m to the left

(2) 0.2m to the right

(3) 1.5 m to the left

(4) 1.5 m to the right

Support (fulcrum)

0.3 m

5N

Στ = 0

(5N)(0.3m) - (1N)L = 0

L = 1.5m

Clicker!

Page 9: Torque τ = F·ℓ F and ℓ must be perpendicular. Units: N*m (enter them this way into computer) F Axis ℓ.

A downward force of 5N is applied to a uniform bar 0.3m to the left of the fulcrum. Another force of 1 N is applied downward 1.5 m to the right of the fulcrum. How much force is applied by the fulcrum to the bar? The bar is in equilibrium.

(1) 2 N.

(2) 3 N.

(3) 6 N

(4) 3.5N

Support (fulcrum)

0.3 m

5N

ΣFy = 0

F - 5N - 1N = 0

F = 6N

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1.5 m

1N

Page 10: Torque τ = F·ℓ F and ℓ must be perpendicular. Units: N*m (enter them this way into computer) F Axis ℓ.

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

Clicker!

Page 11: Torque τ = F·ℓ F and ℓ must be perpendicular. Units: N*m (enter them this way into computer) F Axis ℓ.

Two symmetric objects: a uniform bar and a square box.

Where do you draw the weight?

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A B C

D

E

F1) A, C, 2) B, D 3) C, E, F 4) B, E 5) A, C, D, F

Page 12: Torque τ = F·ℓ F and ℓ must be perpendicular. Units: N*m (enter them this way into computer) F Axis ℓ.

An asymmetric object. Where do you think the weight will be

centered?

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A B C D1) A 2) B 3) C 4) D

Page 13: Torque τ = F·ℓ F and ℓ must be perpendicular. Units: N*m (enter them this way into computer) F Axis ℓ.

Center of Gravity (Center of Mass)

• In extended objects, the location where the weight is considered acting on it is the center of gravity.

• If the object is symmetric, then the "real center“ of the object is the center of gravity.

• If irregular, one can find the center of gravity using the formula:

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WW

xWxWxcg

center of gravity

Page 14: Torque τ = F·ℓ F and ℓ must be perpendicular. Units: N*m (enter them this way into computer) F Axis ℓ.

An asymmetric object.

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2211

WW

xWxWxcg

Center of Gravity

1 2 3 4

x1x2

Page 15: Torque τ = F·ℓ F and ℓ must be perpendicular. Units: N*m (enter them this way into computer) F Axis ℓ.

Roughly where is the center of gravity located? Point 1, 3, or 5 or region 2

(between 1 and 3) or 4 (between 3 and 5)?

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WW

xWxWxcg

75.08/6

kg2kg2kg4

m2kg2m1kg2m0kg4

ggg

gggxcg

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Page 16: Torque τ = F·ℓ F and ℓ must be perpendicular. Units: N*m (enter them this way into computer) F Axis ℓ.

What are the forces do you need to consider if the bar is stationary?

Hinge

F

F Fsin

W W

Fsin

WFcos(1) (2) (3)

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Fx = 0, and Fy = 0

Page 17: Torque τ = F·ℓ F and ℓ must be perpendicular. Units: N*m (enter them this way into computer) F Axis ℓ.

Torque

Hinge

F

Fsin

If the F is not perpendicular to ℓ , then find the F component perpendicular to ℓ .

= (F sin . ℓ

Note: Check where the angle is given. It could be Fcos , ifthe angle is given in other side.

Page 18: Torque τ = F·ℓ F and ℓ must be perpendicular. Units: N*m (enter them this way into computer) F Axis ℓ.

What is the torque on the fishing pole? Use the end where it is held as the axis of

rotation. (1) 80N (cos 20°) 1.2m

(2) 80N (sin 20°) 1.2m

(3) 80N*1.2m

(4) -80N (cos 20°) 1.2m

(5) -80N (sin 20°) 1.2m

(6) -80N*1.2m

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Page 19: Torque τ = F·ℓ F and ℓ must be perpendicular. Units: N*m (enter them this way into computer) F Axis ℓ.

A cook holds a 2.00kg 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

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Page 20: Torque τ = F·ℓ F and ℓ must be perpendicular. Units: N*m (enter them this way into computer) F Axis ℓ.

The bridge shown must be in static equilibrium. A truck and a car are crossing the bridge. Which of the following force diagrams most correctly describe the forces?

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(1) (2)

(3) (4)

Page 21: Torque τ = F·ℓ F and ℓ must be perpendicular. Units: N*m (enter them this way into computer) F Axis ℓ.

Write the equation for the net torque.

FT FcF W

Example

Page 22: Torque τ = F·ℓ F and ℓ must be perpendicular. Units: N*m (enter them this way into computer) F Axis ℓ.

What is the correct force diagram? Assume the gray line as the person.

Clicker!(1)

(2)

(3)

(4)