first condition of equilibrium: fnet = 0

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Equilibrium: the state of no acceleration– balanced forces and balanced torques First condition of equilibrium: F net = 0 Second condition of equilibrium: τ net = 0 Static Equilibrium: Forces and Torques balance and the object is at rest Center of Mass (cm)– the point around which gravitational torques balance– it is where the F w can be considered to be “located” at For a uniform object, the cm is at the center: F w C

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Solving Statics Problems 1) Forces are balanced horizontally and vertically 2) Torques are balanced as counterclockwise (+) and clockwise (-) around an arbitrarily placed pivot point F3 F1 CC PP C F2 C

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Page 1: First condition of equilibrium: Fnet = 0

Equilibrium: the state of no acceleration– balanced forces and balanced torques

First condition of equilibrium: Fnet = 0Second condition of equilibrium: τnet = 0

Static Equilibrium: Forces and Torques balance and the object is at restCenter of Mass (cm)– the point around which gravitational torques balance– it is where the Fw can be considered to be “located” atFor a uniform object, the

cm is at the center:Fw

C

Page 2: First condition of equilibrium: Fnet = 0

Solving Statics Problems

1) Forces are balanced horizontally and vertically2) Torques are balanced as counterclockwise (+) and clockwise (-) around an arbitrarily placed pivot point

F1

F2

F3 PP CCC

C

Page 3: First condition of equilibrium: Fnet = 0

F1

F2

F3

PP

CC

C

CC

F1

F2

F3

PPC

CC

C

Page 4: First condition of equilibrium: Fnet = 0

A uniform meterstick with a weight of .150 N is balanced at the 75.0 cm mark by a mass hung at 90.0 cm mark. How much is the mass that is producing equilibrium? How much does the stand push up?

C

Fw

.500 m

Fs.750 m

Fh

.400 m

PP

CCC

Ø

Page 5: First condition of equilibrium: Fnet = 0

Fw = .150 Nmh = ?

Tc = Tcc

(Fh)(.150 m) = (.150 N)(.250 m)Fh = (.150 N)(.250 m)

.150 m= .250 N

m = Fh

g

= .250 N

9.80 m/s2

= .0255 kg

Fup = FdownFdown = .150 N + .250 N

= .400 N

Page 6: First condition of equilibrium: Fnet = 0

A large wooden beam weighs 820 N and is 3.2 m long. The beam’s center of gravity is 1.4 m from one end. Two workers begin carrying the beam away. If they lift the beam at its ends, how much weight must each worker support?

Page 7: First condition of equilibrium: Fnet = 0

Fw = 820 N

1.4 m

Fu Fu 2

PP

1.8 m

C

CCØ

Page 8: First condition of equilibrium: Fnet = 0

Tc = T cc

(820 N)(1.4 m) = (Fu2)(3.2 m)

Fu2 = (820 N)(1.4 m)

3.2 m

= 360 N

Fup = Fdown

Fu + Fu2 = 820 N

Fu = 820 N - 360 N = 460 N

Page 9: First condition of equilibrium: Fnet = 0

A cat walks along a uniform plank that is 4.00 m long and has a mass of 7.00 kg. The plank is supported by two sawhorses, one 0.500 m from the left end of the board and the other 1.50 m from its right end. When the cat reaches the right end, the plank just begins to tip. What is the mass of the cat?

Page 10: First condition of equilibrium: Fnet = 0

1) A 10.0 N meterstick is suspended from two spring scales, one at the 8.00 cm mark and the other at the 90.0 cm mark. If a weight of 5.00 N is hung at the 20.0 cm mark and a weight of 17.0 N is hung at the 55.0 cm mark, what will each scale read?2) An 850 N painter stands 1.20 m from one end of a 3.00 m scaffold plank supported at each end by a stepladder. The scaffold plank weighs 250 N and there is a 50.0 N paint bucket .50 m from the other end of the plank. How much upward force is exerted by each of the two ladders?

Page 11: First condition of equilibrium: Fnet = 0

3) A 650.0 N boy and a 490.0 N girl sit on a 150.0 N porch swing that is 1.70m and supported at each end by chains. What will be the tension in each chain if the boy sits .750 m from one end and the girl sits .500 m from the other end?

4) A uniform 2.50 N meterstick is hung from the ceiling by a single rope. A 500.0 g mass is hung at the 25.0 cm mark and a 650.0 g mass at the 70.0 cm mark. What is the tension in the rope and at what mark is the rope located on the meterstick?

Page 12: First condition of equilibrium: Fnet = 0

45.0˚

100.0 N

150.0 N

2.00 m

60.0˚

3.00 m1.00 m

100.0 N200.0 N

What force must be added to the right end to produce translational and rotational equilibrium?