Chapter 9 Statics

Two 50 lb. children sit on a see-saw. Will they balance in all three cases?

Review of Forces: Ex 1

A dentist places braces on a person’s teeth that exert 2.00 N in each direction as shown. Calculate the net force on the teeth.

Look at each force separately and resolve it into x and y components.

F x = (2.00 N)(cos 20 o ) = 1.88 N F y = (2.00 N)(sin 20 o ) = 0.684 N For the other force: F x = -1.88 N F y = 0.684 N 70 o 20 o 2.0 N

The x-components cancel and the y components add S F x S F y = 1.88 N – 1.88 N = 0 2.0 N = 0.684 N + 0.684 N = 1.37 N 2.0 N F net = 1.37 N

Review of Forces: Ex 2

Calculate the force exerted by the traction device shown below.

Calculate the x and y-components: Top force: F x = (200 N)(cos 37 o ) = 160 N F y = (200 N)(sin 37 o ) = 120 N Bottom force: F x = (200 N)(cos 37 o ) = 160 N F y = (200 N)(sin 37 o ) = 120 N (downward) S F y S F x = 120 N –120 N = 0 = 160 N + 160 N = 320 N Why doesn’t the patient slide out of the bed?

Review of Torque

Mr. Saba can’t quite budge this rock. How can he increase his torque so he can move it?

 = FR

Conditions for Static Equilibrium

S  = 0 S F = 0

S

F = 0: Ex 1

A 90.0 kg man cannot do a pull-up. His best efforts produce a scale reading of 23 kg. What force is he exerting?

S F x S F y = 0 = F B + F s – mg Since he is not moving S F = 0

0 = F B + F S – mg F B = mg – F S F B = (90 kg)(9.8 m/s 2 ) – (23 kg)(9.8 m/s 2 ) F B = 660 N

S

F = 0: Ex 2

Calculate the F 1 not moving.

and F 2 assuming the chandelier is S F x S F y S F x S F y = 0 = 0 = 0 = F 2 – (F 1 )(cos 60 o ) = 0 = (F 1 )(sin 60 o ) - mg

0 = F 2 – (F 1 )(cos 60 o ) 0 = (F 1 )(sin 60 o ) – mg two eqns, two unknowns 0 = (F 1 )(sin 60 o ) – (200 kg)(9.8 m/s 2 ) 0 = (F 1 )(0.866) – 1960 N 1960 N = (F 1 )(0.866) F 1 = 2260 N 0 = F 2 F 2 – (F 1 )(cos 60 o ) = (F 1 )(cos 60 o ) F 2 = (2263)(cos 60 o ) = 1130 N

A 200 g mass is hung by strings as shown. Calculate the tension in string A and string B.

(A = 1.13 N, B = 1.51 N)

S

t = 0: Example 1

Two children sit on a see-saw as shown. The board of the see-saw has a mass of 2.00 kg centered at the pivot. Where should the 25.0 kg child sit so that they are in perfect balance?

30.0 kg 2.50 m ?

25.0 kg

Let’s pick the pivot as the origin S = 0 0=(245 N)(x) – (294 N)(2.5 m) – (F N )(0) + (mg)(0) 0 = (245 N)(x) – (294 N)(2.5 m) (245 N)(x) = 735 m-N x = 3.0 m 2.5 m F N x mg 294 N 245 N

A crane lifts a 1000.0 kg car as shown in the picture below, 8.00 m from the pivot. Calculate the force that the motor of the crane must provide 2.00 m from the pivot. (39,200 N)

A teacher (56.0 kg) sits at the end of a 4 m long see saw. Where should a 203 kg owlbear sit to perfectly balance the seesaw? (55 cm)

S

F = 0 and

S

t = 0: Example 1

A heavy printing press is placed on a large beam as shown. The beam masses 1500 kg and the press 15,000 kg. Calculate the forces on each end of the beam.

S F y S F y = 0 and = 0 = F 1 0 = F 1 + F 2 S = 0 (we’ll ignore F x ) + F 2 - (1500 kg)(g) – (15,000 kg)(g) – 1.617 X 10 5 N F 1 + F 2 = 1.617 X 10 5 N We’ll choose F 1 S = 0 as the pivot 0 = (F 1 )(0 m) – (1500kg)(g)(10m) - (15,000 kg)(g)(15 m) + (F 2 )(20 m) 0 = -2.352 X 10 6 m-N + (F 2 )(20 m) (F 2 )(20 m) = 2.352 X 10 6 m-N

F 1 + F 2 = 1.617 X 10 5 N (F 2 )(20 m) = 2.352 X 10 6 m-N (F 2 ) = 2.352 X 10 6 m-N/(20m) = 1.176 X 10 5 N F 1 F 1 + 1.176 X 10 5 N = 1.617 X 10 5 N = 4.41 X 10 4 N

S

F = 0 and

S

t = 0: Example 3

The beam below has a mass of 1200 kg. Calculate F 1 and F 2 for the cantilever as shown. Assume the center of gravity is at 25 m.

F 1 F 2 20.0 m 30.0 m (1200 kg)(g)

S F y S F y = 0 and = 0 = F 1 0 = F 1 + F 2 S = 0 + F 2 - (1200 kg)(g) – 11,760 N F 1 + F 2 = 11,760 N We’ll choose F 1 S = 0 as the pivot 0 = (F 1 )(0 m) + (F 2 )(20m) - (11,760 N)(25 m) 0 = (F 2 )(20m) – 294,000 N (F 2 )(20m) = 294,000 N

F 1 + F 2 = 11,760 N (F 2 )(20m) = 294,000 N F 2 = 294,000 N/20 m = 14,700 N F 2 = 14,700 N F 1 = 11,760 N - F 2 F 1 = 11,760 N - 14,700 N F 1 = -2940 N (we picked the wrong direction for Force 1)

The following system is completely in balance. Calculate the mass of Mass C and the value of the force at the support point (S).

(217 g, 3.34 N)

S

F = 0 and

S

t = 0: Example 4

A sign of mass M = 280 kg is suspended from a 25.0 kg beam that is 2.20 m long. The angle between the sign and the wire is 30 o . Calculate F H and F T , the forces at the hinge and the tension in the wire.

S F x = 0 0 = F H cos q

0 = F H cos

q - F T cos30 o

– 0.866F

T

S F y = 0 0 = F H sin q

0 = F H sin

q + F T sin30 o

+ 0.5F

T

– (25 kg)(g) – (280 kg)(g)

– 2989 N

S = 0 (choose hinge as origin) 0 = (25kg)(g)(1.1m) + (280kg)(g)(2.2m) – (F T )(sin30 o )(2.2m)

T hree equations, three unknowns 0 = F H cos

q

0 = F H sin

q

– 0.866F

+ 0.5F

T T – 2989 N 0 = 6306 – 1.1F

T F T = 5730 N, F H = 4960 N,

q

= 1.43

o

S

F = 0 and

S

t = 0: Example 5

Calculate the force on the hinge and the tension in the cord for the following sign. The beam is 2.00 m long, though the sign is hung at 1.80 m.

S F x = 0 0 = F H cos q  =

F H cos

q – F T cos20 o

– 0.940F

T

S F y = 0 0 = F H sin q

0 = F H sin

q + F T sin20 o

+ 0.342F

T

– (20 kg)(g) – (150 kg)(g)

– 1666N

S = 0 (choose hinge as origin) 0 = (20kg)(g)(1m) + (150kg)(g)(1.8m) – (F T )(sin30 o )(2m)

T hree equations, three unknowns

 =

F H cos

q

0 = F H sin

q

– 0.940F

T + 0.342F

T – 1666N 0 = 2842 – 0.684F

T F T = 4155 N, F H = 3914 N,

q

= 3.59

o

S

F = 0 and

S

t = 0: Example 6

5.0 m A crane is designed to hold a maximum of 10,000 kg (M). The top crossbeam has a mass of 450 kg. Calculate the forces exerted by the post (F p ) and the angled beam (F A ).

F p 15.0 m 35 o F A M

F p F px F py F Ay F Ax q F A

S F x =0 0 = F A cos35 o

0 = 0.819F

A

- F P sin q

- F P sin

q S F y =0 0 = F A sin35 o – F p cos q – mg – Mg 0 = F A sin35 o

0 = 0.574F

A

– F p

– F p

cos

cos

q q – (450 kg)(g) –(10,000kg)(g)

– 1.02 X 10 5 N

S = 0 (choose top of post as origin) 0 = (450 kg)(g)(10m) – (10,000 kg)(g)(20m) + (F A )(sin35 o )(15m)

0 = 2.00 X 10 6 – 8.6 F A Three Equations, three unknowns 0 = 0.819F

0 = 0.574F

A A 0 = 2.00 X 10 - F P sin

q

– F p cos

q

6 – 8.6 F A – 1.02 X 10 5 N F A = 2.34 X 10 5 N, F p = 1.95 X 10 5 N,

q

= 80.7

o

The Ladder: Example 1

A 5-m long ladder leans against a wall at a point 4 m above the ground. The ladder has a mass of 12.0 kg and is uniform. Assume the wall is frictionless, but the ground is not. Calculate the force from the wall (F w ) and the force from the ground (F G ).

Working with the Triangle 5 2 = 4 2 + x 2 x 2 = 5 2 – 4 2 x = 3 sin ?

= 4/5 ?

= 53 o 5 m 4 m ?

x 5 m 37 o 4 m 53 o 3 m

S F x = 0 0 = F Gx – F w

0 = F g cos

q

- F w

S F y = 0 0 = F Gy – mg

0 = F g sin

q

- 118

Back to the ladder Tilting the Ladder F G 37 o mg F w F G 53 o 37 o mg 53 o F w 53 o 37 o

Calculating the Torque forces F G 53 o F w 37 o mgsin37 o mg Choose the point at ground as the pivot S = 0 0 = (5m)(F w )(sin53 o ) – (2.5m)(mg)(sin 37 o )

0 = 3.99F

w - 177 m-N

F w sin53 o

Three equations, three unknowns

0 = F g cos

q

- F w 0 = F g sin

q

- 118 0 = 3.99F

w - 177

q

= 69.3

o F g = 126 N F w = 44.3 N

The Ladder: Example 2

Mr. Fredericks (56.0 kg) leans a 3.00 m, 20.0 kg ladder against his house at an angle of 65.0

o with the ground. He can safely climb 2.50 m up the ladder before it slips. Calculate the coefficient of friction between the ground and the ladder.

S F x = 0

0 = F g cos

q

- F w

S F y = 0 0 = F g sin q

0 = F g sin

q – (20.0 kg)(g) – (56.0 kg)(g)

– 744.8

Choose the point at ground as the pivot S = 0 0 = (20)(g)(1.5)(sin25 o ) + (56)(g)(2.5m)(sin 25 o ) - (F W )(3)(sin 65 o )

0 = 704 – 2.72F

w

Three equations, three unknowns

0 = F g cos

q

0 = F g sin

q

- F w – 744.8 0 = 704 – 2.72F

w F W = 259 N, F g = 788 N,

q

= 70.8

o

Dealing with Friction

F gx = F g cos 70.8

o F gx = 259 N = F fr F gy = F g sin 70.8

o F gy = 744 N = F N F fr = m F N 259 = m 744 m

= 0.35

The Ladder: Example 3

A man who weighs 800 N climbs to the top of a 6 meter ladder that is leaning against a smooth wall at an angle of 60°. The non uniform ladder weighs 400 N and its center of gravity is 2 meters from the foot of the ladder. What must be the minimum coefficient of static friction between the ground and the foot of the ladder if it is not to slip?

S F x = 0

0 = F w - F g cos

q S F y = 0 0 = F g sin q

0 = F g sin

q – 800 – 400

– 1200

Working with the Triangle Tilt the Ladder F w 60 o 30 o 800 N F w 60 o 30 o F G 30 o 400 N 800 N 400 N 30 o 60 o F G

Calculating the Torque forces F w sin60 o 800sin30 o F w 60 o 30 o 800 N 400sin30 o F G 30 o 400 N Choose the ground as the pivot S = 0 0 =(2)(400sin30 o ) + (6)(800sin30 o ) - (6)(F w sin60 o )

0 = 2800 – 5.2F

w

Three equations, three unknowns 0 = F w - F g cos

q

0 = F g sin

q

– 1200 0 = 2800 – 5.2F

w F w = 538 N, F g = 1315 N,

q

= 65.9

o

Dealing with Friction

F gx = F g cos65.9

o F gx = 537 N = F fr F gy = F g sin65.9

o F gy = 1200 N = F N F fr = m F N 537 = m 1200 m

= 0.45

Stable vs. Unstable Equilibrium

A body will fall if its center of gravity is no longer above the base

Humans adjust their posture to keep CG above their base Try leaning against a wall and lifting one leg.

Elasticity: Hooke’s Law

• Hooke’s Law – usually used with a spring • Can consider

anything

to be like a spring • F = k D L (F=kx spring ) • k = proportionality (spring) constant • Can’t stretch things forever

Elastic region – material will still bounce back Plastic region – material will not return to original length (but has not broken) F = k D L This is only linear in the proportional region

Elastic Region: Young’s Modulus(E)

Stress = Force Area = F A Strain = Change in length Original length = D L L o

E = stress strain E = F/A D L/L o or F = E D L A L o

Young’s Modulus: Example 1

A 1.60 m long steel piano wire has a diameter of 0.20 cm. How great is the tension in the wire if it stretches 0.30 cm when tightened?

A = p r 2 = p (0.0010 m) 2 F = E D L = 3.1 X 10 -6 m 2 A L o F = AE D L L o

F = AE D L L o F = (3.1 X 10 -6 m 2 )(200 X 10 11 N/m 2 )(0.0030 m) 1.60 m F = 1200 N

Young’s Modulus: Example 2

A steel support rod of radius 9.5 mm and length 81 cm is stretched by a force of 6.2 X 10 4 N (about 7 tons). What is the stress? What is the elongation?

Area = p r 2 = p (0.0095 m 2 Stress = Force = 6.2 X 10 -4 m Area 2.84 X 10 -4 m 2 =

2.2 X 10 8 N/m 2

F = E D L A L o D L = FL EA = (6.2 X 10 4 (200 X 10 9 N)(0.81m) N/m 2 )(2.84 X 10 -4 m 2 ) D L = 8.84 X 10 -4 m = 0.89 mm

A 100 kg block hangs at the end of a 20 cm copper wire. The wire has a diameter of 5.05 X 10 -4 m.

a) Calculate the area of the wire b) Calculate the stress on the wire.

c) If the wire stretches 8.9 mm, calculate the Young’s Modulus for copper.

(1.1 X 10 11 N/m 2 )

The Three Types of Stress

Stretching Squeezing Horizontal

Other Modulus’

Shear Modulus – Used for shear stress Bulk Modulus – Used for even compression on all sides (an object when submerged)

Fracture

• Breaking Point • Uses – Tensile Strength – Stretching – Compressive Strength – under a load – Shear Stress – Shearing • Safety Factor – reciprocal that is multiplied by the tensile strength • Ex: A safety factor of 3 means you will only use 1/3 of the maximum stress

Fracture: Example 1

A concrete column 5 m tall will have to support 1.2 X 10 5 N (compression). What area must it have to have a safety factor of 6?

Max stress = (1/6)(compressive strength) Max stress = (1/6)(20 X 10 6 N/m 2 )= 3.3X10

6 N/m 2 Stress = F A

Stress = F A Area = F Stress = (1.2 X 10 5 N) = 3.64 X 10 -2 m 3.3 X 10 6 N/m 2

How much will the column compress under the load?

F = E D L A L o D L = FL EA = (1.2 X 10 5 N)(5 m) (20 X 10 9 N/m 2 )(3.64 X 10 -2 m 2 ) D L = 8.3 X 10 -4 m = 0.83 mm

Fracture: Example 2

Spider-man’s webbing has a tensile strength of 600 X 10 6 N/m 2 and he wishes to use a safety factor of 3. What is the diameter of the webbing if the maximum force at the bottom of a swing is 1500 N?

Maximum Stress (1/3)(600 X 10 6 N/m 2 ) = 200 X 10 6 N/m 2 Stress = Force Area Area = Force Stress = 1500 N = 7.5 X 10 -6 m 2 200X10 6 N/m 2 Area = p r 2 r = (A/ p ) 1/2 = 1.55 X 10 -3 m or 1.55 mm Diameter = 3.10 mm

Concrete

• Concrete is

much

stronger under compression than tension – Tensile Strength – 2 X 10 6 N/m 2 – Compressive Strength – 20 X 10 6 N/m 2 • Prestressed concrete – rods or mesh are stretched when the concrete is poured. Released after concrete dries. • Now under compression

2. 1.7 N 4. 1.7 m 6. 6.52 kg 8. 6130 N, 8570 N 10. 3.0 m from the adult 12. F T2 = 3900 N, F T1 = 3400 N 14. 560 N 16. 48 N (only vertical components balance wt.) 18. 1.1 m (pivot on the side of lighter boy) 20. F 1 = -2100 N(down), F 2 = 3000 N (up)

24. a) 210 N b) 2000 N 26. F T = 542 N, F hingeH = 409 N, F hingeV =-5.7 (down) 28. q min = tan -1 (1/2 m ) 44. 1.91 cm 46. 0.029 mm 50. 9.6 X 106 N/m 2 56. F Tmax = 390 N 58. a) 4.4 X 10 -5 m 2 b) 2.7 mm