Chapter 9 statics
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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.

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Chapter 9 Statics

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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.

Fx = (2.00 N)(cos 20o) = 1.88 N

Fy = (2.00 N)(sin 20o) = 0.684 N

For the other force:

Fx = -1.88 N

Fy = 0.684 N

20 o

70 o

2.0 N


The x-components cancel and the y-components add

SFx = 1.88 N – 1.88 N = 0

SFy = 0.684 N + 0.684 N = 1.37 N

2.0 N

2.0 N

Fnet = 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:

Fx = (200 N)(cos 37o) = 160 N

Fy = (200 N)(sin 37o) = 120 N

Bottom force:

Fx = (200 N)(cos 37o) = 160 N

Fy = (200 N)(sin 37o) = 120 N (downward)

SFy = 120 N –120 N = 0

SFx = 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

St = 0

SF = 0


SF = 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?

SFx = 0

SFy = FB + Fs – mg

Since he is not moving SF = 0


0 = FB + FS – mg

FB = mg – FS

FB = (90 kg)(9.8 m/s2) – (23 kg)(9.8 m/s2)

FB = 660 N


SF = 0: Ex 2

Calculate the F1 and F2 assuming the chandelier is not moving.

SFx = 0

SFy = 0

SFx = 0 = F2 – (F1)(cos 60o)

SFy = 0 = (F1)(sin 60o) - mg


0 = F2 – (F1)(cos 60o)two eqns, two unknowns

0 = (F1)(sin 60o) – mg

0 = (F1)(sin 60o) – (200 kg)(9.8 m/s2)

0 = (F1)(0.866) – 1960 N

1960 N = (F1)(0.866)

F1 = 2260 N

0 = F2 – (F1)(cos 60o)

F2 = (F1)(cos 60o)

F2 = (2263)(cos 60o) = 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)


St = 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?

?

2.50 m

30.0 kg

25.0 kg


Let’s pick the pivot as the origin

St = 0

0=(245 N)(x) – (294 N)(2.5 m) – (FN)(0) + (mg)(0)

0 = (245 N)(x) – (294 N)(2.5 m)

(245 N)(x) = 735 m-N

x = 3.0 m

FN

2.5 m

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)


SF = 0 and St = 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.


SFy = 0 and St = 0 (we’ll ignore Fx)

SFy = 0 = F1 + F2 - (1500 kg)(g) – (15,000 kg)(g)

0 = F1 + F2 – 1.617 X 105 N

F1 + F2 = 1.617 X 105 N

We’ll choose F1 as the pivot

St = 0

0 = (F1)(0 m) – (1500kg)(g)(10m) - (15,000 kg)(g)(15 m) + (F2)(20 m)

0 = -2.352 X 106 m-N + (F2)(20 m)

(F2)(20 m) = 2.352 X 106 m-N


F1 + F2 = 1.617 X 105 N

(F2)(20 m) = 2.352 X 106 m-N

(F2) = 2.352 X 106 m-N/(20m) = 1.176 X 105 N

F1 + 1.176 X 105 N = 1.617 X 105 N

F1 = 4.41 X 104 N


SF = 0 and St = 0: Example 3

The beam below has a mass of 1200 kg. Calculate F1 and F2 for the cantilever as shown. Assume the center of gravity is at 25 m.

F1

F2

20.0 m

30.0 m

(1200 kg)(g)


SFy = 0 and St = 0

SFy = 0 = F1 + F2 - (1200 kg)(g)

0 = F1 + F2 – 11,760 N

F1 + F2 = 11,760 N

We’ll choose F1 as the pivot

St = 0

0 = (F1)(0 m) + (F2)(20m) - (11,760 N)(25 m)

0 = (F2)(20m) – 294,000 N

(F2)(20m) = 294,000 N


F1 + F2 = 11,760 N

(F2)(20m) = 294,000 N

F2 = 294,000 N/20 m = 14,700 N

F2 = 14,700 N

F1 = 11,760 N - F2

F1 = 11,760 N - 14,700 N

F1 = -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)


SF = 0 and St = 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 30o. Calculate FH and FT, the forces at the hinge and the tension in the wire.


SFx = 0

0 = FHcosq - FTcos30o

0 = FHcosq – 0.866FT

SFy = 0

0 = FHsinq + FTsin30o – (25 kg)(g) – (280 kg)(g)

0 = FHsinq + 0.5FT – 2989 N


St = 0(choose hinge as origin)

0 = (25kg)(g)(1.1m) + (280kg)(g)(2.2m) – (FT)(sin30o)(2.2m)

T hree equations, three unknowns

0 = FHcosq – 0.866FT

0 = FHsinq + 0.5FT – 2989 N

0 = 6306 – 1.1FT

FT = 5730 N, FH = 4960 N, q = 1.43o


SF = 0 and St = 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.


SFx = 0

0 = FHcosq – FTcos20o

  • = FHcosq – 0.940FT

    SFy = 0

    0 = FHsinq + FTsin20o – (20 kg)(g) – (150 kg)(g)

    0 = FHsinq + 0.342FT– 1666N


St = 0(choose hinge as origin)

0 = (20kg)(g)(1m) + (150kg)(g)(1.8m) – (FT)(sin30o)(2m)

T hree equations, three unknowns

  • = FHcosq – 0.940FT

    0 = FHsinq + 0.342FT– 1666N

    0 = 2842 – 0.684FT

    FT = 4155 N, FH = 3914 N, q = 3.59o


SF = 0 and St = 0: Example 6

5.0 m

15.0 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 (Fp) and the angled beam (FA).

35o

FA

Fp

M


FAx

q

Fpy

Fp

FAy

FA

Fpx


SFx =0

0 = FAcos35o - FPsinq

0 = 0.819FA - FPsinq

SFy =0

0 = FA sin35o – Fp cosq – mg – Mg

0 = FA sin35o – Fp cosq – (450 kg)(g) –(10,000kg)(g)

0 = 0.574FA – Fp cosq – 1.02 X 105 N


St = 0 (choose top of post as origin)

0 = (450 kg)(g)(10m) – (10,000 kg)(g)(20m) + (FA)(sin35o)(15m)

0 = 2.00 X 106 – 8.6 FA

Three Equations, three unknowns

0 = 0.819FA - FPsinq

0 = 0.574FA – Fp cosq – 1.02 X 105 N

0 = 2.00 X 106 – 8.6 FA

FA = 2.34 X 105 N, Fp = 1.95 X 105 N, q = 80.7o


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 (Fw) and the force from the ground (FG).


5 m

Working with the Triangle

52 = 42 + x2

x2 = 52 – 42

x = 3

sin ? = 4/5

? = 53o

4 m

?

x

37o

5 m

4 m

53o

3 m


SFx = 0

0 = FGx – Fw

0 = Fgcosq - Fw

SFy = 0

0 = FGy – mg

0 = Fgsinq - 118


Fw

53o

37o

Back to the ladder

Tilting the Ladder

FG

37o

53o

mg

Fw

FG

53o

37o

mg


Fw

FG

Fwsin53o

Calculating the Torque forces

Choose the point at ground as the pivot

St = 0

0 = (5m)(Fw)(sin53o) – (2.5m)(mg)(sin 37o)

0 = 3.99Fw - 177 m-N

53o

37o

mgsin37o

mg


Three equations, three unknowns

0 = Fgcosq - Fw

0 = Fgsinq - 118

0 = 3.99Fw - 177

q = 69.3o

Fg = 126 N

Fw = 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 65.0o 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.


SFx = 0

0 = Fgcosq - Fw

SFy = 0

0 = Fgsinq – (20.0 kg)(g) – (56.0 kg)(g)

0 = Fgsinq – 744.8


Choose the point at ground as the pivot

St = 0

0 = (20)(g)(1.5)(sin25o) + (56)(g)(2.5m)(sin 25o)

- (FW)(3)(sin 65o)

0 = 704 – 2.72Fw


Three equations, three unknowns

0 = Fgcosq - Fw

0 = Fgsinq – 744.8

0 = 704 – 2.72Fw

FW = 259 N, Fg = 788 N, q = 70.8o


Dealing with Friction

Fgx = Fgcos 70.8o

Fgx = 259 N = Ffr

Fgy = Fg sin 70.8o

Fgy = 744 N = FN

Ffr = mFN

259 = m744

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?


SFx = 0

0 = Fw - Fgcosq

SFy = 0

0 = Fgsinq – 800 – 400

0 = Fgsinq – 1200


Fw

60o

Working with the Triangle

Tilt the Ladder

30o

800 N

30o

400 N

FG

60o

Fw

FG

60o

30o

30o

800 N

400 N


Fw

Fwsin60o

Calculating the Torque forces

Choose the ground as the pivot

St = 0

0 =(2)(400sin30o) + (6)(800sin30o) - (6)(Fwsin60o)

0 = 2800 – 5.2Fw

FG

60o

30o

30o

400sin30o

800sin30o

800 N

400 N


Three equations, three unknowns

0 = Fw - Fgcosq

0 = Fgsinq – 1200

0 = 2800 – 5.2Fw

Fw = 538 N, Fg = 1315 N, q = 65.9o


Dealing with Friction

Fgx = Fgcos65.9o

Fgx = 537 N = Ffr

Fgy = Fg sin65.9o

Fgy = 1200 N = FN

Ffr = mFN

537 = m1200

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 = kDL (F=kxspring)

  • 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 = kDL

This is only linear

in the proportional

region


Elastic Region: Young’s Modulus(E)

Stress = Force=F

AreaA

Strain = Change in length=DL

Original lengthLo


E = stress

strain

E = F/AorF = E DL

DL/LoA Lo


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 = pr2 = p(0.0010 m)2 = 3.1 X 10-6 m2

F = E DL

A Lo

F = AE DL

Lo


F = AE DL

Lo

F = (3.1 X 10-6m2)(200 X 1011N/m2)(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 104 N (about 7 tons). What is the stress? What is the elongation?


Area = pr2 = p(0.0095 m2

Stress = Force = 6.2 X 10-4m = 2.2 X 108 N/m2

Area 2.84 X 10-4 m2

F=EDL

A Lo

DL = FL=(6.2 X 104 N)(0.81m)

EA(200 X 109 N/m2)(2.84 X 10-4 m2)

DL = 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.

  • Calculate the area of the wire

  • Calculate the stress on the wire.

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

    (1.1 X 1011 N/m2)


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 105 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 106 N/m2)= 3.3X106 N/m2

Stress = F

A


Stress = F

A

Area = F=(1.2 X 105 N) = 3.64 X 10-2 m

Stress3.3 X 106N/m2


How much will the column compress under the load?

F=EDL

A Lo

DL = FL=(1.2 X 105 N)(5 m)

EA(20 X 109 N/m2)(3.64 X 10-2 m2)

DL = 8.3 X 10-4 m = 0.83 mm


Fracture: Example 2

Spider-man’s webbing has a tensile strength of 600 X 106 N/m2 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 106 N/m2) = 200 X 106 N/m2

Stress = Force

Area

Area = Force=1500 N= 7.5 X 10-6m2

Stress 200X106N/m2

Area = pr2

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 106N/m2

    • Compressive Strength – 20 X 106N/m2

  • 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. FT2 = 3900 N, FT1 = 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. F1 = -2100 N(down), F2 = 3000 N (up)


24. a) 210 Nb) 2000 N

26. FT = 542 N, FhingeH = 409 N, FhingeV=-5.7 (down)

28. qmin = tan-1(1/2m)

44. 1.91 cm

46. 0.029 mm

50. 9.6 X 106 N/m2

56. FTmax = 390 N

58. a) 4.4 X 10-5 m2b) 2.7 mm


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