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12.1

x = 1.3 m x = 1.3 m

12.2

(b), (c)

12.3

316.7 g; 5.8 N

12.4

T = 1963 N ; F = 1732 N T = 1963 N ; F = 1732 N

12.5

μ s < 0.5 cot β μ s < 0.5 cot β

12.6

F door on A = 100.0 N i ^ 200.0 N j ^ ; F door on B = −100.0 N i ^ 200.0 N j ^ F door on A = 100.0 N i ^ 200.0 N j ^ ; F door on B = −100.0 N i ^ 200.0 N j ^

12.7

711.0 N; 466.0 N

12.8

1167 N; 980 N directed upward at 18°18° above the horizontal

12.9

206.8 kPa ; 4.6 × 10 −5 206.8 kPa ; 4.6 × 10 −5

12.10

5.0 × 10 −4 5.0 × 10 −4

12.11

63 mL

12.12

Fluids have different mechanical properties than those of solids; fluids flow.

Conceptual Questions

1.

constant

3.

magnitude and direction of the force, and its lever arm

5.

True, as the sum of forces cannot be zero in this case unless the force itself is zero.

7.

False, provided forces add to zero as vectors then equilibrium can be achieved.

9.

It helps a wire-walker to maintain equilibrium.

11.

(Proof)

13.

In contact with the ground, stress in squirrel’s limbs is smaller than stress in human’s limbs.

15.

tightly

17.

compressive; tensile

19.

no

23.

It acts as “reinforcement,” increasing a range of strain values before the structure reaches its breaking point.

Problems

25.

46.8 N · m 46.8 N · m

27.

4,472 N, 153.4°

29.

23.3 N

31.

80.0 kg

33.

40 kg

35.

right cable, 444.3 N; left cable, 888.5 N; weight of equipment 156.8 N; 16.0 kg

37.

784 N, 132.8 N

39.

a. 539 N; b. 461 N; c. do not depend on the angle

41.

tension 778 N; at hinge 778 N at 45°45° above the horizontal; no

43.

1500 N; 1620 N at 30°30°

45.

1.2 mm

47.

32.9 cm

49.

4.0×102N/cm24.0×102N/cm2

51.

11.9 μm 11.9 μm

53.

0.57 mm

55.

8.59 mm

57.

1.35 × 10 9 Pa 1.35 × 10 9 Pa

59.

259.0 N

61.

0.01%

63.

1.44 cm

65.

0.63 cm

Additional Problems

69.

tan −1 ( 1 / μ s ) = 51.3 ° tan −1 ( 1 / μ s ) = 51.3 °

71.

a. at corner 66.7 N at 30°30° with the horizontal; at floor 177 N at 109°109° with the horizontal; b. μs=0.346μs=0.346

73.

a. 1.10×109N/m2;1.10×109N/m2; b. 5.5×10−3;5.5×10−3; c. 11.0 mm, 31.4 mm

Challenge Problems

75.

F = Mg tan θ ; f = 0 F = Mg tan θ ; f = 0

77.

with the horizontal, θ=42.2°;θ=42.2°; α=17.8°α=17.8° with the steeper side of the wedge

79.

W ( l 1 / l 2 1 ) ; W l 1 / l 2 + m g W ( l 1 / l 2 1 ) ; W l 1 / l 2 + m g

81.

a. 1.1 mm; b. 6.6 mm to the right; c. 1.11×105N1.11×105N

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