351_PartUniversity Physics Solution

351_PartUniversity Physics Solution - 11-10 Chapter 11 (b)...

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11-10 Chapter 11 (b) 21 τ > since 2 F has a larger moment arm; the net torque is clockwise. (c) See Figure 11.21b. 11 1 (8.00 N) Fl l =− 2 0 = since 2 F ! is at the axis Figure 11.21b 6.40 N m z gives (8.00 N) 6.40 N m l −= 0.800 m, l = same as in part (a). EVALUATE: The force couple gives the same magnitude of torque for the pivot at any point. 11.22. IDENTIFY: 0 lF Y Al = Δ SET UP: 24 2 50.0 cm 50.0 10 m A == × . EXECUTE: relaxed: 4 42 2 (0.200 m)(25.0 N) 3.33 10 Pa (50.0 10 m )(3.0 10 m) Y −− × ×× maximum tension: 5 2 (0.200 m)(500 N) 6.67 10 Pa Y × EVALUATE: The muscle tissue is much more difficult to stretch when it is under maximum tension. 11.23. IDENTIFY and SET UP: Apply Eq.(11.10) and solve for A and then use 2 Ar π = to get the radius and 2 dr = to calculate the diameter. EXECUTE: 0 Y = Δ so 0 A Yl = Δ ( A is the cross-section area of the wire) For steel, 11 2.0 10 Pa Y (Table 11.1) Thus 62 11 2 (2.00 m)(400 N) 1.6 10 m . (2.0 10 Pa)(0.25 10 m) A × 2 , = so 4 /1 . 6 1 0 m / 7 . 1 1 0 m rA ππ × = × 3 2 1.4 10 m 1.4 mm ==× = EVALUATE: Steel wire of this diameter doesn&t stretch much; 0 / 0.12%. ll Δ= 11.24. IDENTIFY: Apply Eq.(11.10). SET UP: From Table 11.1, for steel, 11 Y and for copper, 11 1.1 10 Pa Y . 2 ( /4) 1.77 10 m Ad × . 4000 N F = for each rod. EXECUTE: (a) The strain is 0 l F lY A Δ = . For steel 4 11 4 2 0 (4000N) 1.1 10 . (2.0 10 Pa)(1.77 10 m ) l l Δ × Similarly, the strain for copper is 4 2.1 10 . × (b) Steel: 45 (1.1 10 )(0.750 m) 8.3 10 m ×= × . Copper: 44 (2.1 10 )(0.750 m) 1.6 10 m × . EVALUATE: Copper has a smaller Y and therefore a greater elongation. 11.25. IDENTIFY: 0 Y = Δ SET UP: 2 0.50 cm 0.50 10 m A × EXECUTE: 11 2 (4.00 m)(5000 N) (0.50 10 m )(0.20 10 m) Y × EVALUATE: Our result is the same as that given for steel in Table 11.1. 11.26. IDENTIFY: 0 Y = Δ SET UP: 23 2 5 2 (3.5 10 m) 3.85 10 m × . The force applied to the end of the rope is the weight of the climber: 2 (65.0 kg)(9.80 m/s ) 637 N F . EXECUTE: 8 52 (45.0 m)(637 N) 6.77 10 Pa (3.85 10 m )(1.10 m) Y × × EVALUATE: Our result is a lot smaller than the values given in Table 11.1. An object made of rope material is much easier to stretch than if the object were made of metal.
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Equilibrium and Elasticity 11-11 11.27. IDENTIFY: Use the first condition of equilibrium to calculate the tensions 1 T and 2 T in the wires (Figure 11.27a). Then use Eq.(11.10) to calculate the strain and elongation of each wire. Figure 11.27a SET UP: The free-body diagram for 2 m is given in Figure 11.27b. EXECUTE: y y Fm a = 22 0 Tm g = 2 98.0 N T = Figure 11.27b SET UP: The free-body-diagram for 1 m is given in Figure 11.27c EXECUTE: y y a = 12 1 0 TT m g −= TTm g =+ 1 98.0 N 58.8 N 157 N T =+= Figure 11.27c (a) stress strain Y = so stress strain F YA Y == upper wire: 3 1 72 1 1 157 N strain 3.1 10 (2.5 10 m )(2.0 10 Pa) T AY = × ×× lower wire: 3 2 1 1 98 N strain 2.0 10 T AY =
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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351_PartUniversity Physics Solution - 11-10 Chapter 11 (b)...

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