Trabalho 4

(Ex 16) • F: 90 i - 40 j - 20 k • A: (1.5, 0, 0) • F x r = | i j k | |90 -40 -20| |20 20 50| = (1(1.50i - 2.20j): - (2.20 - 2.20i): -1.5.20 - (-80i)k) = -280i + 90j + 180k • My = F x r = Ey: My1 = 20 - . : 20 z My2 = 60: 150 = [110 k.m] (Ex 20) • ME = 1150 - 0.4): + 1200.0.q) - (500.0.4) [ME = 460 k.m] (Ex 21) • M = 1.5 KII.wq ?) • M = rx F 1.5(0,3,2) x (F,0,9) [F = 2.32 K.I.m] (Ex 22) • M R = ? • Fq: Fx = F; • M = ?·2 F: Fq: 10.i. -1 F; • II . q) . -8.q . 20 M = 20 KN.m ) Fq(4.3,.8) (F.E. 42) - w1 = \frac{2}{3}x^3 - Fe \int_0^4 x^2 dx = \frac{x^3}{3} \Bigg|_0^4 = 160 N - M_{eq} \int_0^6 x^2.5 dx = \frac{x^3.5}{3.5} \Bigg|_0^6 = 512 Nm - d = \frac{512}{160} = \boxed{3.20 \ m} (F.E. 32) - w(x) = 10 - \frac{x}{3} - w(10) = 10 - \frac{10}{3} - Fe \int_0^{12} (10 - \frac{x}{3})dx = \frac{x^2}{2} - \frac{x^3}{3 \times 9} \Bigg|_0^{12} = 9 \times 12 - 15.6 = 27.5 \times 2 - Meq \int_0^5 x^{1.5} dx = \frac{x^{2.5}}{2.5} \Bigg|_0^5 = 45 - 120 - 275.5 - dFe \frac{9 \times 2x}{2} + (9-15.6x) d=\boxed{4.13 km} (F.E. 41) - w(2) = 6x + 3 - F = \int_0^{4.5} (6x + 3) dx + \int_{4.5}^{12.5} 2 dx = \frac{6x^2}{2} + 3x \Bigg|_0^{4.5} + 43 (13,23) : 4.5 = 21 Km (\mu_l) - Meq \int_{4.5}^{12.5} \frac{6x^2}{9} + \frac{3x}{3} = \frac{30}{2} - \frac{2x^3}{2} \Bigg|_0^{12.5} = 2x \ - d = \frac{43.23}{21} = 2.1 m \boxed{\text{2.1 m}}