Question

Please help with these questions.

2. If the position vector r= 151 - 10j +20k (m) and force F = 250i +400j - 600k (N). a) Find the angle between r and F in degrees. 8 = degrees (5 points) b) Calculate vector M = r XF= k (5 points) 3. For the following composite area shown below. Shaded regions have material while white regions are empty. Include proper units. a. Find the location of the centroid measured from the shown X and Y axes, b. Calculate the moment of inertia and radius of gyration about the indicated axes +Yc centroidal Ixe = bh/12 and lyc = hb/12 h - Хc b tyc centroidal IX = n 14/4 and Iye = r 4 Xc Ky = ได้ Ky = Lil = (5 points) (5 points) 899 Ix = (5 points) 49 ly= (5 points) KX= 3" (5 points) 2

Answer 1

2 q=15a logt + 20 i (m) 7 = 250 î+400ģ- Gook (N) Let angle between 7 and F be d. Å F = 1 All Flead [By det product Al 15²+10² + 202 = 26.926 2 TFT 250²+ 400 + 6002 = 763.22 of . F = Q5x250) -(10x400) - (2exbod - - 12250 0.596 : Caso = -12 250 26.926 4763-22 0.- /26.6° To find a XP (By eres product rule le.] k 15. -10 20 250 400 600 - ( 6000 - 2000) – (-9000-5 3 000) - î (-9000 - 500) + Ê (6000 +25 ( = - 2000 ? + 14000 7 + 8500 h Nom

6" 2 * I!" 21" // 3 fexs ch Total area of the shaded fertion is A = D + (8 x 6) Fixers] <quaverian = 48.86 sq. in. Let distances of centreid be x fraut arisalay right. Taking moment of area about Yaxis, 48.86 xE7 (10x1x5) (3x6% 3)-(x2'xs Х 7 » ã os = 5 in 7, -[ib xoxlo't be xo'x2)x17 = 226.5479 in 4 Let T be distance of centroid from Xaxis. Similarly, taking moment about Xaxis, 425410) = (* 183.5+ (2xvx4)-(2 - Ý = 5.412 no in

IM Ix xixo 10x1x (8.5-5.412)² +(4x8'46) + 8x6x (5.412-4) 2 x (5.412-4) + in >> 4 (26.19 +351.699 -7.049] in" = 440.34 in 4 :: Kx 440,84 7x А in = 3 in 48.86 X = 5 in Y = 5.412 in 4 Ix = 440.84 in Iy = 226.5479 4 Kx = 3 in