Question

F13-4. The 2-Mg car is being towed by a winch. If the winch exerts a force of T = (100s) N on the cable, where s is the displacement of the car in meters, determine the speed of the car when s = 10 m, starting from rest. Neglect rolling resistance of the car. F13-4

Answer 1

13.4 Mass of Car = 2 kg = 2000 kg. Force T=(1008) NI when s=10m, Force T= 100x10 = 1000 m N. i Force - Mass & Acceleration. , looo = 2000x a a = 0.5 m/s² from Newton's law of Motion 1² = u²+ aas & but u=0 initial velocity ress) 2 - 2x0.52 x 10 Tv = 2.24 ms As Ik 2 q= 15k/ft = 271 b=8,75 in RAK 22 – Re RA+ RB = 12 + 115*22= 45 Moment about A RAXO + RBX22 - Px9 – 9x22 x 22 = 0 Re = 12x9 +1°5X22X!! = 21.4091 K. RA= 23.59K. for Portion BC (Taking a distance from B), Moment, Ma-RB X - 9. 2. 20 – 22

1.5 At B x=0, M₂ = 0 At x=13m Rotht c, Mc = 21.48 x 13 - 105 x 132 Mc = 151.58 kft for Maximum, d (Me) =0 a (Rox- q. 22 ) =0 RB- quo x= Re - 21.48 = 14,273 ft. Mmax = 21.41 x 14.273 - 105 x 14,2732 = 157.8 kft for Portion AB (a distance from B) Ma = Rpx - (2x 13 x (2 – 65)] – P (2-13) – EX (X-1372 2= 13, Mc = 21.41813- (105X13 * (13-6.5) - o-0=151.58 kst x=22, 22 / 21141899 – (195x13x62-65)] – 12 (22-13) – 105 (22 MA = MA = 0 for maximum, red (Mx)=0 RB - 138 - P- 2 (0-1)=0 2741+ 13x1.5-12 = 1:5 (x-13) x = 6.273 which is not lies on Portion Ac. 2 2

Thus Maximum Bending Stress, 4 s= b = 8.75427 I = 14352.19 int 1 - Neutral anes k-by Maximum compress stress, at Top layer of beam, in (15718X12) x(23) Mmax 8 14382.19 (2) &= 1.78116 klinz (compressive) .. Maximum Compressive stress at top layer, z = 2 (oc) max = 1.78116 klin An & Manimum Tensile stress at bottom tager = 1/2 (2) max = Mmax.y = (157.8 x12) x442) K ED mon = 1.78116 klin) Tensile Au. 14352.19