Question

1. Old MacDonald had a farm, and on that farm he had a gate (see below). The gate is 3m wide and 1.8m high, with hinges attached to the top and bottom. The guy wire makes an angle of 30° with the top of the gate and is tightened by a turn buckle to a tension of 200 N. The mass of the gate is 32kg. 30.0° 1.80 m 3.00 m (a) Determine the horizontal force exerted on the gate by the bottom hinge. (b) Find the horizontal force exerted on the gate by the upper hinge. (c) Determine the combined vertical force exerted by both hinges. Hints: You should start by drawing extended free-body diagram for the gate. Label the forces clearly. Write down three equilibrium conditions for the gate before you try to answer any questions. Reaction forces from the hinges may have both horizontal and vertical components. Calculate torque relatively to the upper hinge as one of your equilibrium conditions (this simplifies algebra to be done).

Answer 1

Extended F.BD M 32 kg, T=200N Tsinoo 30 Exa T60530 1.8m - 3m ng b) from 0 FM - Fzx — - 94.6 - 200x3 Equilibrium conditions {F = 0 {fy = 0 {T=0 {Fzzo → Fiz to TCA30° = Fox - 0 {Fy = 0 Fiy + Fzy + TS n30 = mg Pret =0 → Fox x 1.8 + TS4530 X3= wy xl-5 - 4). Ⓡ E. = 32x9.81.5 – 200 X X3 1.8 = 94.6 N And - in - 78.6 N direction is of F2x = 7806N - So Fix C) From Ans Figt Fry = ng Tsin30 = 32x9.8-200x1 = 213.6N Ans Ans.