RI Y5 Remedial Forces Soln

Raffles Institution Physics Department Y5 H2 Physics Term 3 Remedial 5 Chapter 4: Forces Solutions 1 A by A on mass by B on mass 3 3 xFF k x k y y    2 D additional upthrust required to carry oil = weight of oil water  Vtankerl  g = oil  Voil  g 1030  36 500  h = 930  34000  22 h = 18.5 m Worked Example 1 3 A The lines of action of the three forces must meet at a common point P so that resultant moment is zero, 0 . (Option A or C) To cancel the rightward horizontal component of force F, the horizontal component of force exerted by the wall on rod must point leftward. (Option A) Alternatively, the resultant force on the rod is zero, 0F , implies that the forces, when taken in order, form a closed triangle. (Option A) 4 A by floor on his feet by each ar mrest on his arms by seat on him by floor on his feet by each ar mrest on his arms by seat on him Isolate the man who is in static equilibrium: 0 20 26 0 0 2 2 yF FF F W FW F F             .50 5 0 0 5 0 N Worked Example 2 Taking moments about pivot X, clockwise moments = anti-clockwise moments W  0.80 + 180  1.5 = 220  3.0 W = 488 N Answer: B W F A W Nwall fwall Nfloor ffloor C

Raffles Institution Physics Department Y5 H2 Physics Term 3 Remedial 6 5 A  oo o taking moments about P, anticlockwise moment = clockwise moment cos30 200 sin30 2 100 tan30 58 N 0 0 static frictional force 58 N y x LCL C FN W Ff C           6 (a) Consider the two blocks as a single body Fs = W // Fs = 2 (mg sin 40) = 2 (6  9.81  sin 40) = 75.7 N By Newton’s Third law, compressive force on the spring = Fs = 75.7 N B1 B1 (b) F = kx 75.7 = 400 x x = 0.189 m B1 B1 7 (a) T AB = mg = 500  9.81 = 4910 N B1 (b) At the hook, TAB = 2  T sin 30 4905 = 2  T sin 30 T = 4910 N M1 A1 (c) Taking moments about the right corner of the cab, T AB  9.0 + 2500  9.81  4.5 = 20000  9.81  2.0 TAB = 3.13  104 N M1 A1 P 30.0o W C f N