HCI 04 Forces Solutions (Discussion Questions)

1 Tutorial 4 Forces Discussion Questions (Suggested Solutions) D1. On static equilibrium of a ladder W : weight of ladder N1 : normal contact force by floor on ladder N2 : normal contact force by wall on ladder f1 : frictional force, by floor on ladder f2 : frictional force, by wall on ladder In scenarios A and B, the ladder will definitely slip, as there is a net force: the horizontal forces are not balanced. D2. Application of the Principle of Moments Initially, the centre of gravity (c.g.) is at O. After the mass is tied to the handle, is 27 cm to the right of O, producing a clockwise moment about O. Similarly, the mass is initially 95 cm to the left of O. After the broom is moved, it is 95 – 27 = 68 cm to the left of O, producing an anticlockwise moment about O. Taking moments about O, by the principle of moments, sum of anticlockwise moments = sum of clockwise moments (0.200 x 9.81) (0.68) = (9.81m) (0.27) The mass of the broom m = 0.504 kg = 500 g (to 2 s.f.) D3. On static equilibrium (a) The component of X perpendicular to the bar is cos40X  . Taking moments about A: (36)( CW mo 0.45c ment by 36 os60 N ACW mo ) ( cos ment b 40 )(1.2) 8.81 N y X X X  =  = = bi) Approach 1 For translational equilibrium, the net force acting on the bar must be zero. The vector sum of the weight and X is a leftward and downward force. There must be a rightward and upward force at A. A B C D

2 Approach 2 For rotational equilibrium, the net moment acting on the bar about any point must be zero. Taking moments about the C.G., X exerts an ACW moment. There must be a force at A that produces a CW moment. bii) Method 1 Vertically, net force must be zero: sin70 36yFX+ = N (8.81)sin70 36 27.7 N y y F F + = = Horizontally, net force must be zero: cos70xFX= cos70 3.01 NxF = = 2 2 2 2 3.01 27.7 27.9xyF F F= + = + = N Method 2 Using cosine rule: 2 2 2 36 8.81 2(36)(8.81)cos20F = + −  27.9 NF = biii) See drawing. COMMENT: Part (b)(ii) should be used to guide the drawing of F. The force’s line of action must pass above the bar (because it is supposed to produce a CW moment about the CG to counter the ACW moment of X) You may also recall that in a 3-force system, the lines of action of all 3 forces should intersect. 20° X=8.81 N 36 N F

3 D4. Application of Hooke’s Law. Static equilibrium of a rigid extended body. Taking moments about P, sum of clockwise moments = sum of anticlockwise moments T x = (8.0 x 9.81) (0.25) cos 30° (500 e) x = (8.0 x 9.81) (0.25) cos 30° ----- (1) where e is the extension of the spring. From the blue dashed triangle, using trigonometry, tan 30° = (0.20 + e) / x x tan 30° = 0.20 + e e = x tan 30° – 0.20 ----- (2) Combining (1) and (2), 16.99 = 500 (x tan 30° – 0.20) x 16.99 = 500 x2 tan 30° – 100 x 289x2 – 100x – 16.99 = 0 x = 0.47