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 m (accept) or x = -0.12 m (reject) D5. On static equilibrium of a rigid extended body. [RJC/2009/Prelim/P3/Q1] (a) The moment of a force about a point is the product of the (magnitude of the) force and the perpendicular distance from the line of action of the force to the point. (b) (i) Taking moments about the hinge, sum of clockwise moments = sum of anticlockwise moments (ii) Analyse vertical components of forces: Analyse horizontal components: o 0 cos30 2400 900N 900N (downwards) y y y y F FT F F = += =− =  0 0 sin30 1910N (right) x x F FT = =  2000 N Fx Fy 400 N T T cos 30° 30° support cable 60° 30° load T sin 30°

4 D6. On static equilibrium of a rigid extended body. [HCI/BT/05/P2/Q3(b)] (i) Taking moments about the hind wheels, sum of anticlockwise moments = sum of clockwise moments, (22.0) (9.81) (0.40) = F1 (1.00) F1 = 86.3 N Taking moments about the front wheels, sum of anticlockwise moments = sum of clockwise moments, F2 (1.00) = (22.0) (9.81) (0.60) F2 = 129 N (ii) If the stroller topples, it will topple about point P. At the instant of toppling, the normal contact force from the ground acting on the front wheels (F1) is zero. Taking moments about the hind wheels, sum of anticlockwise moments = sum of clockwise moments, (22.0) (9.81) (0.40) = W (0.30) The weight of the groceries is W = 288 N. Because the perpendicular distance from the handle to point P is 3/4 of the perpendicular distance from the centre of mass to the handle, the load can be 4/3 times as large as the combined weight of the stroller and the baby. (iii) The baby may shift the centre of gravity towards P, which reduces the anticlockwise moments due to the baby about P. Parents may lean on the handle and inadvertently press it down, creating an additional clockwise moment about P. On an upward slope, the anticlockwise moment due to the weight of the stroller would decrease, but the clockwise moment due to the weight of the groceries would increase.

5 D7. Application of the principle of moments. Answer: B Method 1 Let A be the cross-sectional area of the rod. Taking moments about the pivot, Mwoodg(1.45L) = M’woodg(0.55L) + Mrubberg(1.60L) wood(2.90L)A(1.45L) = wood(1.10L)A(0.55L) + rubber(L)A(1.60L) 3.600wood = 1.60rubber  Ratio = 2.25 Method 2 Note that the centre of gravity of the rod is at the pivot. Hence, the pivot exerts an upwards force of magnitude Wrod at the pivot. Taking moments about the left end of the rod, Wrod (2.90L) = Wwood (2.00L) + Wrubber (4.50L) (4ρwood + ρrubber) (2.90L) = (4ρwood) (2.00L) + ρrubber (4.50L) ρrubber / ρwood = 3.6 / 1.6 = 2.25 D8. On upthrust. Assume that in air, the upthrust by the air on the solid is negligible. Hence, W1 is the weight of the solid. When immersed in a liquid, let the upthrust by the liquid on the solid be U. Thus, W2 = W1 - U Since the solid is totally immersed in the liquid, by Archimedes’ Principle, the upthrust is equal to the weight of the liquid displaced by the solid, U = ρ Vsolid g. Thus volume of the solid, Vsolid = (W1 - W2) / (ρg) D9. On upthrust. Answer: C Upthrust acting on bubble = weight of air + viscous drag force By Archimedes’ principle, the upthrust acting on the bubble is equal to the weight of the water displaced, weight of water displaced = weight of air + viscous drag force (1000) (2.370 x 10-8) (9.81) = (1.290) (2.370 x 10-8) (9.81) + viscous drag force Hence, the viscous drag force is FVD = 2.322 x 10-4 N Rubber handle wooden section Mwood M’wood Mrubber

6 D10. On upthrust. (a) By the principle of flotation, upthrust acting on wood = weight of wood 𝑈 = 𝑚𝑔 𝜌water𝑔𝐴ℎin water = 𝜌wood𝑔𝐴ℎwood ℎin water = 𝜌wood 𝜌water ℎwood = 0.65 × 103 1.00 × 103 20 = 13 cm ℎabove water = 20 − 13 = 7.0 cm (b) By the principle of flotation, upthrust = weight of wood + weight of lead 𝑈 = 𝑚wood𝑔 + 𝑚lead𝑔 𝜌water 𝑔 𝐴 ℎ = 𝜌wood 𝑔 𝐴 ℎ + 𝑚lead𝑔 𝑚lead = 2.8 kg

7 D11. On upthrust. Answer: D Identify the forces X, Y, Z. X: weight of the beaker and water Y: weight of the solid Z: weight of water displaced by object immersed in water = upthrust on object (by water) Sketch the FBD of the object immersed in water. T: tension by the spring balance on object (determine the spring balance reading) 𝑇 = 𝑌 − 𝑍 Sketch the FBD of the beaker & water. N: normal force by weighing machine on Beaker & water (determine the weighing machine reading) N = X + Z Normal, N Weight of beaker & water, X Force by object on water, Z

8 D12. On upthrust. ∑ 𝐹𝑥 = 0 → 𝐹current − 𝑇 sin 20 = 0 → 𝐹current = 𝑇 sin 20 (1) ∑ 𝐹𝑦 = 0 → 𝑇 cos 20 + 𝑈 − 𝑊 = 0 → 𝑇 = (𝑊 − 𝑈) tan 20 (2) Substituting equation (2) into equation (1), 𝐹current = (𝑊 − 𝑈