DHS 04 Forces (Tutorial Solutions)

DUNMAN HIGH SCHOOL 9749 H2/ 8867 H1 PHYSICS Year 5(2024) 1 1. C 2. B 3. (a) When the boat is in equilibrium, By Principle of Floatation: U = msg = 1.2 ×107 (9.81) = 1.18 ×108 N (b) By Archimedes’s Principle: U = ρwVdisplacedg = 1.18 ×108 (1000)Vdisplaced(9.81) = 1.18 ×108 Vdisplaced = 1.20 ×104 m3 Volume of boat below the water line = Vdisplaced = 1.20 × 104 m3 4. (a) kA = 6.0 N m-1 kB = 3.0 N m-1 Tension in A and B = 0.6 N extension in spring A = (0.6 / 6) = 0.10 m extension in spring B = (0.6 / 3) = 0.20 m (b) Total extension = 0.20 + 0.10 = 0.30 m Hence, the effective spring constant of A and B = (0.60 / 0.3) = 2.0 N m-1 5. Acceleration of both trolleys = 0 . 40 . 5 20 = m s-2 Consider the 3.0 kg trolley: 120 . 4 0 . 3 '= × =F N Compression of spring = 0 . 30 . 4 12 = cm Ans: B 3.0 kg force by spring, F’

2 6. Method 1 Let the spring constant of each spring be k. In the first case, the downward force on each spring is W/3. So, the spring constant is given by: kxW =3 ⇒ x Wk 3= In the second case, the force on each spring is (2W)/2 = W Extension e is given by: keW = ⇒ xx W W k We 3) 3/( = = = Method 2 Let the spring constant of each spring be k. Using the formula for parallel springs, In the first case, k eff = k1 + k2 + k3 = k + k + k = 3k Using Hooke’s Law, “F 1 = k1x1” W = 3k x ---------------- (1) In the second case, k eff = k1 + k2 = k + k = 2k Using Hooke’s Law, “F 2 = k2x2” 2W = 2k x2 ---------------- (2) Sub (1) into (2) 2(3k x) = 2k x2 x2 = 3x

3 7. Method 1 Let k be the spring constant of the spring. By Hooke’s Law, 1 2 m N 100010 0 . 2 20 − − =×= =x Fk Elastic potential energy (PE) stored in the spring when it is compressed by 3.0 cm: 45 . 0 ) 030 . 0 )( 1000 (2 1 2 1 22 == =kxU J When the toy reaches its maximum height h above its point of release, Gain in gravitational PE = Loss in elastic PE mgh = 0.45 J ⇒ h = 0.917 m Position 1 Position 2 Position 3 (At max compression) (when toy just leaves spring) (At max height) Method 2 Energy conversion from Position 1 to Position 2 Gain in GPE + Gain in KE = Loss in EPE mgx + ½ mu2 = ½ kx2 (0.050) (9.81) (0.030) + ½ (0.050)u2 = ½ (1000)(0.030)2 u = 4.17 m s-1 (upwards) where u is the velocity of the toy which just leaves the spring Using equation of motion between Position 2 and Position 3, Taking upwards as positive, v 2 = u2 + 2as 02 = (4.17)2 + 2(-9.81)s s = 0.886 m h = x + s = 0.030 + 0.886 = 0.916 m x = 0.030 m s h

4 8. (a) When the block is submerged in water, there is a pressure difference between the top and bottom surfaces of the block, with the water pressure being greater at the bottom surface of the block. Since pressure is force per unit area, the force the water exerts on the bottom surface of the block is greater than the force it exerts on the top surface of the block. This results in a net upward force the water exerts on the block which is the force of upthrust. 8. (bi) By Archimedes’ Principle, 361.0 10 (27.8 10 )(9.81) 0.272718 N 0.27 N w blockU Vgρ − = = ×× = ≈ 8. (bii) From Fig. 8.1: Since the system is in equilibrium, taking moments about the pivot Sum of clockwise moments = Sum of anticlockwise moments 0.18𝐹𝐹 = 0.083𝑊𝑊 𝑊𝑊(0.083) = 𝐹𝐹(0.180) ----- (1) From Fig. 8.2: Since the system is in equilibrium, Sum of clockwise moments = Sum of anticlockwise moments 𝑊𝑊(0.078) = (𝐹𝐹 − 𝑈𝑈)(0.19) ----- (2) Sub (1) into (2): 𝐹𝐹 = 2.49 𝑁𝑁

5 9. Method1: Using sine rule: N 75. 7 120sin ) 81 . 9 )( 00 . 2 ( 20sin = = T T  N 6 . 14 120sin ) 81 . 9 )( 00 . 2 ( 40sin = = N N  Method 2: Let x represent the direction along the slope; as positive. Let y represent the direction perpendicular to the slope; as positive. For translational equilibrium, ∑ = 0xF and ∑ = 0yF . ∑ = 0xF N 75 . 730cos 20sin 0 20sin30cos == =−    WT WT ∑ = 0yF N 6 . 14 0 20cos30sin = =− + N WT N  10. (a) X: Resultant of normal contact force and frictional force by wall on ladder. Friction is upwards as ladder tends to slide down. Y: Resultant of normal contact force and frictional force by ground on ladder. Friction is to the right as ladder sliding down and to the left. W: Gravitational force of Earth on ladder, called the weight. (b) W Y X X Y W

6 (c) Using the cosine rule, X2 = W2 + Y2 - 2WYcos200 X2 = 2002 + 1502 -2(200)(150)cos200 X = 78.2 N 11. (a) Considering the free body of the picture: Considering forces acting along the vertical axis: Since in equilibrium, 2(25 sin α) = 5 α = 5.74o (b) 2 T sin α = weight of picture = 5 N So when the tension T = breaking tension which is the maximum tension, sin α and hence α will be a minimum. {This is a minimum value as when α gets smaller, the tension will be greater than 25 N which will break the cord.} Mathematical Proof 2 T sin α = 5 T = αsin2 5 ≤25 sin α ≥ 0.10 α ≥ 5.739 12. (a) 20o Y = 150 N X W = 200 N 70o θ 25 N 25 N 5 N α

7 (b) (c) Take moments about the hinge, 700(1) + 200(3) + 80(6) = T(6 sin 600) T = 343 N At the hinge, Rx = T cos 600 = 172 N Ry + T sin 600 = 700 + 200 + 80 Ry = 683 N Let the maximum distance be y Take moments about the hinge, 700(y) + 200(3) + 80(6) ≤ 900(6 sin 60 0) y ≤ 5.14 m 13. (a) What reason does the paragraph give for the construction with many cables? (b) When a civil engineer designs the tower, he needs to consider the maximum total mass which the tower may need to support. Calculate that maximum total mass. (c) Calculate the mass of 10 m of the roadway and the maximum mass of traffic which the 10 m of roadway may have to support. (d) Calculate the angle, θ, (in degree) between a cable and the horizontal. The bridge will be better as the load of the structure and traffic will be distributed amongst the cables instead of being concentrated on a cable. [2] Maximum total mass = 3.5 x 105 + 2.9 x 105 + 6.8 x 104 [1] = 7.08 x 105 kg [1] mass of 10 m of roadway = 10 100 x 3.5 x 105 kg = 3.5 x 104 kg [1] mass of traffic on 10 m of roadway = 10 100 x 2.9 x 105 kg = 2.9 x 104 kg [1] tan θ = 4.5 10 [1] θ = tan-1   4.5 10 = 24.2 o [1]

8 (e) Draw a force diagram of a fully laden 10 m section of road at segment A given that the following forces act at the segment. W : Weight of 10 m of roadway N : Force exerted by traffic on 10 m of roadway T : Tension in cable R : Net force exerted by the rest of the roadway (f) By considering the conditions necessary for translational equilibrium, state the equations relating W, N, T and R. (g) Hence, or otherwise, calculate the tension in a cable when the bridge is fully laden. (The tension in all cables is assumed to be the same.) (h) When the civil engineer designs the bridge, he needs to consider the possibility of vibration of the bridge. Under what condition will the bridge vibrate with maximum amplitude? Suggest two possible external sources that will give rise to this condition. T = NW 44+ 2.9×10 ×9.81+3.5×10 ×9.81=sin θ sin 24.2 [1]