H1.04. 2022 Forces Tutorial Solutions

1 Types of Forces Tutorial – Solution Q1 Ans: A Compression of air, the fall of water droplets and the spreading of petrol all have very little resistance to its motion. But when applying paint to the surface of a wall it would actually take more effort to ‘push’ the paint across the wall. Q2 (a) Friction is the force which opposes the relative motion of the two bodies in contact or tends to oppose one body from moving relative to the other. (b) (i) 1. Friction is useful in braking when the brakes of a moving car is stepped. 2. Friction is a useful force to enable walking by the floor exerting a forward force on the shoe/foot. 3. Friction provides the driving force of a car through the frictional force exerted by the road surface on the tire in the forward direction of the car. 4. In unloading heavy wooden crates from a lorry on an inclined plank, friction slows down the crates as the crates slide down the plank. (ii) Without friction during braking, the car will skid and may result in accident. Without friction between the floor and the shoe/foot, a person may slip and falls. Without friction to provide driving, it is difficult to control a vehicle on the road which will result in accident. Without friction during unloading heavy wooden crates on the incline plank, the crates may slide down too fast and cause injury to workers. 1 2 1 Q3 Ans: B For parallel combinations, keff = k1 + k2 (make sure you know how to derive this) F = keff x 10 = 2k (0.2) k = 25 Nm-1 Q4 When W = 0 N, the length L is the natural length (Lo) of the spring. From the graph, when W = 0 N, L = Lo = 6 cm Using Hooke’s Law, Fs = ke where Fs = force on spring, e = extension of spring, k = spring constant Therefore, sFk e Using the point ( 5 N, 10 cm), o Wk N mLL 1 2 5 () (10 6) 10     = 125 N m-1

2 Q5 First, find the resultant force FR of the 3 coplanar forces: y-component of the resultant force, Fy = 40 N x-component of the resultant force, Fx = 50-20 = 30 N tan θ = (30/40)  θ = 36.9° Resultant = 504030 22  N To maintain equilibrium, the counter force must be of same magnitude but in opposite direction to FR. Hence, F = 50 N at bearing of 220° (2 s.f.) 1 1 1 Q6 As this is an equilibrium system, resultant force is zero and resultant torque is zero. Hence all three lines of force must intersect at a common point, and, when the force vectors are joined head-to-tail they should form a closed triangle with arrows pointing in the same direction (anticlockwise in this case). Answer: B Q7 Essential questions 1. What is this force that the question speaks of?  force of gum on tooth 2. What does it mean that the tooth is in equilibrium?  no net force acting on body, and no net moment about ANY POINT on the tooth. 3. Given the directions of the two 2.5 N forces, which is the only way that force can act to put the tooth in equilibrium?  Since the two 2.5 N forces give a resultant force that acts in the direction of X, the 3rd force must act opposite in direction (i.e. direction Z) and be of equal magnitude to the resultant of the two 2.5 N forces. 4. For 3 coplanar forces that are in equilibrium, what other special result exists?  they form a closed triangle since no net force exists. By elimination, answer is D. How to verify that magnitude is 1.3 N?  By sine rule, 𝐹𝑛𝑒𝑡 𝑜𝑓 2.5𝑁 𝑓𝑜𝑟𝑐𝑒𝑠 sin 30° = 2.5 𝑁 sin 75° 𝐹𝑛𝑒𝑡 𝑜𝑓 2.5𝑁 𝑓𝑜𝑟𝑐𝑒𝑠 = 1.294 = 1.3𝑁 Answer: D Fy Fx FR θ

3 Q8 Draw a Free-body diagram of system comprising the Trailer & its load of 20 kN, indicating all the external forces acting on the system and their respective perpendicular distances from the pivot. Let NA and NX be the upward forces exerted on the trailer at points A and X respectively. Aim: To find NX By Principle of Moments, taking moments about point A, Sum of anticlockwise moments = Sum of clockwise moments NX (10 + 5 + 5) = 20 (5) + 30 (5 + 5) NX = 20 kN (1 s.f.) 1 1 Q9 (a) Moment of a force about any point is defined as the product of the force and the perpendicular distance from that point to the line of action of the force. A couple are two forces of equal magnitude and acting in opposite directions whose lines of action are parallel but separate from one another. 1 1 1 1 (b) For a body to be in equilibrium, the following two conditions must be met: 1. No net force acts on the body. 2. No net moment of force about any point acts on the body. OR 1. The vector sum of all forces acting on the body must be equal to zero 2. The vector sum of all moments of force acting on the body about any point must be equal to zero 1 1 5 m 5 m 10 m 20 kN 30 kN NA NX A X

4 Note: “Total clockwise moments = Total anticlockwise moments” is NOT acceptable because this only covers 2-D cases. (c) (i) 2 key words : labelled and vector diagram Let T be the pull exerted by the cable, W the weight of the section S and FB the force exerted by B on S which is specified as a horizontal force in the question. Note:  The length of the arrows drawn must reflect the magnitude of the forces. Indicate the directions.  Vector diagram shows a closed polygon since forces are in equilibrium.  Question did not ask for a free-body diagram but a vector diagram. Thus a free-body diagram is not acceptable here. 1 1 (ii) From the vector triangle, T = W / sin 25° = (3.0 x 105)/ sin 25º = 7.1 x 105 N (2 s.f.) FB = W / tan 25° = (3.0 x 105)/ tan 25° = 6.4 x 105 N (2 s.f.) 1 1 1 1 (iii) All three lines of force will meet at the same point. If the free-body diagram is drawn, since locations of T and W are given by the question, location of FB is deduced to act along the upper surface as shown: 1 W T FB

5 Q10 (a) It is always helpful to draw out the diagram of what you are dealing with if the diagram is not given. RB is a normal contact force (the force is perpendicular to the wall) since the wall is smooth and there is no frictional force. Let RA be the resultant contact force on ladder by ground. RA consists of both vertical and horizontal components because the ground is rough and thus both normal contact force and friction act on end A. Free Body Diagram of Ladder with Man: ** Check that you have drawn in the Man in the free -body diagram & showed clearly that the weight of the man acts on the man, not the ladder. Since ladder is in equilibrium, Taking moments about A, by Principle of Moments, Sum of clockwise moments = Sum of anti-clockwise moments   3(20)(9.81) 1.5 (70)(9.81) (3) (4)4 460 B B R RN    1 1 (b) Balancing vertical forces, Vertical component of RA = (20 x 9.81) + (70 x 9.81) = 883 N Balancing horizontal forces, Horizontal component of RA = RB = 460 N Magnitude: 1 RB Wladder Wman RA A B 5 m 4 m 3 m 3 m θ

6 NRA 1000883460 22  (2 s.f.) Direction: 883tan 460 63 , anticlockwise above the horizontal       OR For vertical equilibrium of forces, RA sin θ = (20 x 9.81) + (70 x 9.81) RA sin θ = 883 ---------(1) For horizontal equilibrium of forces, RA cos θ = RB RA cos θ = 460 --------(2) (1)/(2): 883tan 460 63 , anticlockwise above the horizontal       Sub into (1) or (2): RA = 1000 N (2 s.f.) 1 1 Q11 Ans: D PX = Patm + ρgh (h: vertical height of liquid column) = 100kPa + (1020)(9.81)(1.00 sin 30°) = 105 kPa Q12 Ans: D Upthrust = Difference in forces acting on the top and bottom surfaces = F2 – F1 = P2A