Fig. 2.1 shows a computer resting on a tabletop that is hunged at A. The tabletop has a mass of 5.0kg and its centre of gravity is 0.40 m from the axis of the hinge at A. The computer has a weight of 200 N acting through a point 0.25m from the hinge at A. The tabletop is supported to maintain it in a horizontal position by a force of F acting vertically at B. The distance AB is 0.80 m. a) Calculate the weight if the tabletop. weight = ................ N (1 Mark) b) On Fig. 2.1 draw and label an arrow to reprsent the weight W of the tabletop. (1 Mark) c) Apply the principle of momentd about the hinge at A to determine the vertical force F applied at B that is required to maintain the tabletop in equilibrium. force = F .................... N (3 Marks) d) The tabletop must experience a resultant force of zero in order to be in equilibrium. Explain how the forces acting on the tabletop fulfill this condition. (2 Marks) (Marks available: 7) Fig. 4.1 shows the open bonnet of a car. The bonnet is held open at an angle of 60° to the horizontal by a vertical force V applied at one end of the bonnet (shown on the diagram). The bonnst is 0.90 m long, has a weight of 25 N and its centre of gravity G is 0.35 m from the hinge at O. a) On Fig. 4.1, draw and label the two forces other than V acting on the bonnet. (2 Marks) b) By taking moments about O, show that the vertical force V applied at the end of the bonnet is 9.7 N. (2 Marks) c) Calclate the magnitude of the force acting at the hinge O. Show your working. force at hinge = ............................. N (2 Marks) (Marks available: 6) Answer Answer outline and marking scheme for question: a) W = 5 x 9.81 = 49 N (allow g = 10) (1 Mark) b) arrow acting down (labelled W) drawn approximately halfway from A to B (1 Mark) c) any correct moment F x 0.8 = 200 x 0.25 + 49 x 0.4 F = 87 N (3 Marks) d) upward force acts at the hinge So F and force at hinge equals weight of table and computer (allow one mark for the upward forces equal the downward forces) (2 Marks) (Marks available: 7) a) W vertically down at G Force at O vetical (2 Marks) b) By taking moments about O, show that the vertical force V applied at the end of the bonnet is 9.7 N. V x 0.9 x cos60 = W x 0.35 x cos60 V = (25 x 0.35) / 0.9 = 9.7 (22) (N) (2 Marks) c) Total force is zero stated or implied / 25 - 9.7 force at hinge = 15.3 (N) (or may take moments about G or V) (2 Marks) (Marks available: 6)