HCI Assignment 06 Answer Turning effect of forces

Turning effects of forces 1 Name: Index No: Date: Class: Assignment 06: Turning effect of forces 1. The diagram below shows a man of weight 800 N standing in the middle of a plank supported at the sides. The plank weighs 1000 N. Which one of the following diagrams shows the forces acting on the plank? ( A ) 2. The diagram shows a balancing toy pivoted on a triangular stand . When the toy is tilted slightly, it stays at the new position. Where is the centre of gravity of the toy? ( D ) 3. A thin uniform sheet of metal is cut into an irregular shape. It is 380 mm long, weighs 6.00 N and is freely suspended a t point P which is halfway along it s length. A 5.30 N weight hung at Y makes it balance and in equilibrium as shown in the diagram. What is the perpendicular distance from the line of action of its weight to the point P? A. 3.7 mm B. 130 mm C. 170 mm D. 540 mm ( C ) HWA CHONG INSTITUTION Sec 3 Physics 5.30 N D B C

Turning effects of forces 2 4. A person exerts a horizontal force on a rectangular book such that it is just tilted about point O as shown in the di agram below. If the book has a weight of W, what is the moment of the force that is produced by the person? A.  W a 2 B.  W b 2 C.  Wa D.  Wb ( B ) 5. A force F of magnitude 8.20 N acts on a non -uniform rod of weight 10.0 N and length 1.00 m freely hinged to a wall at pivot P, such that the rod remains horizontal. F 45O pivot weight What is d, the distance of the centre of gravity of the rod from P? A. 0.18 m B. 0.42 m C. 0.58 m D. 0.82 m ( C ) 6. For each of the following cases, calculate the moment of the force, F about the pivot P. [4] (a) (b) d O 0.80 m F = 12.3 N 55  P F = 300 N 60  0.75 m P [1]: Moment of F about P = 12.3 X 0.80 sin 55o [1]: = 8.1 N m, clockwise Common mistake for (a) and (b): • Type of moment (i.e. clockwise or anticlockwise) is not mentioned [1]: Moment of F about P = 300 sin 60o X 0.75 [1]: = 190 N m, anti-clockwise a b

Turning effects of forces 3 7. A uniform metre rule of weight 0.9 0 N is suspended horizontally by two vertical loops of thread s A and B placed at 20.0 cm and 30.0 cm from its ends respectively as shown in the diagram below. Calculate the distance d from the centre of the rule, C, at which a 2.0 N weight should be suspended to balance the metre rule and keep it in equilibrium while (a) making loop A slack, [2] (b) making loop B slack. [2] State if the distance d is to be on the left side or the right side of C. (a) If loop A is slack, there is no tension in the loop. Upward forces acting on rule: Tension in B Downward forces acting on rule: Weight of metre rule; 2.0 N weight If B is pivot, 2.0 N weight must be on right side of B to balance rule. Taking moments about B, sum of clockwise moments = sum of anticlockwise moments 2.0 X (d – 20.0) = 0.90 X 20.0 [1] d = 29 cm, on the right of C [Accept: 29.0 cm] [1] (b) If loop B is slack, there is no tension in the loop. Upward forces acting on rule: Tension in A Downward forces acting on rule: Weight of metre rule; 2.0 N weight If A is pivot, 2.0 N weight must be on left side of A to balance rule. Taking moments about A, sum of clockwise moments = sum of anticlockwise moments 0.90 X 30.0 = 2.0 X (d – 30.0) [1] d = 43.5 cm d = 44 cm, on the left of C [Accept: 43.5 cm] [1] Common mistakes: • d in (a) is not added to 20.0 cm, thus answer obtained was 9.0 cm. Likewise, d obtained for (b) was 13.5 cm. • d in (a) and (b) were swapped as student wrongly deduced zero tension in slack loop. 20.0 cm 30.0 cm A B 2.0 N d C 20.0 cm 20.0 cm 30.0 cm A B 2.0 N d C

Turning effects of forces 4 8. The diagram below shows the mechanism of a weighing machine. When the 50 N weight is placed on the plate , the spring extends until the beam is horizontal. A pointer attached to the spring moves along the scale. When the weight is removed from the plate , the spring returns to its original length (suggesting plate’s and beam’s weights are negligible). (a) Determine the moment of the force about the pivot due to a 50 N weight placed on the plate . [2] Moment of force about pivot = 50 N x 5.0 cm = 250 N cm (or 2.5 N m) [1], clockwise [1] Common mistake for (a): • wrong unit or missing type of moment (b) What is the direction of the force acting on the beam due to the spring? _Upwards_______ [1] (c) What is the direction of the force acting on the spring due to the beam? _Downwards_____ [1] Common mistake for (b) and (c): • wrongly described force as clockwise / anticlockwise; a force is a vector which acts in one direction, thus it does not rotate CW or ACW. Do not confuse force with torque! (d) Is the magnitude of the force in the spring greater than, smaller than or equal to 50 N? (Ignore the weight of the beam and the plate.) Explain your answer. [2] For rotational equilibrium (OR for clockwise moment to be equal to anticlockwise moment about same pivot), [1]: spring is further away from pivot than 50 N weight OR perpendicular distance between spring force and pivot is greater than that of 50 N weight and pivot, [1]: thus spring force is smaller than 50 N. Common mistake: • missing details in description, e.g. ’50 N’, ‘perpendicular’, ‘pivot’ (e) The plate is now moved along the beam towards the spring. The rest of the weighing machine remains unchanged in design. Explain why the pointer reading obtained is higher than it should be when the 50 N weight is placed on the plate for weighing. [2] [1]: Moving the plate results in a larger clockwise moment of weight about pivot. [1]: To achieve rotational equilibrium (OR to produce a larger anticlockwise moment about the same pivot to balance this), the spring force increased since distance between spring and pivot is unchanged. 5.0 cm 50 N weight plate pivot beam pointer scale fixed point spring 0

Turning effects of forces 5 9. The diagram shows a uniform rectangular block maintained at an angle  by a horizontal force F. (a) What is the maximum angle, max, the block can rotate in the direction of F without toppling? Explain your answer. [2] [1]: 90o – 30.96o = 59o (to 2 s.f.) [1]: At this angle, the line of action of weight passes through pivot OR the block’s centre of gravity is directly above pivot. Common mistakes: • use ‘gravity’ instead of ‘weight’ • use ‘centre of mass’ instead of ‘centre of gravity’ • poor description e.g. centre of gravity is perpendicular to pivot. (b) From the position shown in the above diagram , describe and explain the motion of the block if F were to be removed suddenly. [2] [1]: Block rotates clockwise about pivot and falls back to original position , with its 5.0 cm side touching the table surface. [1]: When F was removed, the block’s weight, which acts from its centre of gravity, lies within its base area and produces a clockwise moment about the pivot. Common mistake: • misinterpret ‘toppling’ as ‘falling back to original position’; ‘toppling’ refers to scenario where object falls further away from origin