Forces Question Bank

Forces Question Bank Question 1. Distinguish between mass and weight. Answer:   Question 2. Define pressure. Answer: Pressure is the force acting normally on unit area of the surface.   Question 3. Define the term 'torque'. Answer: The turning effect produced by a force on a rigid body about a point, pivot or fulcrum is called the moment of a force or 'torque'. Question 4. Calculate the resultant moment of forces about O and state its direction in figure below.   Answer: Sum of the clockwise moments = (10 x 5 + 4 x 6) = 74 N m Sum of the anticlockwise moments = 15 x 4 = 60 N m Moment of the force due to 100 N = 0 (distance from the line of action of the force to the point O being zero). Resultant moment = 74 - 60 = 14 N m in the clockwise direction. Question 5. State the principle of moments. A metre scale is pivoted at the 30 cm mark and it is in equilibrium, when a mass of 40g is suspended from the 10 cm mark. Calculate the mass of the ruler. Answer: Weight of the scale acts at the centre of gravity of the scale i.e., at the mid point 50 cm mark. Taking moments about X, 40 x (30 - 10) = w x (50 - 30) mass of scale = 40 g The principle of moments states that if a body is in equilibrium under the action of a number of forces, then the algebraic sum of the moments of the forces about any point is equal to zero. In other words, the sum of clockwise moments equals the sum of the anticlockwise moments when the body is in equilibrium. Question 6. A metre scale is pivoted it its mid point when the various masses are suspended on it as shown in figure below, From which point will you suspend a 50g mass in order to keep the ruler in equilibrium? Anticlockwise moment = 80 x (50 - 20) = 2400 dyne cm Clockwise moment = 40 x (60 - 50) = 400 dyne cm Answer: Since the anticlockwise moment is greater, the 50g mass should be suspended on the same side of the pivot as the 40gf. to keep the ruler in equilibrium. Suspend the 50g mass at a distance 'x' cm away from the 50cm mark. Then 80 x 30 = 40 x 10 + 50 x 2400 - 400 = 50 x = 40 cm. The 50g has to be suspended at the (50 + 40) = 90 cm mark. Question 7. Define 'Centre of gravity'. Answer: Centre of Gravity of a body is the point through which the weight of the body acts irrespective of the position of the body. Question 8. What are the different methods by which you can increase the stability of a body? Answer: The stability of a body an be increased (i) by increasing the area of the base of the body, so that the vertical line through the centre of gravity should fall within the base of support even when the body is tilted. (ii) The centre of gravity should be kept as low as possible. This can be achieved by increasing the mass of the body at the bottom. Question 9. (i) What do you understand by the term couple of forces? (ii) Calculate the moment of a couple shown in figure.   Answer: (i) Two equal and opposite parallel forces acting along different lines, on a body, constitute a couple. A couple produces a turning effect on the body. This turning effect is called the Moment of the couple. (ii) Moment of a couple = Force (N) x Perp. distance (m) between the forces, Moment of the couple = 4 x 10 = 40 N m. Question 10. Give scientific reasons for the following:- (i) Even though the tower of Pisa is leaning through a greater angle it does not fall. (ii) While climbing a hill you will try to bend your body forward. (iii) In a moving bus, the standing passenger stands keeping both his legs apart. (iv) In a double-decker bus passengers are not allowed to stand in the upper deck. Answer: (i) (a) The base of the tower is larger than the rest of the tower. Hence the centre of gravity falls within the base of support. (b) The centre of gravity is also low, because the mass of the lower part of the tower is more than that of the upper part. (ii) The sides of a hill are sloping. By leaning forward, you adjust your centre of gravity, so that the vertical line through the C.G falls within your feet, which is the base of support. (iii) Keeping both his legs apart increases the base of support, so that the vertical line through the C.G will always falls within it thus increasing the passengers stability. (iv) If passengers stand in the upper deck, the C.G will move up making the bus unstable, thus creating a tendency to topple when taking a sharp turn. When the C.G. is raised, the vertical line through the C.G will fall outside the base of support, when the bus is tilted, as when taking a sharp turn. Question 11. Three forces A, B and C are acting on a rigid body which can turn about O in figure below. If all the three forces are applied simultaneously, in which direction will the body move? Explain.   Answer: A will give an anticlockwise moment = 60 x 1 = 60 N m B will give an anticlockwise moment = 20 x 2 = 40 N m C will give a clockwise moment = 40 x 3 = 120 N m Since the clockwise moment of 120 N m is greater than the sum of the anticlockwise movement of (60 + 40) = 100 N m, the body will turn in the clockwise direction. Question 12. The diagram shows a uniform metre scale weighing 100 N pivoted at its centre. Two weights of 500 N and 300 N are hung from the ruler as shown in figure. (i) Calculate the total clockwise and anticlockwise moments. (ii) Calculate the difference in clockwise and anticlockwise moments. (iii) Calculate the distance from O where a 100N weight should be suspended to balance the metre scale. Answer: (i) Since the scale is pivoted at the centre of gravity, the weight of the scale of 100N has no moment. Clockwise moment = 300 x (80 - 50) = 9000 N m Anticlockwise moment = 500 x (50 - 40) = 5000 N m (ii) Difference between moments = 9000 - 5000 = 4000 N m (iii) Since the anticlockwise moment is less, the 100 N weight should be on the same side of the pivot as the 500 N weight. Let the 100 N weight be suspended at a distance of 'x' cms from O. Then 100(50 - x) + 5000 = 9000 100x = 10,000 - 9000 = 1000 100 N should be suspended at a distance of 10 cm from O. Question 13. A metre scale is pivoted at 10 cm mark and is balanced by suspending 400 g from 0 cm mark (Figure below). Calculate the mass of the metre scale. Answer: The weight of the metre scale acts at the centre i.e., 50 cm mark 400 x 10 = w x (50 - 10) = 100 gf. Mass of scale = 100g Question 14. Figure below shows a uniform metres scale weighing 200 gf pivoted at its centre. Two weights 300gf and 500gf are suspended from the ruler as shown in the diagram. Calculate the resultant torque of the ruler and hence calculate the distance from the mid point where a 100gf should be suspended to balance the meter scale. Answer: Moment due to 200gf is zero since it is acting at the 50 cm mark. Anticlockwise moment = 300 x (50 - 10) = 12000 Nm Clockwise torque = 500 x (70 - 50) = 10000 Nm. Resultant torque = 12000 - 10000 = 2000 N m in the anticlockwise direction. Since the clockwise moment is less, the 100gf should be suspended on the same side as the 500gf. Say 'x' cms from the pivot. Then 500 x 20 + x x 100 = 300 x 40. 100 x = 2000 x = 20 cms. 100 gf weight should be suspended at the 70 cm mark. Question 15. A metre scale is pivoted at its mid point and a 50g mass suspended from the 20 cm. Mark what mass balances the rule when suspended from the 65m mark? Answer: Let mass to be suspended = m(g) m x 15 = 50 x 30 [Clockwise moment = Anticlockwise moment] mass to be suspended = 100g. Question 16. What are the conditions for stable equilibrium? Answer: The conditions for stable equilibrium are (i) The body should have a broad base (ii) The centre of gravity of the body should be as low as possible. (iii) The vertical line drawn through the centre of gravity should fall within the base of support. Question 17. Figure below shows the dimensions of an acute angled . By geometrical construction mark the C.G of the D.   Answer: The point of intersection of the medians is the centre of gravity of the triangle.   Question 18. A right angled triangular card board piece is placed as shown in figure below, Redraw the diagram showing the relative position of the vertices of the D when it is suspended by a pin from the hole A. Explain why the position changes.   Answer: The position of the vertices of the will be as shown in figure below. This is because when suspended from the vertex A, the vertical line through A should pass through the C.G which is the point of interaction of the medians of the .   Question 19. Give scientific reasons for the following: (i) It is easier to push a boy standing on one leg than on both legs. (ii) When a man climbs a slope he bends forward. (iii) There are chances to toppling when a truck takes a sharp turn specially when it is fully loaded. Answer: (i) A boy standing on both feet has a larger base area and hence is in more stable equilibrium than a boy standing on one foot. Therefore it is easier to push a boy standing on one leg than on both legs. (ii) The man has to bend forward in order to keep the vertical line passing through the C.G between his feet all the time. (iii) When a truck is fully loaded, its centre of gravity will be raised, thus making it unstable. When taking a sharp turn the vertical line through the C.G will fall outside the base of support, and the truck may topple over. Question 20. Define Force. State its unit in the C.G.S and S.I systems. Is it a scalar a vector quantity? Answer: Force is that physical quantity which changes or tends to change the dimensions or state of rest or of uniform motion in a straight line or in the direction of motion of the body. Units in C.G.S system = Dyne (g cm s-2) In S.I System = Newton (kg m s-2) Force is a vector quantity. Question 21. State the effects of a force when applied on an object, giving an example of each effect. Answer: (i) Force can produce a change in the dimension of the object. Ex., Stretching a spring by a load produces a change in length. (ii) It can start or stop the motion of an object. Ex., A ball lying on the ground moves when kicked. A fielder stops a moving ball by applying force with his hands. (iii) It can change speed or direction of motion or both. Ex., A player applies force by a hockey stick to change the speed and direction of motion of the ball. Question 22. Define the Newton. Derive its relationship to the Dyne. One Newton is the force which produces an acceleration 1 m s-2 in a body of mass 1kg. Answer: 1 Newton = 1kg x 1m s-2 = 1000 g x 100 cm s-2 1 Newton = 105 Dyne Question 23. What is the gravitational unit of Force in S.I system? How is it related to the Newton? Answer: The gravitational unit of force is the Kilogramme force (kgf) 1kgf = 9.8 newton Question 24. What is 'Friction'? Answer: The opposing force which acts when a body moves or tends to move over another is known as friction. Question 25. State two factors on which the force of friction depends. Answer: The force of friction depends upon:- (i) The nature of the two surfaces in contact (ii) The normal reaction. Question 26. Name the different types of friction and arrange them in descending order or magnitude. Answer: The three types of friction are Static friction, Kinetic or Dynamic friction and Rolling friction. In descending order of magnitude:- Static friction > Kinetic or Dynamic friction > Rolling friction. Question 27. Give two advantages of friction. Answer: (i) Friction helps us to walk. (ii) It helps in supporting ladders inclined to a wall Question 28. Give two disadvantages of friction. Answer: (i) Friction reduces the efficiency of a machine. (ii) There will be greater wear and tear in the different parts of a machine due to friction. Question 29. What are the ways of reducing friction? Answer: (i) By using Lubricants (ii) By using ball - bearings. (iii) By polishing the surfaces (iv) By using anti-friction metals (v) By separating the surfaces by an air cushion (vi) By streamling the shape of the body. Question 30. Explain the role of lubricants in reducing friction. Answer: Friction is caused because of the unevenness of the surfaces in contact. The lubricants fill up the depressions in the surfaces making them smooth and thus reducing friction. Question 31. State two ways of increasing friction. Answer: (i) By increasing the roughness of the surfaces in contact. (ii) By increasing the weight of the moving body Question 32. Why is it easier to tie a knot in a jute rope as compared to a silk rope? Answer: A jute rope being rough offers more friction than a silk rope which is smooth and hence offers less friction. Thus it is easier to knot a jute rope. Question 33. On what does the speed of a body moving through a fluid, depend? Answer: The speed of the body depends on the nature of the fluid and the shape of the body. Question 34. Explain why a hovercraft (a vessel which floats on a cushion of air) travels much faster than a streamer pushing through water. Answer: A hovercraft travels much faster than a steamer because the resistance offered by air is lesser than the resistance offered by water, which is more viscous than air. Question 35. What is streamlining? What is the benefit of streamlining bodies? Answer: Streamlining means to give a body a tear drop shape with its front end pointed. Streamlining helps bodies to move faster in air or water, as the pointed end minimises fluid friction. Question 36. Why is it easier to ride a bicycle when its tyres are fully inflated? Answer: When the tyre is fully inflated it is circular in shape with very little flattering at the point of contact with the road, hence there is rolling friction. When it is not fully inflated, there is a larger flattened surface in contact with the road, producing sliding friction. Since rolling friction is lesser in magnitude than sliding friction, a fully inflated tyre makes riding easier. Question 37. State two factors on which the moment of a force about a point depends. Answer: The moment of a force depends upon (i) The magnitude of the force (ii) the perpendicular distance of the line of action of the force from the given point. Question 38. The hand flour grinder is provided with a handle near its rim, explain the reason. Answer: The handle is placed at the rim so that the perpendicular distance of the line of action of the force from the axis of rotation will be maximum, so as to get the largest turning effect. Question 39. Give two examples of couple action in daily life. Answer: (i) Opening the cap of an ink bottle (ii) Closing a water tap Question 40. State the kind of equilibrium in the following: (i) A sleeping man (ii) A man standing on one leg (iii) A ball on the ground (iv) A cone resting on its base (v) A funnel lying on its side (vi) A bottle lying on its side Answer: (i) Neutral equilibrium (ii) Unstable equilibrium (iii) Neutral equilibrium (iv) Stable equilibrium (v) Unstable equilibrium (vi) Neutral equilibrium.