Jurong Junior College 2014 J1 H1 Physics (8866) Tutorial 3: Forces (Solutions) Take g = 9.81 m s -2, P atm = 1.0 x 10 5 Pa unless otherwise stated Learning Outcomes (a) Sub-Topic recall and apply Hooke s law to new situations or to solve related problems deduce the elastic potential energy in a deformed material from the area under the force-extension graph describe the forces on mass, charge and current in gravitational, electric and magnetic fields, as appropriate. Tutorial Questions 6(a), 15(a) 6 Not directly tested (d) solve problems using the equation p = ρgh 7 (e) show a qualitative understanding of frictional forces and viscous forces including air resistance (No treatment of the coefficients of friction and viscosity is required) 9(a),, (f) use a vector triangle to represent forces in equilibrium 8, 9 (g) (h) (i) (j) (k) show an understanding that the weight of a body may be taken as acting at a single point known as its centre of gravity show an understanding that a couple is a pair of forces which tends to produce rotation only define and apply the moment of a force and the torque of a couple show an understanding that, when there is no resultant force and no resultant torque, a system is in equilibrium apply the principle of moments to new situations or to solve related problems 9(a) 9(a) 9(a) 2(a), 3, 4, 5, 9(a) 2, 3, 4, 5, KPL/2014 Page 1 of 11
1. Draw the free body diagram / force diagram for the following : (a) Drag Skydiver Upthrust (negligible) weight Coin Book θ of coin Normal contact force by book on coin Normal contact force by coin on book Normal contact force by table on book of book 110 N mass man s left leg w 2 mass Tension 1 Tension 2 Tension 1 Tension 2 40 α 110 N w 1 w 2 (d) 1.0kg mass 2.0 kg mass 3.0 kg mass N 1 Tension Contact force by 1kg on 2kg N 2 Tension N 3 42 N Contact force by 2kg on 1 kg 2 3 (e) 5.0 kg mass 4.00 kg mass 3.00 kg mass Tension 1 Tension 1 Tension 2 4 kg 3kg of 5 kg of 4 kg Tension 2 of 3 kg (f) Left Box Right Box F N N F Friction (force on box by floor) Friction (force on box by floor) KPL/2014 Page 2 of 11
(g) 2.0 kg mass 3.0 kg mass 10.0 kg mass Normal Contact force by 3 kg on 2 kg T 2 T 1 of 2 kg Friction between 2 kg & 3 kg Normal Contact force by floor on 3 kg of 10 kg T 1 Friction between 3 kg & floor of 3 kg T 2 Force on 3 kg by 2 kg 2. (a) State the conditions necessary for a rigid body to be in equilibrium. [2] Fig. 2.1 shows an object M of mass 20 kg is supported by a hinged uniform rod of mass 10 kg and string. wall string Fig. 2.1 45 0 hinge 1.5 m M = 20 kg 2.0 m (i) Two of the forces acting on the rod are shown in Fig.2.2. Sketch the remaining two forces acting on the rod by drawing & labelling them clearly on Fig.2.2. [2] A F, force on rod by wall W, weight of the rod Fig.2.2 By taking moment about hinge A, calculate (ii) the tension in the string, [2] (iii) both the horizontal component and the vertical component of the force F acting on the rod by the wall. [4] KPL/2014 Page 3 of 11
Solutions 2. (a) The conditions are Vector sum of all forces acting on a rigid body must be zero. Vector sum of all torques acting on a rigid body must be zero. (i) tension T F, force on rod by wall W, weight of rod weight of mass M (ii) Taking moment about the hinge, 10 x 9.81 x 1.0 + 20 x 9.81 x 2.0 = T sin45 0 x 1.5 T = 462 N (iii) Net force horizontally = 0 F x = T cos 45 0 = 462 x cos 45 0 = 327 N F y F x 45 0 T Net force vertically = 0 462 sin 45 0 = 20 x 9.81 + 10 x 9.81 + F y F y = 32.4 N 3. (a) A small ball of weight W is suspended by a light thread. When a strong wind blows horizontally exerting a constant force F on the ball, the thread makes an angle θ to the vertical as shown in the diagram. State an equation relating θ, F and W. θ F W A body of mass 1.50 kg is placed on a plane surface inclined at 30 o to the horizontal. Calculate the friction and normal reaction forces which the forces must exert if the body is to remain at rest. [7.36 N; 12.7 N] A uniform rod XY of weight 10.0 N is freely hinged to a wall at X. It is held horizontal by a force F acting from Y at an angle of 60 o to the vertical as shown in the diagram. Find the value of F. [10 N] KPL/2014 Page 4 of 11
Solutions 3 For objects in equilibrium, F = 0 (a) F x = 0 Tsin θ = F (1) 30 o F y= 0 Tcos θ 30 o = W (2) T F (1)/(2): F tanθ = W F = W tanθ W For objects in equilibrium, F = 0 F x = 0 (x-axis along slope) F f = W sin30 o = (1.50)(9.81)sin30 o = 7.36 N F f F N W sin30 o F x = 0 (axis perpendicular to slope) F N = W cos30 o = (1.50)(9.81) cos30 o = 12.7 N W cos30 o 30 o 30 o Let x = length of rod Taking moments about hinge, anticlockwise moment = clockwise moment Fcos60 o (x) = 10 (x/2) F = 10 N F N F f F cos60 o x/2 60 o 10 N F sin60 o 4. A trailer of weight 30 kn is hitched to a cab at the point X as shown in the diagram below. If the trailer carries a weight of 20 kn at the position shown in the diagram, calculate the upward force exerted by the cab on the trailer at point X. [20 kn] Solutions 4. Taking moments about F, anticlockwise moment = clockwise moment F x (20) = 20k(5) + 30k(10) F x = 20 kn F 5 m 10 m F x 20kN 30kN KPL/2014 Page 5 of 11
2010 / RI / H2 / P2 Q2 5 (a) A mass hanging from a spring balance in air gives a reading of 50 N. When the mass is completely immersed in water, the reading on the balance is 40 N. It is now completely immersed in another liquid, giving a reading of 34 N. Calculate the density of this liquid. Assume that the density of water is 1000 kg m -3. In Fig. 5 below, a uniform beam of length 10.0 m and weight 500 N is hinged to a wall at point O. Its far end is supported by a cable that makes an angle of 53.0 with the horizontal. A 70.0 kg worker stands on the beam. cable O s Fig. 5 53.0 beam (i) Draw a labelled diagram showing the forces acting on the beam. (ii) (iii) The worker walks towards the far end of the beam from O. Calculate the furthest distance s he can walk if the maximum possible tension in the cable is 1000 N. Calculate the magnitude of the force exerted by the hinge on the beam when the tension in the cable is 1000 N. Solution 5 (a) U = W -T 1 1 = 50-40 = 10 N 10 = ρvg V 1 10 = 1000 9.81 U = 1.02 10 m 2 =50-34 = 16 N -3 3 16 ρ = 2-3 1.02 10 9.81 = 1600 kg m -3 KPL/2014 Page 6 of 11
(i) reaction force tension weight of beam force by worker on beam (ii) (iii) Taking moments about O, 5.00 500+(70 9.81 s)=1000sin53.0 10.0 s =7.99 m At equilibrium, the net vertical and horizontal forces must be zero. Rx = Tcos53.0 = 602 N R = 500+(70 9.81)-Tsin53.0 = 388 N y R = 602 +388 =716 N 2 2 6 (a) A spring obeying Hooke s law has an unstretched length of 20 cm and a spring constant of 400 N m -1. What is the tension in the spring when its overall length is 40 cm? [80 N] Some weight-lifters use a chest expander, consisting of a strong spring with a handle at each end, to exercise chest and arm muscles. For one such chest expander obeying Hooke s law, the force F applied by the weight-lifter is proportional to the extension x of the spring. Given that the spring constant k is 400 N m -1 and assuming that the limit of proportionality is not exceeded, i) plot a graph showing how W, the work done depends on the extension x. ii) calculate the work done required to extend its length by 10 cm with a force of 40 N. Solution: 6 x = 0.2 m (a) F = kx = 400 (0.2) = 80 N *remember to convert the units of x to metres since k has units of N m -1 b (i) W = ½ kx 2 W/J x/m b (ii) x = 0.1 m, F = 40 N, k = 400 N m -1 W = ½ kx 2 = ½ (400)(0.1) 2 = 2 J W = ½ Fx = ½ (40)(0.1) = 2 J OR KPL/2014 Page 7 of 11
7 Calculate the actual pressure at the bottom of an open barrel of diameter 0.500 m filled with 300.0 kg of water and 200.0 kg of olive oil. Assume the 2 liquids are perfectly immiscible (i.e. insoluble). (Take density of water = 1000 kg m -3, density of olive oil = 920 kg m -3 ) [1.25 x 10 5 Pa] Solution: 7. The pressure at the bottom of the barrel due to the Olive oil & Water only, Force weight of both liquids P = = Area Area (300 + 200) g 500(9.81) = = 2 π d ( 4 ) π 0.500 ( ) 2 4 = 25 kpa Actual pressure at the bottom of the barrel = P + atmospheric pressure = 25 kpa + 100 kpa = 125 kpa Olive oil water Area 8. A cable car travels along a fixed support cable and is pulled along this cable by a moving draw cable. For the situation shown, where the cable car can be considered to be stationary and the draw cable exerts negligible force on it, the weight, W of the cable car and the passengers is 8.0 x 10 4 N. (a) Sketch a vector triangle to show the weight, W of the cable car and passengers and T 1 and T 2, the two cables. Find the magnitude of T 1. [136 kn] Solutions 13. (a) 26 o T 2 T 1 116 o 32 o W Cable car is stationary. i.e. in equilibrium, W, T 1, T 2 should form a closed. is isosceles with T 2 = W T 1 = 2W cos32 o = 2(8.0 x 10 4 )cos32 o = 1.36 x 10 5 N [8866 / Nov 2011 / Paper 2, Q5] 9. (a) Explain what is meant by the following terms when used in the context of forces. KPL/2014 Page 8 of 11
(i) equilibrium (ii) friction (iii) centre of gravity (iv) moment (v) torque of a couple A truck of mass 39 000 kg is stationary on a hill with a gradient of 15. The three forces acting on the truck are its weight, a frictional force parallel to the road and a reaction force normal to the road. (i) Draw a triangle of forces to show that the truck is in equilibrium. Label each force. [2] (ii) Determine the numerical value of each of the three forces, using the same labels as you used in (i). [3] Friction is often regarded as a nuisance. (i) State two different situations where friction is of critical importance. [2] (ii) For one of your examples explain why friction is so important. [1] Solutions 9. (a) (i) An object is said to be in static equilibrium if it is at rest, without any translational or rotational motion. The resultant force acting on the object and the resultant torque on the object are zero. (ii) Friction is a force that opposes relative motion. It acts along the common surface in contact between the two bodies. (iii) Centre of gravity of a rigid body is the single point through which the weight of the object appears to act. (iv) Moment of a force about a pivot is the product of the force and the perpendicular distance from the line of action of the force to the pivot. (v) Torque of a couple is the product of one of the force and the perpendicular distance between the two forces. (i) Friction Reaction force normal to road Friction 15 o 15 o Forces on car (not required by question) 15 o Reaction force normal to road Force Diagram (ii) = mg = (39 000)(9.81) = 383 kn Reaction force normal to road = mg cos 15 = (383 x 10 3 ) cos 15 = 370 kn Since Friction is Reaction force : By Pythagoras theorem : Friction = mg sin 15 = 99 kn (i) Braking of car to avoid an accident. To enable movement, e.g. walking, car moving on the road. Lighting of matches KPL/2014 Page 9 of 11
(ii) Braking of car Friction produces retardation to the motion of the car, so that its speed is reduced. This will enable the car to stop in time to avoid an accident. Enable movement (e.g. walking) When we want to walk forwards, the feet must push the floor back. If there is friction, the frictional force by the floor on the feet will be forwards & this force is what propels the body forward. However, if there is no friction, there will be no force that can act on the feet to move it forward. Lighting of matches Matches are lit when the match head is struck against a rough surface. This action produces enough heat, due to friction, to cause the chemicals in the match head to combust & light the matchstick. Optional Questions [Extra Practice] 1. A water wheel has eight buckets equally spaced around its circumference, as illustrated in the figure on the right. The distance between the centre of each bucket and the centre of the wheel is 1.6 m. When a bucket is at its highest point, the bucket is filled with a mass of 40 kg of water. The wheel rotates and the bucket is emptied at its lowest point. (a) (d) Define the moment of a force. Write down the number of the bucket that provides the largest moment about the axle of the wheel. [3] Write down the numbers of those buckets containing water that cause a moment about the axle. [2, 3, 4] Calculate, for the wheel in the position shown in the figure, the total resultant moment about the centre of the wheel of the water in the buckets. [1520 N m] Solution 1. (a) The moment of a force is defined as the product of the force and the perpendicular distance between the line of action of the force and the pivot. (d) Bucket 3. [Perpendicular distance from pivot is the largest] Buckets 2, 3 and 4. [perpendicular distance from pivot zero] Taking moments about the centre of the wheel, Total clockwise moment = Sum of moments due to buckets 2, 3 and 4 = mgrsinθ 2 + mgrsinθ 3 + mgrsin θ 4 = mgr ( sinθ 2 + sinθ 3 + sinθ 4 ) = (40)(9.81)(1.6)(sin45 o + sin90 o + sin135 o ) = 1515.7 = 1520 N m KPL/2014 Page 10 of 11
2. A uniform horizontal beam of mass 30.0 kg, 5.00 m long is attached to a wall by a hinge that allows the beam to rotate. Its far end is supported by a cable that makes an angle of 53.0 o with the horizontal. Wall Rope Box Hinge 53.0 º beam 5.00 m If a box of mass 60.0 kg is placed 1.50 m away from the wall, find the tension in the cable and the force exerted by the hinge on the beam. [405 N; 610 N] Solution 2. Taking moments about hinge, anticlockwise moment = clockwise moment T y (5) = W box (1.5) + W beam (2.5) T sin 53 o (5) = 60(9.81)(1.5) + 30(9.81)(2.5) T = 405.35 = 405 N F y 53 o T y F x = 0 F x = T x = 405.35 cos 53 o = 243.95 N F x T x 1.50 m 2.50 m W box F y = 0 F y + T y = W box + W beam F y = (30 + 60)(9.81) 405.35 sin 53 o = 559.17 N W beam F = F + F 2 2 x y 2 2 243.95 559.17 610 N = + = KPL/2014 Page 11 of 11