Mechanics 2. Revision Notes


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1 Mechanics 2 Revision Notes November 2012
2
3 Contents 1 Kinematics 3 Constant acceleration in a vertical plane... 3 Variable acceleration... 5 Using vectors Centres of mass 8 Centre of mass of n particles... 8 Centres of mass of simple laminas... 8 Centres of mass of composite laminas... 9 Laminas suspended freely under gravity Toppling on a slope Centres of mass of wire frameworks Work, energy and power 13 Definitions Work done by a (constant) force Forces parallel to the displacement Forces at an angle to the displacement Work done by gravity Workenergy equation Potential energy Power Collisions 19 Impulse and momentum using vectors Impulse = change in momentum Conservation of linear momentum Newton s law of restitution Coefficient of restitution Collisions with a plane surface Multiple collisions Kinetic energy and impulses/collisions Statics of rigid bodies 23 Moment of a force Equilibrium Three nonparallel forces in equilibrium Triangle of forces /04/2013 Mechanics 2 SD 1
4 ppendix 27 Centre of mass of n particles Medians of a triangle Centre of mass of a triangle /04/2013 Mechanics 2 SD
5 1 Kinematics Constant acceleration in a vertical plane We can think of the horizontal and vertical motion separately and use the formulae: v = u + at, s = ut + at2, v 2 = u 2 + 2as, s = (u + v)t Example 1: stone is thrown at a speed of 20 ms 1 at an angle of 35 o to the horizontal. Find (a) the greatest height reached (b) the direction in which it is moving after 1 second (c) the height of the stone after it has travelled a horizontal distance of 25 m. Solution: + (a) Vertical motion u = 20 sin 35 o, a = 9.8, v = 0, s = h v 2 = u 2 + 2as 0 = (20 sin 35 o ) h h = greatest height reached is 6 71 m to 2 S.F. 35 o 20 (b) Horizontal motion + u = 20 cos 35 o Vertical motion + u = 20 sin 35 o, a = 9.8, t = 1, v =? v = u + at = 20 sin 35 o = v tan 1 67 θ 20 cos 35 θ = stone is moving at 5 8 o above the horizontal, 2 S.F. (c) Horizontal motion + u = 20 cos 35 o, s = 25, a = 0 t = Vertical motion + u = 20 sin 35 o, a = 9.8, t = , s =? s = ut + at2 = 20 sin 35 o (1 525 )2 s = stone is at a height of 6 1 m after it has travelled 25 m horizontally 2 S.F. 14/04/2013 Mechanics 2 SD 3
6 Example 2: ball is projected from a point on horizontal ground with a speed of U, making an angle of α with the horizontal. (a) Show that the ball moves along a parabola. (b) Find the range. (c) If α is allowed to vary, find the maximum range. Solution: (a) Take the point of projection as the origin. Horizontal motion u = U cos α, s = x, t = T x = (U cos α) T Vertical motion u = U sin α, s = y, a = g, t = T s = ut + at2 y = (U sin α) T g T 2 I note that T is the same for both directions II + U α + I and II y = (U sin α) g y = x tan α which is a quadratic function of x, and so the ball moves in a parabola. (b) When the ball hits the ground, y = 0 range is solution of 0 = x tan α x tan α 0 x = 0 (the start), or x = α range is = (c) The range is = and, since the maximum value of sin θ is 1, the maximum range is, and occurs when α = 45o. 4 14/04/2013 Mechanics 2 SD
7 Variable acceleration When a is given as a function of t v = do not forget the +c s = do not forget the +c When s (or v) is given as a function of t v = and a = or a = Note that s is the displacement (the distance from the origin), which is not necessarily the same as the distance travelled (the particle may have moved forwards and backwards). Example 1: particle is moving along the xaxis with an acceleration 5 2t ms 2. t time t = 0, the particle moves through the origin with speed 6 ms 1 in the direction of the positive xaxis. (a) Find the displacement of the particle after 9 seconds. (b) Find the distance travelled in the first 9 seconds. Solution: (a) a = = 5 2t v = 5 2 = 5t t 2 + c v = 6 when t = 0, c = 6 v = 5t t s = 65 6 s = 0 when t = 0, c = 0 s = 6 When t = 9, s = = 13 5 The displacement after 9 seconds is 13 5 m. 14/04/2013 Mechanics 2 SD 5
8 (b) Note: the particle could have gone forwards then backwards, in which case the distance travelled would not be the same as the final displacement. We must first find t when the velocity is zero. v = 5t t = (6 t)(1 + t) v = 0 when t = ( 1) or 6. The particle is moving away from the origin for 0 t < 6, and towards the origin for 6 < t 9, so we want the sum of the two distances d 1 + d 2. When t = 6, s = d 1 d 1 d 1 = d 2 d 1 = 54 d 2 = = 40 5 O t = 9 t = 6 x total distance travelled = d 1 + d 2 = 94 5 m Using vectors This is just combining horizontal and vertical motion into one expression. Example 1: particle moves with velocity v = 3 4 ms1. It is initially at the point (6, 3). Find its acceleration after 2 seconds, and its displacement at time t. Solution: a = = = 6 4 at t = 2, a = 12 4 ms2 s = = 3 4 = 2 Particle is initially at (6, 3) = 6 3 s = m 6 14/04/2013 Mechanics 2 SD
9 Example 2: particle is initially at (1, 2) moving with velocity 4 3 ms1. second particle,, is initially at the origin, moving with velocity 2 2 ms1. Investigate whether and collide. Solution: For s = 4 2 = 3 3 s = 1 2 at t = s = For s = 2 = 2 s = 0 0 at t = s = If they collide, both x and y coordinates must be equal for the same value of t. The y coordinates are equal when 2 3t = t 2 t 2 3t + 2 = 0 (t 1)(t 2) = 0 t = 1 or 2. When t = 1, s = 3 1 and s = 1 1 s s When t = 2, s = 9 1 and s = 4 1 s s and do not collide. 14/04/2013 Mechanics 2 SD 7
10 2 Centres of mass Centre of mass of n particles The centre of mass, (, ), of n particles, which have masses m 1, m 2,, m n, at points (x 1, y 1,), (x 2, y 2 ),, (x n, y n ) is given by M = M = and M = where M is the total mass, M = or M g = Centres of mass of simple laminas 1) The centre of mass of a uniform rod is at its mid point. 2) The centre of mass of a uniform rectangular lamina (sheet) is at its point of symmetry. G 3) The centre of mass of a uniform triangular lamina (a) G is at the point where the three medians meet. G divides each median in the ratio 2:1 F G E (b) If, and C are the points (a 1, a 2 ), (b 1, b 2,) and (c 1, c 2 ), then the centre of mass, G, is at the point C D G ( (a 1 + b 1 + c 1 ), (a 2 + b 2 + c 2 )) or g = (a + b + c) 4) The centre of mass of a uniform sector of a circle with angle 2α Circle centre O with radius r r ngle of sector is 2α G lies on the axis of symmetry and O α α G OG = r 8 14/04/2013 Mechanics 2 SD
11 Example 1: uniform triangular lamina has mass 3 kg, and its vertices are O (0, 0), (5, 0) and (4, 3). Masses of 2, 4 and 5 kg are attached at O, and respectively. Find the centre of mass of the system. Solution: First find the centre of mass of the triangle, G T. y (4, 3) 5 g T = (a + b + c) = G T is (3, 1). O 2 3 G T (3, 1) 4 (5, 0) x We can now think of 4 masses, 2 kg at O, 4 kg at, 5 kg at and 3 kg (the triangle) at G T. Mg = ( ) g = g = G is at,. Centres of mass of composite laminas In the following examples ρ is the surface density, or mass per unit area. Example 3: Find the centre of mass of a rectangle attached to a semi circle, as shown in the diagram. O is the centre of the semicircle G 1 3 O G 2 C E D Solution: The centre of mass will lie on the line of symmetry, OC, so we only need to find the horizontal distance of the centre of mass from O. y symmetry the centre of mass of the rectangle is at G 1, as shown. The semicircle is a sector with angle 2 OG 2 = 2 2 = Note OG 1 = 3 and OG 2 = negative and positive 14/04/2013 Mechanics 2 SD 9
12 Rectangle Semicircle Composite mass 24ρ 2πρ (24 + 2π)ρ distance of G from O = M 24ρ ( 3) + 2πρ = (24 + 2π)ρ = = = = 2 20 The centre of mass lies on the line of symmetry, 2 20 on the left of O (inside the rectangle). Example 4: 20 Find the centre of mass of the uniform lamina CDE. FC is an axis of symmetry. C = CD. E = 10, = 20, FC = 14. F E G 2 C G H 5 D Solution: Think of the triangle CD combining with this shape to form the rectangle DE. Mass of rectangle = 200ρ, mass of triangle = 10 6 = 30ρ mass of shape = 170ρ The centre of mass lies on the line of symmetry, FH. Let G 2 be the centre of mass of the rectangle FG 2 = 10 Let G 1 be the centre of mass of the triangle. G 1 divides CH in the ratio 2 : 1 and CH = 6 CG 1 = 4 and G 1 H = 2 FG 1 = = 18 mass 170ρ 30ρ 200ρ distance of G from E g = M 170ρ g + 30ρ 18 = 200ρ g = = 1460 g = = = 8 6 to 2 S.F. The centre of mass lies on FC at a distance 8 6 from E /04/2013 Mechanics 2 SD
13 Laminas suspended freely under gravity Example 5: The lamina in example 4 is suspended from, and hangs in equilibrium. Find the angle made by with the downward vertical. Solution: G must be vertically below. This must be stated in any solution (method). From example 4 we know that FG = and F = 5 The angle made by with the downward vertical is G = FG = arctan = angle made by with the downward vertical is 30 2 o to the nearest 0 1 o. 5 F G Toppling on a slope Example 6: What angle of slope would cause a 4 6 uniform rectangular lamina to topple (assuming that the friction is large enough to prevent sliding). Solution: G must be vertically above when the lamina is on the point of toppling. 4 C This must be stated in any solution (method). G y angle theory GM = θ, the angle of the slope. 6 θ 3 tan θ = θ = θ 2 M The lamina will topple when the angle of slope exceeds 33 7 o, to nearest 0 1 o. Centres of mass of wire frameworks. In the following examples ρ is mass per unit length. 1) The centre of mass of a uniform straight wire is at its centre. 2) The centre of mass of a uniform circular arc, of radius r and angle at the centre 2α, lies on the axis of symmetry and OG =. O r α α G 14/04/2013 Mechanics 2 SD 11
14 Example 7: uniform wire framework is shown in the diagram. OC is a rectangle and the arc is a semicircle. G 3 O = 10 and OC = 4. Find the position of the centre of mass. G 1 G 4 O G 2 C Solution: With frameworks it is often easier to use vectors. Take the origin at O. The semicircle is an arc of angle 2 centre of mass is at G 4, where OG 4 = 2 2 semicircle has mass 5πρ at 4 10,5. = We now consider each straight wire as a point mass at the midpoint of the wire. O OC semicircle mass 10ρ 4ρ 4ρ 5πρ centre of mass Mg = (18 + 5π)ρ g = 10ρ ρ ρ 2 + 5πρ g = 66 = 3 = 3 82 to 3 S.F /04/2013 Mechanics 2 SD
15 3 Work, energy and power Definitions v 2 = u 2 + 2as mas = mv2 mu2 rearranging and multiplying by m Fs = mv2 mu2 We define The kinetic energy, K.E., of a body of mass m moving with speed v is mv2. Work done by a (constant) force F is the magnitude of the force distance moved in the direction of the force. The units of kinetic energy and work are Joules, J. Work done by a (constant) force. Forces parallel to the displacement (a) Work done by a force of 6 N moving from to ( = 3 m) in the direction of the force, is 6 3 = 18 J (b) Work done by a force of 6 N moving from to in the opposite direction to the force, is 6 ( 3) = 18 J 6 N 3 m 6 N 3 m Forces at an angle to the displacement Work done can be calculated in two ways (a) s a body moves from to, we can think that it has moved through the distance N in the direction of the force. N = s cos θ, work done is F s cos θ = Fs cos θ. (b) Or, we can resolve the force F parallel and perpendicular to the displacement. The component F sinθ is perpendicular to the displacement, and so does no work. The component F cosθ is parallel to the displacement, work done is F cosθ s = Fs cos θ, as in part (a). F N θ s m F sinθ N N F cosθ N s m 14/04/2013 Mechanics 2 SD 13
16 Work done by gravity If a particle of mass m falls a vertical distance h, then the work done by gravity is mgh, force and displacement are in the same direction If a particle of mass m rises a vertical distance h, then the work done by gravity is mgh, force and displacement are in opposite directions When a particle is moving on a slope, it is usual to consider the vertical distance moved and multiply by mg to calculate the work done by gravity. s From to the particle moves a distance s, but its vertical movement is h = s sinθ θ mg h work done by gravity = mgh = mgs sinθ Workenergy equation The equation Fs = mv2 mu2 can be rearranged as mv2 = mu2 + Fs which can be thought of as Final K.E. = Initial K.E. ± Work done Notice the ± If a force increases the K.E. then the work done is positive If a force decreases the K.E. then the work done is negative 14 14/04/2013 Mechanics 2 SD
17 Example 1: particle of mass 5 kg is being pulled up a rough slope by a force of 50 N parallel to the slope. The coefficient of friction is 0 2, and the slope makes an angle of α = tan 1 with the horizontal. The particle is observed to be moving up the slope with a speed of 3 ms 1. Find its speed when it has moved 12 m up the slope. v Solution: = 12 h = 12 sin α = 6 Resolve R = 5g cos α = 4g 3 F R 5g 50 h Moving F = μr = 0.2 4g = 0 8g Work done by R = 0 ( to motion) α Work done by F = 0 8g 12 = 9.6g J reduces K.E. so negative Work done by 50 N = = 600 J increases K.E. so positive Work done by gravity = mgh = 5g 6 = 30g uphill, reduces K.E. so negative Workenergy equation Final K.E. = Initial K.E. ± Work done 5v2 = g g v 2 = v = = 9 7 to 2 S.F. Particle is moving at 9 7 ms 1 after it has travelled 12 m. Example 2: Tarzan swings on a rope, lets go and falls to the ground. If Tarzan was initially 7 m above the ground and not moving, and if he lets go when he is 3 m above the ground, with what speed does he hit the ground? Solution: The only forces acting on Tarzan are the tension in the rope, T, and gravity, mg. The work done by T is 0, since T is always perpendicular to the motion 7 m mg T 3 m Work done by gravity = mgh = mg 7 downwards, increases K.E. so positive Workenergy equation Final K.E. = Initial K.E. ± Work done mv2 = 0 + 7mg v 2 = 14g v = = 12 ms 1 to 2 S.F. 14/04/2013 Mechanics 2 SD 15
18 Note that it does not matter when he lets go of the rope. This will only affect the direction in which he is moving when he hits the ground, not his speed. Example 3: Tarzan now goes skiing in Switzerland. It is much colder than frica so he is wearing lots of warm clothes and his mass is 90 kg. He starts with a speed of 3 ms 1 and skis along a path as shown in the diagram when he comes to a cliff. The total length of his path is 150 m, and he experiences a constant resistance of 95 N. Find his speed as he launches himself into thin air. 95 N R 30 m mg 13 m Solution: Height lost between start and finish is = 17 m Work done by gravity = mgh = 90 g 17 = 1530g increases K.E. so positive Work done by R = 0 Work done by 95 N = Fs = = R is always perpendicular to the motion decreases K.E. so negative Workenergy equation Final K.E. = Initial K.E. ± Work done 90v2 = g v 2 = 1149 v = = 34 ms 1 to 2 S.F. (Super Tarzan) Potential energy When a body falls it gains K.E. The higher its starting point the greater the gain in K.E. We say that the Potential Energy, P.E., of a body depends on its height above some fixed point. When a body falls a distance h, the loss in P.E. is the work done by gravity, mgh. mv2 = mu2 + mgh mgh = mv2 mu2 or Loss in P.E. = Gain in K.E. loss in P.E. increases K.E. so positive Similarly when a body rises a distance h, the gain in P.E. = mgh mv2 = mu2 mgh mgh = mu2 mv2 or Gain in P.E. = Loss in K.E. gain in K.E. decreases K.E. so negative 16 14/04/2013 Mechanics 2 SD
19 You are expected to understand the terms loss in P.E. and gain in P.E., but all you can always use work done by gravity if you wish. Example 4: block, of mass 20 kg, is pulled up a rough slope of angle α, tan 1, by a rope. The block starts from rest and is moving at 6 ms 1 when it has moved a distance of 10 m. If the coefficient of friction is μ = 0 3, find the tension in the rope. Solution: R R = 20g cosα = 16g moving F = μr = g = 4 8g Work done by R = 0 ( to motion) Work done by F = Fs = 4 8g ( 10) = 48g Gain in P.E. = mgh = 20g ( 10 sinα) = 120g Work done by tension = Ts = 10 T F α 10 m T R 20g decreases K.E. so negative decreases K.E. so negative increases K.E. so positive Workenergy equation Final K.E. = Initial K.E. ± Work done = 0 48g 120g + 10T T = = 200 N to 2 S.F. Notice that we could have thought of work against by gravity, mgh, instead of gain in P.E. 14/04/2013 Mechanics 2 SD 17
20 Power Power is the rate of doing work. The units are Watts = Joules/second. For a constant force, F, moving a distance s the work done is W = Fs the power, P = (Fs) = F = Fv since F is constant the power developed by a constant force F moving its point of application at a speed v is P = Fv. Example 5: car of mass 900 kg is travelling up a slope of 5 o at a constant speed. ssume that there is no resistance to motion other than gravity. (a) If the engine of the car is working at a rate of 20 kw, find the speed of the car. (b) The car later travels up a slope of 8 o at the same speed. Find the power developed by the engine. Solution: (a) Power, P = = Dv v = Constant speed 5 o R 900g v D R D = 900g sin 5 o v = = = 26 ms1 to 2 S.F. (b) v = R from part (a) D = 900g sin 8 o R v D power developed = D v P = 900g sin 8 o 8 o 900g P = sin = power developed by the engine is 32 kw to 2 S.F /04/2013 Mechanics 2 SD
21 4 Collisions Impulse and momentum using vectors Impulse = change in momentum I = m v m u Conservation of linear momentum If there is no external impulse during a collision, then the total momentum before impact equals the total momentum after impact. m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2 Example 1: ball,, of mass 3 kg is moving with velocity 3 2 ms1 when it collides with another ball,, of mass 2 kg moving with velocity 3 1 ms1. fter the collision moves with velocity 1 3 ms1. Find the velocity of after the collision, and the impulse on during the collision. Solution: CLM There is no external impulse before = v v = ms1 during 3 kg 2 kg I For I = m v m u I = = 6 3 after I 1 3 v impulse on is 6 3 Ns Note: the impulse on is I = 6 3 Ns Newton s law of restitution This is also known as Newton s Experimental Law, NEL Coefficient of restitution The coefficient of restitution in a collision, e, is defined as 14/04/2013 Mechanics 2 SD 19
22 e = 0 e 1 If e = 1 the collision is perfectly elastic and K.E. is conserved during the collision. s usual it is essential to draw good diagrams and to take care over positive and negative values. Example 2: Particles P and Q with masses 2 kg and 3 kg are moving towards each other with velocities of 7 ms 1 and 5 ms 1 respectively. If the coefficient of restitution is, find the velocities of P and Q after the collision. Solution: Let velocities of P and Q after the collision be v P and v Q. We do not know their directions, so take them as marked in the diagram. No external impulse CLM = 1 = 2v P + 3v Q I before after 7 P 2 kg v P 5 Q 3 kg v Q NEL e = = 9 = v Q v P II I + 2 II 17 = 5v Q v Q = 3 4 and v P = 5 6. P and Q both move in opposite directions, and velocities are 5 6 ms 1, and 3 4 ms 1. Collisions with a plane surface The velocity of the surface before and after is 0 before after NEL, e = = m kg u m kg v 20 14/04/2013 Mechanics 2 SD
23 Multiple collisions Treat each collision as a new problem the final velocities from one collision become the initial velocities for the next collision. Example 3: Two particles, and, are of equal mass and are moving towards each other with speed of 3 ms 1 and 2 ms 1 respectively and collide. Particle then strikes a plane surface which is perpendicular to it direction of motion and rebounds. The coefficient or restitution between the two particles is and between and the plane surface is. Show that collides with a second time, and find the velocities of both particles after this collision. Solution: Let masses of particles be m kg. First collision, and + CLM 3m 2m = mx + my 1 = x + y I before 3 m kg 2 m kg NEL e = = after x y 2 = y x II I + II y = 1 5 and x = 0 5 moves at 0 5 ms 1, and moves at 1 5 ms 1. Second collision, with plane surface NEL e = = v = m kg 0 5 before after 1 5 m kg v fter this second collision, is moving at 1 ms 1 and is still moving at 0 5 ms 1. is moving faster than in the same direction there will be a third collision. Third collision, and + CLM 0 5m + m = ms + mt 1 5 = s + t NEL e = = I II = s t before after 0.5 m kg s 1 m kg t 14/04/2013 Mechanics 2 SD 21
24 + s = and t = moves at ms 1, and moves at ms 1, both moving away from the plane surface. Kinetic energy and impulses/collisions K.E. will be generated if there is an external impulse, but in collisions K.E. will be lost (unless the collision is perfectly elastic). Example 4: rifle of mass 5 kg fires a bullet of mass 25 grams with a muzzle velocity of 800 ms 1. The rifle is pointing in a horizontal direction and is free to move. Find the K.E. generated in firing the rifle. Solution: Linear momentum will be conserved and the rifle will move in the opposite direction to the bullet. Note that the muzzle velocity of the bullet is the velocity relative to the rifle. Let the velocity of the rifle be v ms 1, then the actual velocity of the bullet will be 800 v ms before 0 0 CLM 0 = (800 v) 5v v = = velocity of bullet = K.E. before = 0 after v 800 v K.E. after = 5 (3 98 ) (796 )2 = K.E. generated is 7960 J to 3 S.F. Example 5: Particle, mass 3 kg, and particle, mass 4 kg, are moving towards each other with speeds of 5 ms 1 and 2 ms 1 respectively. If e =, find the K.E. lost in the collision. Solution: There is no external impulse + CLM = 3x + 4y 7 = 3x + 4y NEL e = = 7 = 2y 2x I II before after 5 3 kg x 2 4 kg y I 2 II 7 = 7x x = 1 and y = /04/2013 Mechanics 2 SD
25 K.E. lost = [ ] [ ] K.E. lost = = 31 5 J 5 Statics of rigid bodies Moment of a force The moment of a force F about a point P is the magnitude of the force multiplied by the perpendicular distance from P to the line of action of F. Examples: (i) P N 5 m 50 o 130 o 10 N (ii) 7 m P 15 N 40 o N Moment = 10 PN Moment = 15 PN = 10 5 sin 50 o = 15 7 sin 40 o = 50 sin 50 o Nm = 105 sin 40 o Nm lternative method Resolve the force, F, in two directions one component passing through P, and the other perpendicular to this. (i) P 5 m 50 o 10 N (ii) 7 m P 15 N 40 o N P 5 m 10 sin 50 o 7 m P 15 cos 40 o 10 cos 50 o 15 sin 40 o Moment = 10 sin 50 o 5 Moment = 15 sin 40 o 7 = 50 sin 50 o Nm = 105 sin 40 o Nm as above as above 14/04/2013 Mechanics 2 SD 23
26 Equilibrium system of forces will be in equilibrium if (i) The sum of the resolved forces in any direction is zero. (ii) The moment about any point is zero. Example 1: uniform ladder of mass 20 kg and length 8 m is leaning against a smooth vertical wall on rough ground. The ladder makes an angle of 60 o with the ground, and the coefficient of friction is 0 5. What is the maximum height that a man of mass 80 kg can climb up the ladder before it starts to slip? Solution: The wall is smooth so the reaction, S, will be perpendicular to the wall. t the man s highest point, h up the ladder, the friction will be at its maximum F = μ R S 60 o R R = 80g + 20g = 100g R F = S h 30 o 80g Moments about 20g sin g sin 30 h = S sin g + 40gh = 4 3 S S = F = 0 5 R = 50g 40g + 40gh = g R 4 m 60 o F 30 o 20g h = = The man can climb a distance of 7.7 m (to 2 S.F.) up the ladder before it starts to slip /04/2013 Mechanics 2 SD
27 Example 2: nonuniform rod, of length 4 m, is freely hinged to a vertical wall at. It is held in equilibrium by a string which makes an angle of 40 o with the rod, and is attached to the wall above. The tension in the string is 65 N, the mass of the rod is 6 kg and the rod makes an angle of 70 o with the upwards vertical. Find the position of the centre of mass of the rod, and the magnitude and direction of the reaction at the hinge. Solution: Let the centre of mass, G, be x m from, and let the reaction at the hinge have components V and H as shown. = 4 m. 65 Moments about 6g sin 70 x = 65 sin 40 4 x = V 70 x H 40 G 70 6g the centre of mass is 3 0 m from to 2 S.F. R V + 65 sin 20 = 6g V = R H = 65 cos 20 = magnitude = = N and angle above the horizontal = arctan = o The reaction at the hinge is 71 N at an angle of 31 o above the horizontal, 2 S.F. Three nonparallel forces in equilibrium F 1 If three forces are not concurrent, as shown in the diagram, then the moment about, intersection of F 1 and F 2, can never be zero, and the forces cannot be in equilibrium. F 2 F 3 Thus, if three forces are in equilibrium, their lines of action must pass through one point. F 1 F 3 F 2 14/04/2013 Mechanics 2 SD 25
28 Note that for the three forces to be in equilibrium, the sum of the resolved forces in any direction must be zero. Example 3: nonuniform rod of length 6 m and mass m kg is supported at its ends by two strings, which make angles of 35 o and 50 o with the horizontal, as shown. If the rod is horizontal and in equilibrium, find the position of its centre of mass. T 2 T 1 50 o 35 o mg Solution: The three forces are in equilibrium, and therefore their lines of action must be concurrent. If the directions of T 1 and T 2 meet at P, then mg must pass through P. T 2 x G 50 6 x 35 T 1 Now it is just trigonometry GP = x tan 50 o, and GP = (6 x) tan 35 o P x = = mg Centre of mass is 2 2 m, 2 S.F., from end with the 50 o angle. Triangle of forces If three forces P, Q and R are in equilibrium, then their (vector) sum must be zero. Thus the three forces must form a triangle. Example 4: The three forces shown are in equilibrium. Find the magnitudes of P and Q. Solution: P 115 o Q 45 o P 135 o 110 o o o Q From the diagram we can draw a triangle of forces check that the arrows go round the triangle in the same direction. Sine Rule 26 14/04/2013 Mechanics 2 SD
29 P = and Q = P = 52 N and Q = 78 N to 2 S.F. ppendix Centre of mass of n particles Consider three particles with masses m 1, m 2 and m 3 at points with position vectors r 1, r 2 and r 3. Let the force of m 1 on m 2 be Q 12, of m 2 on m 1 be Q 21. Then Q 12 and Q 21 are internal forces and Q 12 + Q 21 = 0. m 1 P 1 Q 31 Q 21 Q 13 Q 23 m 3 P 3 The other internal forces are defined in a similar way. Q 12 Q 32 Let P 1, P 2 and P 3 be external forces on m 1, m 2 and m 3. Newton s second law for each particle gives P 2 m 2 P 1 + Q 21 + Q 31 = m 1 P 2 + Q 32 + Q 12 = m 2 P 3 + Q 13 + Q 23 = m 3 dding P 1 + P 2 + P 3 = m 1 + m 2 + m 3 I all internal forces, Q, cancel Define g= = = ( II From I and II P 1 + P 2 + P 3 = ( g the point moves as if all the mass was concentrated at that point, and all the external forces acted at that point. This point,, is called the centre of mass. This can be generalised for n particles to give M g = M = M = and M = where M is the total mass, M = 14/04/2013 Mechanics 2 SD 27
30 Medians of a triangle median of a triangle is a line joining one vertex to the midpoint of the opposite side. Let E and CF be medians of the triangle C. F and E are the midpoints of the sides C and FE = C and FE is parallel to C. F E XFE = XC and XEF = XC X triangles XFE and XC are similar in the ratio 2 : 1 FX and XC are corresponding sides FX = XC C also EX = X. Thus X lies on the two medians, dividing each one in the ratio 2 : 1. Similarly, X lies on the third median, D, in the ratio 2 : 1 the medians meet in a point, X, which divides each median in the ratio 2 : 1. Centre of mass of a triangle Divide the triangle into narrow strips parallel to one side. The centre of mass of each strip will be at the centre of each strip the centre of mass of the triangle must lie on the line joining these centres of mass the centre of mass lies on the median of the triangle. Similarly the centre of mass of the triangle lies on the other two medians, and therefore lies at the intersection of the medians. G is the centre of mass of the triangle, and G : GD = G : GE = CG : GF = 2 : 1 F G E D C 28 14/04/2013 Mechanics 2 SD
31 Index cceleration constant acceleration, 3 in a vertical plane, 3 variable acceleration, 5 vectors, 6 Centre of mass composite laminas, 9 laminas hanging freely under gravity, 11 n particles, 8 n particles, proof, 27 simple laminas, 8 toppling on a slope, 11 triangle, proof, 28 wire frameworks, 11 Collisions CLM, 19 coefficient of restitution, 19 loss of K.E., 22 multiple collisions, 21 Newton's law, 19 with a plane surface, 20 Energy kinetic energy, 13 potential energy, 16 workenergy equation, 14 Equilibrium system of forces, 24 three nonparallel forces, 25 Forces equilibrium, 24 moment, 23 power, 18 triangle of forces, 26 work done, 13 Impulse change in momentum, 19 loss or gain in K.E., 22 Kinematics in a vertical plane, 3 variable acceleration, 5 vectors, 6 Laminas centre of mass, 8 composite laminas, 9 suspended freely under gravity, 11 toppling on a slope, 11 Medians of a triangle meet at a point, proof, 28 Moment of a force, 23 Momentum CLM using vectors, 19 impulse using vectors, 19 Power constant force, 18 Wire frameworks, 11 Work constant force, 13 done by gravity, 14 workenergy equation, 14 14/04/2013 Mechanics 2 SD 29
www.mathsbox.org.uk Displacement (x) Velocity (v) Acceleration (a) x = f(t) differentiate v = dx Acceleration Velocity (v) Displacement x
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