ASLevel Maths Mechanics 2 for Edexcel M 2

AS-Level Maths: Mechanics 2 for Edexcel M 2. 6 Statics of Rigid Bodies These icons indicate that teacher’s notes or useful web addresses are available in the Notes Page. This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation. 1 of 19 © Boardworks Ltd 2006

Moments and non-parallel forces So far, all questions requiring the use of moments have involved parallel forces acting perpendicular to a body. This is not always the case. Where forces are not acting perpendicular to a body they can be resolved into two components – perpendicular and parallel to the body. 2 The moment is 2 × 10 cos θ Nm ↻. θ 10 10 cos θ Alternatively, we can multiply the magnitude of the force by the perpendicular distance from the pivot to the line of direction of the force. 2 cos θ The moment is 2 cos θ × 10 Nm ↻. θ 2 2 of 19 10 © Boardworks Ltd 2006

Equilibrium and rigid bodies A rigid body is one that does not bend or change shape when forces are applied to it. Possible examples are rods and ladders. For a rigid body to be in equilibrium two conditions need to be satisfied: i. The vector sum of the forces acting on the body must be zero. ii. The sum of the moments about any point must be zero. Rods are sometimes said to be freely hinged or smoothly hinged. In this case there will be a reaction at the hinge but the direction of this reaction is not known. Hence horizontal and vertical components of the reaction force are used in the solution. 3 of 19 © Boardworks Ltd 2006

Non-parallel forces Question 1: A uniform rod AB of length 6 m and mass 5 kg is resting on a smooth pivot 1 m from A. It is held in equilibrium by means of a light inextensible string attached to the rod 5 m from A. The other end of the string is fixed to a point vertically above A so that the angle it makes with the rod is 30°. Find the tension in the string if the rod is in equilibrium. T sin 30° T R A 1 30 2 2 5 g 1 B Take moments about the pivot: 5 g × 2 = T sin 30° × 4 10 g = 2 T T = 5 g T = 49 N 4 of 19 © Boardworks Ltd 2006

Non-parallel forces Question 2: A uniform rod AB of length 4 m and mass 4 kg is resting in equilibrium. The forces acting on the rod are shown in the diagram. Find X, Y and Z. X B Y 60 1 First we need to find the vertical and horizontal components of X. X sin 30° 1 X X cos 30° 30 2 4 g Z A 30° 60° Resolving the forces vertically gives: X sin 30° = 4 g X = 78. 4 N 5 of 19 © Boardworks Ltd 2006

Non-parallel forces Resolving the forces horizontally gives: Y = Z + X cos 30° Now take moments about B using the components of Y, 4 g and Z perpendicular to the rod: (1 × Y cos 30°) + (2 × 4 g cos 60°) = (4 × Z cos 30°) We can substitute (Z + X cos 30°) for Y: cos 30°(Z + X cos 30°) + (2 × 4 g cos 60°) = 4 × Z cos 30° + 78. 4 cos 2 30° + 8 g cos 60° = 4 Z cos 30° 78. 4 cos 2 30° + 8 g cos 60° = 3 Z cos 30° 2 30°) (8 g cos 60° + 78. 4 cos Z= = 113 N (3 s. f. ) 3 cos 30° Y = 113. 16 + 78. 4 cos 30° = 181 N (3 s. f. ) 6 of 19 © Boardworks Ltd 2006

Non-parallel forces 7 of 19 © Boardworks Ltd 2006

Equilibrium and rigid bodies Question 1: A uniform ladder of length 4 m and weight 6 kg rests on rough horizontal ground against a smooth vertical wall. The ladder is inclined at an angle of 70° to the vertical when it is on the point of slipping. Calculate the coefficient of friction between the ladder and the ground. B S 2 R 2 A 6 g 70 F 8 of 19 We first need to draw a diagram showing the forces acting on the ladder. Since the ladder is uniform, the weight will act through its mid-point. There is a normal contact force from the smooth wall. The total reaction force on the ladder from the rough ground has normal and frictional components, R and F. © Boardworks Ltd 2006

Equilibrium and rigid bodies Resolving vertically gives: S sin 70° S 2 6 g cos 70° R 2 6 g A 70 F R = 6 g Resolving horizontally gives: F = S B In order to take moments we need to use the components of S and 6 g perpendicular to the ladder. Take moments about A: (6 g cos 70° × 2) = (S sin 70° × 4) S= = 3 g cot 70° The ladder is on the point of slipping, so F = µR. (F = S and R = 6 g: ) 3 g cot 70° = 6µg µ = ½ cot 70° = 0. 182 (3 s. f. ) 9 of 19 © Boardworks Ltd 2006

Equilibrium and rigid bodies Question 2: A uniform ladder AB of mass 20 kg stands on rough horizontal ground and leans against a smooth vertical wall. A mass of 10 kg is attached to the ladder ¾ of the way up. The coefficient of friction between the ladder and the ground is ½. If the ladder is on the point of slipping, find the angle it makes with the ground. Resolving vertically, R = 30 g S B Resolving horizontally, S = F 10 g 20 g F 10 of 19 R At the point of slipping F = R: F = ½ × 30 g = 15 g A © Boardworks Ltd 2006

Equilibrium and rigid bodies Let the length of the ladder be a. Take moments about A: ( a × 20 g cos ) + ( a × 10 g cos ) = (a × S sin ) Cancel the a’s. S = F = 15 g, therefore: 17. 5 g cos = 15 g sin tan = = 49. 4 (3 s. f. ) 11 of 19 © Boardworks Ltd 2006

Examination-style question 1 Question 1: A ladder AB of mass 15 kg and length 4 m is resting on rough horizontal ground and is leaning against a smooth vertical wall. The ladder is modelled as a uniform rod. a) When the ladder is inclined at an angle of 30° to the horizontal it is on the point of slipping. Find the coefficient of friction between the ladder and the ground. The ladder is moved so that it now makes an angle of 40 with the horizontal. b) A boy of mass 40 kg climbs the ladder. How far can he climb up the ladder before it starts to slip? 12 of 19 © Boardworks Ltd 2006

Examination-style question 1 B S 2 R 2 A 15 g Resolving : R = 15 g Resolving : S = F Take moments about A: 4 × S sin 30° = 2 × 15 g cos 30° 30 F S = 127. 31 (5 s. f. ) = F When the ladder is on the point of slipping, F = R: 127. 31 = 15 g = 0. 866 (3 s. f. ) Therefore the coefficient of friction between the ladder and the ground is 0. 866 (3 s. f. ). 13 of 19 © Boardworks Ltd 2006

Examination-style question 1 S B 2–x x R 2 Let x be the distance from the midpoint of AB to the point at which the boy makes the ladder slip. 40 g Resolving , R = 55 g 15 g Resolving , S = F 40 A F At the point of slipping, F = R: F = 0. 86603 × 55 g F = 466. 79 (5 s. f. ) = S 14 of 19 © Boardworks Ltd 2006

Examination-style question 1 Taking moments about A: (2 × 15 g cos 40°) + ((2 + x) × 40 g cos 40°) = (S sin 40° × 4) 30 g cos 40° + 80 g cos 40° + 40 gx cos 40° = 466. 79 sin 40° × 4 x(40 g cos 40°) = 1967. 2 sin 40° – 110 g cos 40° x = 1. 25 (3 s. f. ) The boy can climb 1. 25 m past the midpoint of the ladder. Therefore he can climb 3. 25 m up the ladder before it starts to slip. 15 of 19 © Boardworks Ltd 2006

Examination-style question 2 Question 2: A uniform ladder AB of mass m kg and length 2 a rests on smooth horizontal ground and leans against a smooth vertical wall. The ladder is attached to the wall by means of a horizontal light inextensible string secured at the mid-point of the ladder. The ladder is inclined at an angle ° to the horizontal where tan = ¾. If the maximum tension possible in the string before it breaks is 10 mg N, find, in terms of m, the maximum mass of a person that can reach the top of the ladder. 16 of 19 © Boardworks Ltd 2006

Examination-style question 2 Let the maximum mass be M kg. We need to find M at the top of the ladder. S B Mg T a a mg If tan = R A : sin = and cos = Resolving , R = mg + Mg Resolving , S = T Take moments about B: (Tsin × a) + (mgcos × a) = Rcos × 2 a R = mg + Mg, and for the maximum mass, T = 10 mg: (10 mg( ) + The a’s cancel to give: Mg = 6 mg + M= 17 of 19 mg)a = m (mg + Mg) × 2 a mg – mg = mg M = 3. 25 m © Boardworks Ltd 2006

Examination-style question 3 Question 3: A uniform rod AB of weight W N and length 2 m is freely hinged at A and is held in a horizontal position by means of a light inextensible string inclined at an angle of to the horizontal and attached to B as shown in the diagram. a) Find, in terms of W and , the tension in the string. b) b) Find X and Y in terms of W and only. X A T Y 1 1 B W 18 of 19 © Boardworks Ltd 2006

Examination-style question 3 Resolving gives: X + T sin = W Resolving gives: Y = T cos Take moments about B: W× 1=X× 2 X A T Y 1 1 W X = ½W We can now use the equation above to find T in terms of W and : ½W + T sin = W T sin = ½W T= Now we can use T to find Y in terms of W and : Y= 19 of 19 cos = ½W cot © Boardworks Ltd 2006 B