Statics of rigid bodies

Introduction In this chapter you will revisit Moments from M1 You will learn to use these alongside the resolving of forces You will be able to solve more complicated problems regarding rods and laminas in equilibrium As with previous chapters, you will also get more practice at algebraic questions!

Teachings for exercise 5A

Statics of rigid bodies (2) You can calculate the moment of a force acting on a body The moment of a force acting on a body is the product of the magnitude of the force and its perpendicular distance from the point P. Moments are measured in Newton-metres (Nm), and you should always state the direction, clockwise or anti-clockwise Find the sum of the moments about point P in the diagram shown. 65° 1.2m 1.2Sin65m 7N P 0.8m 5N (1) 𝑀𝑜𝑚𝑒𝑛𝑡=𝐹𝑑 Taking moments about P (1) 5×0.8 = 4𝑁𝑚 𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 Moment (2) has a distance labelled which is not the perpendicular Draw a line from point P to the force which meets it at a right angle Use trigonometry to find an expression for this distance (2) 7×1.2𝑆𝑖𝑛65 =7.612…𝑁𝑚 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 Compare the sizes of the forces acting clockwise and anti-clockwise 7.612…−4=3.612𝑁𝑚 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 5A

Statics of rigid bodies 6N You can calculate the moment of a force acting on a body Find the sum of the moments about point P in the diagram shown to the right.  Mark on the perpendicular distances on first (Big diagrams help with this!)  Calculate each moment individually 2m 50° P (3) 3N (2) 2Sin50 4m 4Sin40 40° 6N (1) Taking moments about P (1) 6×4𝑆𝑖𝑛40 =15.43𝑁𝑚 𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 (2) 6×2𝑆𝑖𝑛50 =9.19𝑁𝑚 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 (3) The force of 3N passes through point P, and hence it will not have any turning motion about P (perpendicular distance is 0)  Imagine sitting in the middle of a seesaw! Compare the sizes of the forces acting clockwise and anti-clockwise 15.43−9.19=6.23𝑁𝑚 𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 5A

Statics of rigid bodies 6N You can calculate the moment of a force acting on a body Find the sum of the moments about point P in the diagram shown to the right.  Mark on the perpendicular distances on first (Big diagrams help with this!)  Calculate each moment individually 2m 50° P (3) 3N (2) 2Sin50 4m 4Sin40 40° 6N (1) Taking moments about P (1) 6×4𝑆𝑖𝑛40 =15.43𝑁𝑚 𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 (2) 6×2𝑆𝑖𝑛50 =9.19𝑁𝑚 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 An alternative method which you will sometimes use is to make keep the distance diagonal and make the force perpendicular instead! (3) The force of 3N passes through point P, and hence it will not have any turning motion about P (perpendicular distance is 0)  Imagine sitting in the middle of a seesaw! 15.43−9.19=6.23𝑁𝑚 𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 5A

Teachings for exercise 5b

Statics of rigid bodies You need to be able to solve problems about rigid bodies that are resting in equilibrium If a body is resting in equilibrium: There is no resultant force in any direction, so the horizontal and vertical forces sum to 0 The sum of moments about any point is 0 (the point does not have to be on the body itself) 5B

Statics of rigid bodies You need to be able to solve problems about rigid bodies that are resting in equilibrium A uniform rod AB, of mass 6kg and length 4m, is smoothly hinged at A. A light inextensible string is attached to the rod at a point C where AC = 3m, and the point D, which is vertically above point A. If the string is keeping the rod in equilibrium in a horizontal position and the angle between the string and the rod is 40°, calculate: The tension in the string The magnitude and direction of the reaction at the hinge. T TCos40 (2) V 40° TSin40 40° A B H 2m 3m C 1m (1) 6g Start with a diagram and label on all the forces – split the tension into horizontal and vertical components At the hinge there will be a vertical reaction and a horizontal reaction Take moments about point A (as we have 3 unknown forces, 2 will be eliminated by doing this!) These must be equal as the rod is in equilibrium (1) 6𝑔×2 =12𝑔 𝑁𝑚 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 (2) 𝑇𝑆𝑖𝑛40×3 =3𝑇𝑆𝑖𝑛40 𝑁𝑚 𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 𝑇=61𝑁 3𝑇𝑆𝑖𝑛40=12𝑔 Divide by 3Sin40 𝑇= 12𝑔 3𝑆𝑖𝑛40 Calculate 𝑇=61𝑁 5B

Statics of rigid bodies You need to be able to solve problems about rigid bodies that are resting in equilibrium A uniform rod AB, of mass 6kg and length 4m, is smoothly hinged at A. A light inextensible string is attached to the rod at a point C where AC = 3m, and the point D, which is vertically above point A. If the string is keeping the rod in equilibrium in a horizontal position and the angle between the string and the rod is 40°, calculate: The tension in the string The magnitude and direction of the reaction at the hinge. 61 61Cos40 V 40° 61Sin40 40° A B H 3m C 1m 6g Resolve Horizontally (set left and right forces equal to each other) 𝐻=61𝐶𝑜𝑠40 Calculate 𝐻=46.72𝑁 Resolve Vertically (set upwards and downwards forces equal to each other) 𝑇=61𝑁 𝑉+61𝑆𝑖𝑛40=6𝑔 Subtract 61Sin40 𝑉=6𝑔−61𝑆𝑖𝑛40 Calculate 𝐻=46.72𝑁 𝑉=19.6𝑁 𝑉=19.6𝑁 5B

Statics of rigid bodies You need to be able to solve problems about rigid bodies that are resting in equilibrium A uniform rod AB, of mass 6kg and length 4m, is smoothly hinged at A. A light inextensible string is attached to the rod at a point C where AC = 3m, and the point D, which is vertically above point A. If the string is keeping the rod in equilibrium in a horizontal position and the angle between the string and the rod is 40°, calculate: The tension in the string The magnitude and direction of the reaction at the hinge. 61 61Cos40 V 40° 61Sin40 40° A B H 3m C 1m 6g R R V V 19.6 A A H 46.72 H 𝑇=61𝑁 The resultant force will be somewhere between V and H  Use a right-angled triangle to help 𝑅𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡= 46.7 2 2 + 19.6 2 𝐻=46.72𝑁 𝑉=19.6𝑁 Calculate 𝑅𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡=50.6𝑁 𝑅𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡=50.6𝑁 5B

Statics of rigid bodies You need to be able to solve problems about rigid bodies that are resting in equilibrium A uniform rod AB, of mass 6kg and length 4m, is smoothly hinged at A. A light inextensible string is attached to the rod at a point C where AC = 3m, and the point D, which is vertically above point A. If the string is keeping the rod in equilibrium in a horizontal position and the angle between the string and the rod is 40°, calculate: The tension in the string The magnitude and direction of the reaction at the hinge. 61 61Cos40 V 40° 61Sin40 40° A B H 3m C 1m 6g R R V V 19.6 θ A A H 46.72 H 𝑇=61𝑁 You also need to calculate the angle above the horizontal 𝐴𝑛𝑔𝑙𝑒=𝑇𝑎 𝑛 −1 19.6 46.72 𝐻=46.72𝑁 𝑉=19.6𝑁 Calculate 𝑅𝑒𝑠𝑢𝑙𝑡𝑎𝑛𝑡=50.6𝑁 , 22.8° 𝑎𝑏𝑜𝑣𝑒 𝑡ℎ𝑒 ℎ𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝐴𝑛𝑔𝑙𝑒=22.8° 5B

Statics of rigid bodies You need to be able to solve problems about rigid bodies that are resting in equilibrium A non-uniform rod AB, of mass 3kg and length 5m, rests horizontally in equilibrium, supported by two strings attached at the ends of the rod. The strings make angles of 40° and 55° with the horizontal, a shown in the diagram. Find the magnitudes of the tensions in the two strings Find the distance of the centre of mass from A  Let the centre of mass be a distance ‘x’ from A T1 T2Sin55 T1Sin40 40° x 5 - x 55° T1Cos40 A B T2Cos55 3g Split the 2 tensions into horizontal and vertical components  The horizontal and vertical components will each cancel themselves out Resolve Horizontally 𝑇 1 𝐶𝑜𝑠40= 𝑇 2 𝐶𝑜𝑠55 Divide by Cos40 𝑇 1 = 𝑇 2 𝐶𝑜𝑠55 𝐶𝑜𝑠40 Resolve Vertically 𝑇 1 = 𝑇 2 𝐶𝑜𝑠55 𝐶𝑜𝑠40 𝑇 1 𝑆𝑖𝑛40+ 𝑇 2 𝑆𝑖𝑛55=3𝑔 𝑇 1 𝑆𝑖𝑛40+ 𝑇 2 𝑆𝑖𝑛55=3𝑔 5B

Statics of rigid bodies 𝑇 2 =22.6𝑁 T2 You need to be able to solve problems about rigid bodies that are resting in equilibrium A non-uniform rod AB, of mass 3kg and length 5m, rests horizontally in equilibrium, supported by two strings attached at the ends of the rod. The strings make angles of 40° and 55° with the horizontal, a shown in the diagram. Find the magnitudes of the tensions in the two strings Find the distance of the centre of mass from A  Let the centre of mass be a distance ‘x’ from A T1 T2Sin55 T1Sin40 40° x 5 - x 55° T1Cos40 A B T2Cos55 3g 𝑇 1 𝑆𝑖𝑛40+ 𝑇 2 𝑆𝑖𝑛55=3𝑔 Sub in an expression for T1 𝑇 2 𝐶𝑜𝑠55 𝐶𝑜𝑠40 𝑆𝑖𝑛40+ 𝑇 2 𝑆𝑖𝑛55=3𝑔 Multiply all terms by Cos40 𝑇 2 𝐶𝑜𝑠55𝑆𝑖𝑛40+ 𝑇 2 𝑆𝑖𝑛55𝐶𝑜𝑠40=3𝑔𝐶𝑜𝑠40 Factorise the left 𝑇 2 (𝐶𝑜𝑠55𝑆𝑖𝑛40+𝑆𝑖𝑛55𝐶𝑜𝑠40)=3𝑔𝐶𝑜𝑠40 Divide by the whole bracket! 𝑇 2 = 3𝑔𝐶𝑜𝑠40 𝐶𝑜𝑠55𝑆𝑖𝑛40+𝑆𝑖𝑛55𝐶𝑜𝑠40 Calculate 𝑇 1 = 𝑇 2 𝐶𝑜𝑠55 𝐶𝑜𝑠40 𝑇 1 𝑆𝑖𝑛40+ 𝑇 2 𝑆𝑖𝑛55=3𝑔 𝑇 2 =22.6𝑁 5B

Statics of rigid bodies 𝑇 1 =16.9𝑁 𝑇 2 =22.6𝑁 T2 You need to be able to solve problems about rigid bodies that are resting in equilibrium A non-uniform rod AB, of mass 3kg and length 5m, rests horizontally in equilibrium, supported by two strings attached at the ends of the rod. The strings make angles of 40° and 55° with the horizontal, a shown in the diagram. Find the magnitudes of the tensions in the two strings Find the distance of the centre of mass from A  Let the centre of mass be a distance ‘x’ from A T1 T2Sin55 T1Sin40 40° x 5 - x 55° T1Cos40 A B T2Cos55 3g 𝑇 1 = 𝑇 2 𝐶𝑜𝑠55 𝐶𝑜𝑠40 Sub in T2 (remember to use the exact value!) 𝑇 1 = 22.6𝐶𝑜𝑠55 𝐶𝑜𝑠40 Calculate 𝑇 1 =16.9𝑁 𝑇 1 = 𝑇 2 𝐶𝑜𝑠55 𝐶𝑜𝑠40 𝑇 1 𝑆𝑖𝑛40+ 𝑇 2 𝑆𝑖𝑛55=3𝑔 5B

Statics of rigid bodies You need to be able to solve problems about rigid bodies that are resting in equilibrium A non-uniform rod AB, of mass 3kg and length 5m, rests horizontally in equilibrium, supported by two strings attached at the ends of the rod. The strings make angles of 40° and 55° with the horizontal, a shown in the diagram. Find the magnitudes of the tensions in the two strings Find the distance of the centre of mass from A  Let the centre of mass be a distance ‘x’ from A T1 (2) (2) T2Sin55 T2Sin55 T1Sin40 40° x 5m 5 - x 55° T1Cos40 A B T2Cos55 (1) 3g Take moments about A to create an equation with x in (1) 3𝑔×𝑥 = 3𝑔𝑥 𝑁𝑚 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 (2) 𝑇 2 𝑆𝑖𝑛55×5 = 5 𝑇 2 𝑆𝑖𝑛55 𝑁𝑚 𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 Remember that moment (2) is acting from the end of the rod!  There is no rotation so these forces must be equal 𝑇 1 =16.9𝑁 𝑇 2 =22.6𝑁 3𝑔𝑥=5 𝑇 2 𝑆𝑖𝑛55 Divide by 3g 𝑥= 5 𝑇 2 𝑆𝑖𝑛55 3𝑔 So the centre of mass is 3.1m from A Calculate using the exact value for T2 𝑥=3.1𝑚 5B

Teachings for exercise 5C

Statics of rigid bodies I have decided to miss out the part in exercise 5C (making a ‘triangle of forces’)  Pupils tend to find this method quite tricky and they can just use other methods anyway! 5C

Teachings for exercise 5D

Statics of rigid bodies N B 2m You can solve problems about rigid bodies resting in limiting equilibrium If a body is on the point of moving it is said to be in limiting equilibrium. In this case, the frictional force takes its maximum value, µR, where µ is the coefficient of friction and R is the normal reaction. A uniform rod AB of mass 40kg and length 10m rests with the end A on rough horizontal ground. The rod rests against a smooth peg C where AC = 8m. The rod is in limiting equilibrium at an angle of 15° to the horizontal. Find: The magnitude of the reaction at C The coefficient of friction between the rod and the ground (2) 3m R (1) 5m 40gCos15 40g 15° 15° 40gSin15 A F Draw a diagram and label all the forces: Weight, the normal reactions and friction. Split into components if needed The rod will have a tendency to slide downwards, with the base moving to the left. Hence, friction will oppose this Taking moments about A will mean we can find the normal reaction at the peg. Taking moments about A (1) 40𝑔𝐶𝑜𝑠15×5 =200𝑔𝐶𝑜𝑠15 𝑁𝑚 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 (2) 𝑁×8 = 8𝑁 𝑁𝑚 𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 40gSin15 is NOT included as a moment about A. This is because it actually acts down the rod and through point A (as opposed to the place where it has been drawn), therefore it has a perpendicular distance of 0 and hence can be ignored… 200𝑔𝐶𝑜𝑠15=8𝑁 𝑁=237𝑁 Divide by 8 25𝑔𝐶𝑜𝑠15=𝑁 Calculate 237=𝑁 5D

Statics of rigid bodies 237Sin15 237Cos15 B 15° You can solve problems about rigid bodies resting in limiting equilibrium If a body is on the point of moving it is said to be in limiting equilibrium. In this case, the frictional force takes its maximum value, µR, where µ is the coefficient of friction and R is the normal reaction. A uniform rod AB of mass 40kg and length 10m rests with the end A on rough horizontal ground. The rod rests against a smooth peg C where AC = 8m. The rod is in limiting equilibrium at an angle of 15° to the horizontal. Find: The magnitude of the reaction at C The coefficient of friction between the rod and the ground 237 R 40gCos15 40g 15° 15° 40gSin15 A F Now you can resolve horizontally and vertically to find the remaining forces You will need to split the normal reaction at the peg into horizontal and vertical components The parallel and perpendicular components of the weight will no longer be needed… Resolving Horizontally 𝐹=237𝑆𝑖𝑛15 Resolving Vertically 𝑁=237𝑁 𝑅+237𝐶𝑜𝑠15=40𝑔 Rearrange 𝑅=40𝑔−237𝐶𝑜𝑠15 𝐹=237𝑆𝑖𝑛15 𝑅=40𝑔−237𝐶𝑜𝑠15 5D

Statics of rigid bodies 237Sin15 237Cos15 B 15° You can solve problems about rigid bodies resting in limiting equilibrium If a body is on the point of moving it is said to be in limiting equilibrium. In this case, the frictional force takes its maximum value, µR, where µ is the coefficient of friction and R is the normal reaction. A uniform rod AB of mass 40kg and length 10m rests with the end A on rough horizontal ground. The rod rests against a smooth peg C where AC = 8m. The rod is in limiting equilibrium at an angle of 15° to the horizontal. Find: The magnitude of the reaction at C The coefficient of friction between the rod and the ground 237 R 40g 15° A F As the rod is in limiting equilibrium, friction is at its maximum value Use the formula for FMAX and sub in the values we have calculated 𝐹 𝑀𝐴𝑋 =𝜇𝑅 Sub in values 237𝑆𝑖𝑛15=𝜇(40𝑔−237𝐶𝑜𝑠15) Divide by the bracket 237𝑆𝑖𝑛15 40𝑔−237𝐶𝑜𝑠15 =𝜇 Calculate 𝑁=237𝑁 0.37=𝜇 𝐹=237𝑆𝑖𝑛15 𝑅=40𝑔−237𝐶𝑜𝑠15 5D

Statics of rigid bodies RWSin60 You can solve problems about rigid bodies resting in limiting equilibrium A ladder, AB, of mass m and length 3a, has one end A resting on rough horizontal ground. The other end, B, rests against a smooth vertical wall. A load of mass 2m is fixed on the ladder at point A, where AC = a. The ladder is modelled as a uniform rod and the load is modelled as a particle. The ladder rests in limiting equilibrium at an angle of 60° with the ground. Find the coefficient of friction between the ladder and the ground. 60° RW B (3) Start with a diagram and label all forces – both masses should be split into parallel and perpendicular components 1.5a (2) (1) 0.5a 60° mgCos60 RG C We will now take moments about point A to give us the value of RW 2mgCos60 a 60° 60° A F 2mg mg (1) 2𝑚𝑔𝐶𝑜𝑠60×𝑎 = 2𝑎𝑚𝑔𝐶𝑜𝑠60 𝑁𝑚 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 (2) 𝑚𝑔𝐶𝑜𝑠60×1.5𝑎 = 1.5𝑎𝑚𝑔𝐶𝑜𝑠60 𝑁𝑚 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 (3) 𝑅 𝑊 𝑆𝑖𝑛60×3𝑎 = 3𝑎 𝑅 𝑊 𝑆𝑖𝑛60 𝑁𝑚 𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 𝑅 𝑊 = 3.5𝑚𝑔𝐶𝑜𝑠60 3𝑆𝑖𝑛60 3𝑎 𝑅 𝑊 𝑆𝑖𝑛60=2𝑎𝑚𝑔𝐶𝑜𝑠60+1.5𝑎𝑚𝑔𝐶𝑜𝑠60 Cancel a’s 3 𝑅 𝑊 𝑆𝑖𝑛60=2𝑚𝑔𝐶𝑜𝑠60+1.5𝑚𝑔𝐶𝑜𝑠60 Calculate in terms of mg Group terms 𝑅 𝑊 = 7𝑚𝑔 3 18 3 𝑅 𝑊 𝑆𝑖𝑛60=3.5𝑚𝑔𝐶𝑜𝑠60 Divide by 3Sin60 𝑅 𝑊 = 3.5𝑚𝑔𝐶𝑜𝑠60 3𝑆𝑖𝑛60 5D

Statics of rigid bodies RWSin60 You can solve problems about rigid bodies resting in limiting equilibrium A ladder, AB, of mass m and length 3a, has one end A resting on rough horizontal ground. The other end, B, rests against a smooth vertical wall. A load of mass 2m is fixed on the ladder at point A, where AC = a. The ladder is modelled as a uniform rod and the load is modelled as a particle. The ladder rests in limiting equilibrium at an angle of 60° with the ground. Find the coefficient of friction between the ladder and the ground. 60° RW B (3) Now we can resolve horizontally and vertically  This will allow us to find expressions for RG and F, and hence, the coefficient of friction 1.5a (2) (1) 0.5a 60° mgCos60 RG C 2mgCos60 a 60° 60° A F 2mg mg Resolving Horizontally 𝐹= 𝑅 𝑊 We already know RW so therefore also know F! 𝐹= 7𝑚𝑔 3 18 𝑅 𝑊 = 7𝑚𝑔 3 18 𝐹= 7𝑚𝑔 3 18 5D

Statics of rigid bodies RWSin60 You can solve problems about rigid bodies resting in limiting equilibrium A ladder, AB, of mass m and length 3a, has one end A resting on rough horizontal ground. The other end, B, rests against a smooth vertical wall. A load of mass 2m is fixed on the ladder at point A, where AC = a. The ladder is modelled as a uniform rod and the load is modelled as a particle. The ladder rests in limiting equilibrium at an angle of 60° with the ground. Find the coefficient of friction between the ladder and the ground. 60° RW B (3) Now we can resolve horizontally and vertically  This will allow us to find expressions for RG and F, and hence, the coefficient of friction 1.5a (2) (1) 0.5a 60° mgCos60 RG C 2mgCos60 a 60° 60° A F 2mg mg Resolving Vertically 𝑅 𝐺 =2𝑚𝑔+𝑚𝑔 Simplify 𝑅 𝐺 =3𝑚𝑔 𝑅 𝑊 = 7𝑚𝑔 3 18 𝐹= 7𝑚𝑔 3 18 𝑅 𝐺 =3𝑚𝑔 5D

Statics of rigid bodies RWSin60 You can solve problems about rigid bodies resting in limiting equilibrium A ladder, AB, of mass m and length 3a, has one end A resting on rough horizontal ground. The other end, B, rests against a smooth vertical wall. A load of mass 2m is fixed on the ladder at point A, where AC = a. The ladder is modelled as a uniform rod and the load is modelled as a particle. The ladder rests in limiting equilibrium at an angle of 60° with the ground. Find the coefficient of friction between the ladder and the ground. 60° RW B (3) 1.5a (2) As the ladder is in limiting equilibrium, we can use the formula for friction (1) 0.5a 60° mgCos60 RG C 2mgCos60 a 60° 60° A F 2mg mg 𝐹𝑀𝐴𝑋=𝜇𝑅 Sub in values 7𝑚𝑔 3 18 =3𝑚𝑔𝜇 Cancel mg’s 7 3 18 =3𝜇 𝑅 𝑊 = 7𝑚𝑔 3 18 𝐹= 7𝑚𝑔 3 18 Divide by 3 7 3 54 =𝜇 𝑅 𝐺 =3𝑚𝑔 Calculate 0.22=𝜇 5D

Summary You have had a reminder of moments You have seen how these can be used alongside your other skills in resolving forces You have solved problems involving laminas and rods, and seen some very tricky algebraic questions!