Moments At angles

Starter: Spot the mistake Moments: At an angle KUS objectives BAT solve problems using moments and friction at angles to a rod/ object using sohCahToa Starter: Spot the mistake

Moments The moment of a force measures the turning effect of the force on the body on which it is acting The moment of a force F about a point P is the product of the magnitude of the force and the perpendicular distance of the line of action of the force from the point P Moment of F about P = Fd clockwise The magnitude of the force is measured in newtons (N) and the distance is measured in metres (m), so the moment of the force is measured in newton-metres (Nm) Moment of F about P = 15 Nm anticlockwise

Using Trigonometry You may need to use trigonometry to find the perpendicular distance Moment of F about P = clockwise e.g. Moment of the force about P = = clockwise

WB 1 Calculate the moment about point A of each of these forces acting on a Lamina , 𝐴 10 𝑁 40° 5 𝑚 𝐴 6 sin 25 5 sin 40 6 𝑚 25° 18 𝑁 𝑚𝑜𝑚𝑒𝑛𝑡=10×5 sin 40=32.1 𝑁𝑚 𝑎𝑐𝑤 𝑚𝑜𝑚𝑒𝑛𝑡=18×6 sin 25=45.6 𝑁𝑚 𝑎𝑐𝑤 𝐴 120 𝑁 50° 8 𝑚 70 𝑁 4 sin 65 8 cos 50 𝐴 65° 4 𝑚 𝑚𝑜𝑚𝑒𝑛𝑡=70×4 sin 65 =254 𝑁𝑚 𝑎𝑐𝑤 𝑚𝑜𝑚𝑒𝑛𝑡=120×8 cos 50 =617 𝑁𝑚 𝑐𝑤

WB 2 Given the moment about point A of each of these forces is 20 Nm, Find the magnitude of each force 3 sin 35 𝐴 𝐴 𝐹 1 30° 5 𝑚 5 sin 30 3 𝑚 𝐹 2 35° 𝐹 2 = 20 3 sin 35 = 𝑁 𝐹 1 = 20 5 sin 30 =8 𝑁 𝐴 75° 9 𝑚 𝐹 4 𝐹 3 = 20 2 sin 55 = 𝑁 9 sin 75 𝐴 125° 2 𝑚 𝐹 3 2 sin 55 𝐹 4 = 20 9 sin 75 = 𝑁

WB 3a Two forces act on a lamina as shown WB 3a Two forces act on a lamina as shown. Calculate the resultant moment about the point A a) 𝐴 𝐹 2 =5 50° 10 𝑚 𝐹 1 =8 30° 5 sin 50 8 sin 30 Moment of 𝐹 1 =8 sin 30 ×8=32 𝑐𝑤 Moment of 𝐹 2 =5 sin 50 ×5=19.2 𝑎𝑐𝑤 Moment of 𝐹 1 and 𝐹 2 =32−19.2 =12.8 𝑁𝑚 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒

WB 3b Two forces act on a lamina as shown WB 3b Two forces act on a lamina as shown. Calculate the resultant moment about the point A b) 𝐴 𝐹 2 =15 60° 10 𝑚 𝐹 1 =12 70° 8 𝑚 8 sin 60 10 sin 70 Moment of 𝐹 1 =10 sin 70 ×12=112.7 𝑎𝑐𝑤 Moment of 𝐹 2 =8 sin 60 ×15=103.9 𝑐𝑤 Moment of 𝐹 1 and 𝐹 2 =112.7−103.9 =8.8 𝑁𝑚 𝑎𝑛𝑡𝑖𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒

Moment on a uniform beam / rod WB4 review A light rod AB is 4 m long and can rotate in a vertical plane about fixed point C where AC = 1 m. A vertical force F of 7 N acts on the rod downwards. Find the moment of F about C when F acts a) at A b) at B c) at C a) Taking moments about C acw 7 3 m 1 m A C B 8×𝐴𝐶=8 𝑁𝑚 b) Taking moments about C cw 8×𝐶𝐵=24 𝑁𝑚 c) Taking moments about C cw 8×𝐶𝐶=0

WB 5 The diagram shows a set of forces acting on a uniform rod of mass 3 kg. Calculate the resultant moment about point A (including direction) 𝐴 45° 3 𝑚 5 𝑁 6 𝑁 70° 4 𝑚 4 sin 45 3 sin 70 Taking moments about A clockwise 6×4 sin 45 − 5×3 sin 70 = 2.88 𝑁𝑚 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒

Equilibrium

WB 6 A uniform rod PQ is hinged at point P, and is held in equilibrium at an angle of 50 to the horizontal by a force of magnitude F acting perpendicular to the rod at Q. Given that the rod has a length of 3 m and mass of 8 kg, find the value of F 𝑃 8𝑔 𝑁 50° 𝐹 𝑄 3 2 cos 50 Taking moments about P 𝐹 ×3 = 8𝑔× 3 2 cos 50 𝐹 = 8𝑔× 3 2 cos 50 3 =25.2 𝑁

WB 6 (cont) A uniform rod PQ is hinged at point P, and is held in equilibrium at an angle of 50 to the horizontal by a force of magnitude F acting perpendicular to the rod at Q. Given that the rod has a length of 3 m and mass of 8 kg, find the value of F 𝑃 8𝑔 𝑁 50° 𝐹 𝑄 3 2 cos 50 Taking moments about P 𝐹 ×3 = 8𝑔× 3 2 cos 50 𝐹 = 8𝑔× 3 2 cos 50 3 =25.2 𝑁

WB 7 A uniform rod PQ of mass 40 kg and length 10 m rests with the end P on rough horizontal ground. The rod rests against a smooth peg C where AC = 8 m 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 𝑃 40𝑔 15° 𝑁 𝐶 𝑅 𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛 75° distance 5 cos 15 a) Taking moments about P 𝑁×8 =40𝑔×5 cos 15 reaction at C 𝑁 = 40𝑔×5 cos 15 8 =236.7 𝑁 b) Equilibrium in horizontal direction 𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛 =𝑁 cos 75 =236.7 cos 75 =61.25 Equilibrium in vertical direction 𝑅+𝑁𝑠𝑖𝑛 75=40𝑔 𝑅=40𝑔−236.7 cos 75 =163.4 𝑁 𝜇= 𝐹𝑟 𝑅 = 61.25 163.4 =0.375 Coefficient friction

WB 8 A ladder PQ of mass m kg and length 3a m rests with the end P on rough horizontal ground. The other end Q rests against a smooth vertical wall. A load of mass 2m is fixed on the ladder at point C, where AC = a. The ladder is modelled as a uniform rod in a vertical plane perpendicular to the wall and the load as a particle. The ladder rests in limiting equilibrium at an angle of 60with the ground. Find the coefficient of friction between the rod and the ground N Add all the forces to the diagram 𝑃 60° 𝐶 𝑄 2𝑎 𝑎 Wall smooth No friction Equilibrium in horizontal direction 𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛=𝑁 R Equilibrium in vertical direction 𝑅=2𝑚𝑔+𝑚𝑔=3𝑚𝑔 Friction

(to get an equation with Friction and R) WB 8 (cont) A ladder PQ of mass m kg and length 3a m rests with the end P on rough horizontal ground. The other end Q rests against a smooth vertical wall. A load of mass 2m is fixed on the ladder at point C, where AC = a. The ladder is modelled as a uniform rod in a vertical plane perpendicular to the wall and the load as a particle. The ladder rests in limiting equilibrium at an angle of 60with the ground. Find the coefficient of friction between the rod and the ground Taking moments about Q (to get an equation with Friction and R) 𝑃 60° 𝐶 𝑄 2𝑎 𝑎 Wall smooth No friction Friction R N 𝑅×3𝑎 cos 60 =𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛 ×3𝑎 sin 60 +𝑚𝑔× 3 2 𝑎 cos 60 +2𝑚𝑔×2𝑎 cos 60 Substituting 𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝜇𝑅 and R=3𝑚𝑔 3𝑚𝑔×3𝑎 cos 60 =𝜇3𝑚𝑔 ×3𝑎 sin 60 +𝑚𝑔× 3 2 𝑎 cos 60 +2𝑚𝑔×2𝑎 cos 60 Cancel by mga and rearrange to 𝜇= 3𝑚𝑔×3𝑎 cos 60 −+𝑚𝑔× 3 2 𝑎 cos 60 −2𝑚𝑔×2𝑎 cos 60 3𝑚𝑔 ×3𝑎 sin 60 = 7 2 cos 60 9 sin 60 = 7 3 54 =0.225

WB 9 (exam Q) sin 𝜃 = 3 5 cos 𝜃 = 4 5 3 4 5 𝜃 Mg 3Mg T R Friction A plank, AB, of mass M and length 2a, rests with its end A against a rough vertical wall. The plank is held in a horizontal position by a rope. One end of the rope is attached to the plank at B and the other end is attached to the wall at the point C, which is vertically above A. A small block of mass 3M is placed on the plank at the point P, where AP = x The plank is in equilibrium in a vertical plane which is perpendicular to the wall The angle between the rope and the plank is α, where tan ∝ = 3 4 The plank is modelled as a uniform rod, the block is modelled as a particle and the rope is modelled as a light inextensible string Using the model, show that the tension in the rope is 5𝑀𝑔 3𝑥+𝑎 6𝑎 sin 𝜃 = 3 5 cos 𝜃 = 4 5 3 4 5 𝜃 Q from 2018 June exam Mg 3Mg T R Friction

WB 9a (exam Q cont ) Using the model, show that the tension in the rope is 5𝑀𝑔 3𝑥+𝑎 6𝑎 a) Taking moments about A 𝑇 sin ∝ ×2𝑎 =𝑀𝑔×𝑎+3𝑀𝑔×𝑥 Mg 3Mg R Friction a −𝑇𝑐𝑜𝑠 ∝ 𝑇 sin ∝ Rearranges to 𝑇× 6𝑎 5 =𝑀𝑔×𝑎+3𝑀𝑔×𝑥 Rearranges to 𝑇= 5 6𝑎 × 𝑀𝑔𝑎+3𝑀𝑔𝑥 = 5𝑀𝑔 𝑎+3𝑥 6𝑎 QED

2𝑀𝑔 =𝑇𝑐𝑜𝑠 ∝ WB 9b (exam Q cont ) The magnitude of the horizontal component of the force exerted on the plank at A by the wall is 2Mg Find x in terms of a b) Horizontally forces are in equilibrium 2𝑀𝑔 =𝑇𝑐𝑜𝑠 ∝ Mg 3Mg R=2Mg Friction a −𝑇𝑐𝑜𝑠 ∝ 𝑇 sin ∝ Substituting result from a) gives 2𝑀𝑔 = 5𝑀𝑔 𝑎+3𝑥 6𝑎 𝑐𝑜𝑠 ∝ 2𝑀𝑔 = 5𝑀𝑔 𝑎+3𝑥 6𝑎 × 4 5 Cancel by 2Mg to simplify gives 1 = 𝑎+3𝑥 3𝑎 Rearranges to a+3𝑥 =3𝑎 then to 𝑥= 2 3 𝑎

WB 9c (exam Q cont ) The force exerted on the plank at A by the wall acts in a direction which makes an angle β with the horizontal c) Find the value of tan 𝛽 c) Resolve forces vertically 𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛+𝑇𝑠𝑖𝑛 ∝=4𝑀𝑔 Mg 3Mg R=2Mg Friction a −𝑇𝑐𝑜𝑠 ∝ 𝑇 sin ∝ 𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛+ 5𝑀𝑔 𝑎+3𝑥 6𝑎 × 3 5 =4𝑀𝑔 Rearranges using 𝑥= 2 3 𝑎 𝐹𝑟𝑖𝑐𝑡𝑖𝑜𝑛=4𝑀𝑔− 𝑀𝑔 𝑎+2𝑎 2𝑎 = 5 2 Mg Now find tan  R=2Mg Friction = 5 2 Mg tan 𝛽 = 5 2 Mg 2𝑀𝑔 = 5 4 𝛽

simplify by cancelling and rearrange WB 9d (exam Q cont ) The rope will break if the tension in it exceeds 5 Mg d) Explain how this will restrict the possible positions of P. You must justify your answer carefully previous results: 𝑇= 5𝑀𝑔 𝑎+3𝑥 6𝑎 Mg 3Mg R=2Mg Friction= 2 5 𝑀𝑔 a −𝑇𝑐𝑜𝑠 ∝ 𝑇 sin ∝ we know: 𝑇= 5𝑀𝑔 𝑎+3𝑥 6𝑎 ≤5𝑀𝑔 simplify by cancelling and rearrange 5 𝑎+3𝑥 6𝑎 ≤5 5a+15x ≤30𝑎 So the distance AP must be less than 5 3 𝑎 For the rope NOT to break 𝑥 ≤ 5 3 𝑎

One thing to improve is – KUS objectives BAT solve problems using moments and friction at angles to a rod/ object using sohCahToa self-assess One thing learned is – One thing to improve is –

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