Understanding Moments and Trigonometry in Physics
Explore the concept of moments and trigonometry in physics through examples and problem-solving exercises. Learn how to calculate moments, find forces, and determine resultant moments using trigonometry. Enhance your understanding of the turning effects of forces and their applications in real-world scenarios.
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Moments At angles
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 F P Moment of F about P = Fd clockwise d 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 3 m 5 N P
Using Trigonometry F You may need to use trigonometry to find the perpendicular distance sin d Nm sin Fd Moment of F about P = clockwise P d 10 N e.g. Moment of the force about P = = clockwise Nm 30 10 6 sin 30 30 P 6 m
WB 1 Calculate the moment about point A of each of these forces acting on a Lamina , ? 5 ? ? 40 6sin25 5sin40 6 ? 25 18 ? 10 ? ?????? = 10 5sin40 = 32.1 ?? ??? ?????? = 18 6sin25 = 45.6 ?? ??? 8 ? 70 ? 4sin65 ? 50 120 ? 65 ? 8cos50 4 ? ?????? = 70 4sin65 = 254 ?? ??? ?????? = 120 8cos50 = 617 ?? ??
WB 2 Given the moment about point A of each of these forces is 20 Nm, Find the magnitude of each force 3sin35 ? ? 5 ? 30 3 ? ?2 5sin30 35 20 ?2= 3sin35= 11.6 ? ?1 20 ?1= 5sin30 = 8 ? ?4 20 ?3 ?3= 2sin55= 12.2 ? 9sin75 ? 75 125 9 ? ? 2 ? 20 ?4= 9sin75= 2.30 ? 2sin55
WB 3a Two forces act on a lamina as shown. Calculate the resultant moment about the point A a) ?1= 8 10sin30 Moment of ?1 = 10sin30 8 = ?? ? 30 10 ? 50 Moment of ?2 = 10sin50 5 = ??? 10sin50 ?2= 5 Moment of ?1and ?2== ?? ?????????
WB 3b Two forces act on a lamina as shown. Calculate the resultant moment about the point A b) ?1= 12 70 Moment of ?1 = 10sin70 12 = 112.7 ??? 10 ? 10sin70 ? 8sin60 8 ? Moment of ?2 = 8sin60 15 = 103.9 ?? 60 ?2= 15 Moment of ?1and ?2= 112.7 103.9 = 8.8 ?? ?????????????
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 8 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 R 8 ?? = 8 ?? 1 m 3 m A C B b) Taking moments about C cw 8 ?? = 24 ?? 8 c) Taking moments about C cw 8 ?? = 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) 6cos45 6sin45 5cos70 5sin70 5 ? 6 ? 70 0.5? 45 4 ? 3 ? ? 0 3? Taking moments about A clockwise 5sin70 3 3? 0.5 = 11.8 ?? ????????? 6sin45 4 11.8 ?? ?????????????
WB6 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 A uniform rod PQ is hinged at point P, and is held in equilibrium at an angle of ? ? 8? ? 50 Taking moments about P ? ? 3 = 8? 3 2cos50 8? 3 2cos 50 3 ? = = 25.2 ?
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 a) The magnitude of the reaction at C b) The coefficient of friction between the rod and the ground ? 75 ? ? 40? a) Taking moments about P 15 ? ???????? distance 5cos15 ? 8 = 40? 5cos15 40? 5 cos 15 8 reaction at C ? = = 236.7 ? b) Equilibrium in horizontal direction ???????? = ?cos75 = 236.7cos75 = 61.25 Equilibrium in vertical direction ? + ???? 75 = 40? ? = 40? 236.7cos75 = 163.4 ? ? =?? ?=61.25 Coefficient friction 163.4= 0.375
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 ? Wall smooth No friction Equilibrium in horizontal direction 2? ???????? = ? ? R Equilibrium in vertical direction ? ? = 2?? + ?? = 3?? 60 ? Friction
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 N Taking moments about Q (to get an equation with Friction and R) ? Wall smooth No friction ? 3?cos60= ???????? 3?sin60 +?? 3 2? 2?cos60 + 2?? 2?cos60 Substituting ???????? ?? and R = 3?? ? R 3?? 3?cos60= ?3?? 3?sin60 +?? 3 ? 2?cos60 + 2?? 2?cos60 60 ? Friction Cancel by mga and rearrange to 7 2cos60 9sin60=7 3 3?? 3?cos60 +?? 3 2?cos60 2?? 2?cos60 3?? 3?sin60 = = 0.225 ? = 54
WB 9 (exam Q) 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 , where tan =3 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 4 a) Using the model, show that the tension in the rope is 5?? 3?+? 6? sin? =3 cos? =4 5 5 3 ? T Friction 4 5 R Mg 3Mg
WB 9a (exam Q cont ) a) Using the model, show that the tension in the rope is 5?? 3?+? 6? a) Taking moments about A ?sin 2? = ?? ? + 3?? ? ???? ?sin Friction R Rearranges to ? 6? 5= ?? ? + 3?? ? a Mg 3Mg Rearranges to 5 6? ??? + 3??? =5?? ? + 3? ? = QED 6?
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 2?? = ???? ???? ?sin Friction Substituting result from a) gives R=2Mg 2?? =5?? ? + 3? a ??? 6? Mg 3Mg 2?? =5?? ? + 3? 4 6? 5 ?+3? 3? Cancel by 2Mg to simplify gives 1 = then to ? =2 Rearranges to a + 3? = 3? 3?
WB 9c (exam Q cont ) The force exerted on the plank at A by the wall acts in a direction which makes an angle with the horizontal c) Find the value of tan? c) Resolve forces vertically ???????? + ???? = 4?? ???? ?sin Friction ???????? +5?? ? + 3? 3 5= 4?? R=2Mg 6? Rearranges using ? =2 3? a ???????? = 4?? ?? ? + 2? =5 Mg 3Mg 2Mg 2? Now find tan 52Mg 2?? Friction =5 2Mg =5 tan? = 4 ? R=2Mg
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: ? =5?? ?+3? 6? ???? ?sin Friction=2 5?? we know: ? =5?? ?+3? 5?? 6? R=2Mg a simplify by cancelling and rearrange 5 ?+3? 6? 5a + 15x 30? Mg 3Mg 5 So the distance AP must be less than 5 For the rope NOT to break 3? ? 5 3?
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
