Equilibrium Moments and Coefficient of Friction

Equilibrium Moment at angles and coefficient friction

Moments: At an angle KUS objectives BAT solve EQUILIBRIUM problems using moments and forces at angles to a rod/ object using sohCahToa Starter:

Limiting Equilibrium: This is when the magnitude of the frictional force is just sufficient to prevent relative motion Friction is a Force on objects caused by a touching surfaces resistance to a direction . Rougher surfaces cause greater friction The coefficient of friction is a measure of this roughness of different surfaces. There is a direct relationship between friction and the reaction force when an object is touching a surface when a particle is in equilibrium and static ???????? ?? ???????? = ?? when a particle is moving Remember: An object moving at a constant velocity will still be in equilibrium An accelerating object is subject to F=ma

WB40a A ladder PQ of mass 10 kg and length 6 m rests with the end P on rough horizontal ground. The other end Q rests against a smooth vertical wall. A child with mass 20 kg stands at the top of the ladder at Q. 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 65 with the ground. Find the coefficient of friction between the rod and the ground N ? Use R with the rough surface 3 20g Remember ?? = ? R R Friction is NOT related to N 65 R 3 10g Careful with cos and sin ! 65 65 65 ?? ?

WB40b N ? R( ) R = ??? + ?? ? = ??? 3 20g At equilibrium ?? = ? R R = ???? 65 R 3 10g 65 M(Q) ???????? ? + ?? ??? ?? ? = ???? ?? ? 65 65 ?? ? ???????? + ????? ??? ?? = ???? ??? ?? ? =180 ??? 65 30???65 180 ??? 65 = ?.???

WB41 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 PC = 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 a) M(P) ? ? 40? ? 8 = 40? 5cos15 15 40? 5 cos 15 8 ? 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

WB42a 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 ? ????????

WB42b (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 M(Q) to get an equation with Friction and R ? ? 3?cos60 = ?? 3?sin60+?? 3 2? 2?cos60 + 2?? 2?cos60 ? R Substituting ?? = ?? and R = 3?? 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 3cos60 +3 2cos60 2 2cos60 3 3sin60 = = 0.225 ? = 54

WB44 Exam Q A uniform ladder AB, of length 2d and weight W, has its end A on rough horizontal ground. The coefficient of friction between the ladder and the ground is 1 4 . The end B of the ladder is resting against a smooth vertical wall, as show. A builder of weight 7W stands at the top of the ladder. To stop the ladder from slipping, the builder s assistant applies a horizontal force of magnitude P to the ladder at A, towards the wall as shown. The ladder rests in equilibrium in a vertical plane perpendicular to the wall and makes an angle with the horizontal ground, where tan? =5 as a uniform rod. (a) Show that the reaction of the wall on the ladder at B has magnitude 3W. (5) (b) Find, in terms of W, the range of possible values of P for which the ladder remains in equilibrium. (5) Often in practice, the builder s assistant will simply stand on the bottom of the ladder. (c) Explain briefly how this helps to stop the ladder from slipping. (3) 2. The builder is modelled as a particle and the ladder is modelled a) M(A) ? ??? ? ? + ????? ? ?? = ? ??? ? ?? ? ? ? + ??? = ?? ??? ? ?? ? ??? ???? ? = ?? ? = ?? ? ? Fr 75 ? ?

WB44b Exam Q a) Show that the reaction of the wall on the ladder at B has magnitude 3W. (5) (b) Find, in terms of W, the range of possible values of P for which the ladder remains in equilibrium. (5) ? b) R( ) ? = 8? ? = ?? ?? ? ?? = ? R R =1 4 8? = 2 ? ?? ? ? If the ladder is about to move right due to P then Friction is at its maximum R( ) ????= ? + ?? = 3? + 2? = 5? Fr 75 ? ? If the ladder is about to move left then friction acts in the opposite direction R( ) ????= ? ?? = 3? 2? = ? So W ? 5?

WB44c Exam Q Often in practice, the builder s assistant will simply stand on the bottom of the ladder. (c) Explain briefly how this helps to stop the ladder from slipping. (3) c) Three marks ! ? ? = ?? ?? First, N is unaffected by the assistant standing on the ladder, same moments equation, same value N M(A) ? ?? ? ? ? ??? ? ? + ????? ? ?? = ? ??? ? ?? Fr 75 Second, R increases when the assistant stands on ladder R( ) ? = 8? + ? ? ? Third, R increases so the value of limiting friction (max Friction) increases which helps stop the ladder from slipping

WB44 Two interlocking gears are in equilibrium. The gear on the right has a radius of 10 cm and has a loop 8 cm from the centre. The loop is to the right of, and level with the centre of the gear. A 10 kg mass hangs from the loop. The other gear has a radius of 5 cm and a loop 2 cm from the centre. The loop is to the left of, and level with the centre of the gear. A mass M kg hangs from the left loop. Find the value of M. Q from exemplar materials 2017 (no friction)

WB10Two interlocking gears are in equilibrium. The gear on the right has a radius of 10 cm and has a loop 8 cm from the centre. The loop is to the right of, and level with the centre of the gear. A 10 kg mass hangs from the loop. The other gear has a radius of 5 cm and a loop 2 cm from the centre. The loop is to the left of, and level with the centre of the gear. A mass M kg hangs from the left loop. Find the value of M. Weight of right mass is 10g (N) Moment on right gear is force distance from centre of right gear Moment = 10g 0.08 = 0.8g (N m) =0.8? 0.1= 8? ?????? ???????? Force on left gear by right gear is Moment on left gear is force distance from centre = 8g 0.05 = 0.4g (N m) ?????? ???????? =0.4? 0.02= 20? ???? ? = M = 20g g= 20 (kg)

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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