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Physics chapter 4: Forces in action revision: Newtons 2nd law: The net force-mass & acceleration (For a constant mass, F = ma AIN Fner=2-1=1N in upwards. direction. IN -2 1 newton = the force required to accelerate a mass of 1 kg by 1 ms² Rin -2 | N=1 kg x 1 ms ²² Forces: weight fner = IN m Img Upthrust: Friction: Normal Forces: 2 Friction (8) Fnet-ON M w = mg always acts vertically downwards. g=9.81m5²² ↑N = normal 12N W IN Force - 2 electrons cannot occupy the same space at the same time. &w=mg - wil always be equal + opposite to Tension: gravity assuming that the surface is level. CA rope pulling an object) mg Free body diagrams: IN 2N → draw to scale → use dot (•) for object. Fnet in norizontal direction = 2-1 = IN to the right. Example Q: A helicopter is lifting a mass of 100 kg at terminal velocity. 1) Find the tension in the cable: F=ma net torke =0 100 ×9.81=981 N T= mg 2) The Force is now increased and the load experiences an acceleration of I ms2 Find the new tension in the cable: Fnet = T-W₂₁ ma= =T-W T = ma+w T = 100 × 1 +100 × 9.81 T= 1081N acceleration is decreasing. 3. terminal Drag: The acceleration is not constant. Da v² If an object travels through a fluid, it will experience drag. factors that affect drag in air: → Surface Area →Speed velocity. acceleration is 9 = IF A=0 TEW: 1) mg >no resultant force 21 mg Fnet-mg-D = 9.81*100=981N 3) 60 √ D = mg Fnet = 0 Practical investigation of the motion in a fuuid: 11 Flil a...
iOS User
Stefan S, iOS User
SuSSan, iOS User
tube with viscous Fluid. 21 mark equal consequtive (10cm) on the tube. 31 drop a bau bearing, and record time it takes to reach each interval. 4) take multiple readings and find the mean to reduce the uncertai- nty. bau bearing 110 cm use a magnet to retrieve img ball bearing. Forces on a slope: •N= normal reaction 90-e 90° to surface. V terminal velocity. masin t Parallel to Plane: Perpendicular to piane: mg cos o The center of gravity = The point at which the entire weight of an object seems to act. The centre of mass = The point where the entire mass of the object seems to act How to determine center of gravity: 1. Pin an object + hang a plumbline. 2. Rotate the object and repeat. 3. Again. 4. intersection of those lines = the center of gravity. Moments: The product of the force and the perpendicular distance from the pivot to the line of action of the Force. unit = Nm (Force & distance) pivot 7 m = F x Clockwise = Clockwise moment M=FX m=6×0.1 F2 Anticlockwise = 0.5 Nm 1. 3. SOH CAH TOA x Pivot 0.2m FT:5N 30 0.2m H wall 30 f 'al m= T=5N 0.2m 30 FOO Sin 30=0=x H 0.2 String = x X-0.2 Sin 30 x₂ 0.1m shelt Perpendicular d From pivot to line of action of force. 2. araw 90°From Slope to pivot T-5N 30° ↑ 0.2m we want this length = X sin Sin 30 = 0 H 0.2 x = 0.2 sin 30° x = 0.1m The principle of moments: An object will be in equilibrium if: 7 11 There is no net torce acting on the object. 21 Sum of anticlockwise moments = Sum of clockwise moments. FX = F₂ X₂ Xi Torave: Force couples and torames: A Force couple consists of 2 caual opposite Forces acting in the same. plane F₁ 30 x ₂ T=Fxd F₁ = F₂ = F Triangle of Forces: F kg m-3 ↓ F₂ N VW N 30 W Density and pressure: ↓ R=m j=d F₂ F₁ = F₂ in magnitude but opposite in direction. otherwise there win be acceleration. True only if the object is in equilibrium. -2 NM = Pa F J N 50% W Pressure in a Huid: >A M 1. Archimedes Principle: The upthrust acting on a body submerged in a liquid is equal to the weight of the liquid that has been displaced. Lmg P= F = mg A A rotation. d m = ev = pvg = pang png A A M = W water M = mass water g : M=pVg M = p Axg moments Practice Q's: F↓ upiniust density area height +gravity. X : P=png dz JFz m=Fxd anticlockwise M₂ F₂ xd₂ Clockwise : T=5N (30 0.2m M₁ = Fxd perp : = 5 × a perp 1 / 0.2 a perpendicular sin 30° = 0 = ∞ 5x 0.1=0.5 Nm ✓ 2. CW = ACW 0.2 sin 30x 2:0km 1. State the principle of moments: -Sum or CIOCKwise moments = surm or anticlockwise moments For objects. in equilibrium. 50 x 0.46 W x 0.14 50*0.46 = W 3 0.14 164 N w = mg m = @= 164 = 60.7 kg 99.81
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All the physics equations for aqa!!! I found this and it’s super helpful, I hope it can help you too.
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Topic notes for AQA A Level Physics - Mechanics
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Forces notes for GCSE triple physics
141
Forces notes made from free science lessons aqa physics
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Revision notes covering mechanics and completed exam practise questions.
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Summary of notes for Alevel Physics AQA - Section 4 mechanics
Physics chapter 4: Forces in action revision: Newtons 2nd law: The net force-mass & acceleration (For a constant mass, F = ma AIN Fner=2-1=1N in upwards. direction. IN -2 1 newton = the force required to accelerate a mass of 1 kg by 1 ms² Rin -2 | N=1 kg x 1 ms ²² Forces: weight fner = IN m Img Upthrust: Friction: Normal Forces: 2 Friction (8) Fnet-ON M w = mg always acts vertically downwards. g=9.81m5²² ↑N = normal 12N W IN Force - 2 electrons cannot occupy the same space at the same time. &w=mg - wil always be equal + opposite to Tension: gravity assuming that the surface is level. CA rope pulling an object) mg Free body diagrams: IN 2N → draw to scale → use dot (•) for object. Fnet in norizontal direction = 2-1 = IN to the right. Example Q: A helicopter is lifting a mass of 100 kg at terminal velocity. 1) Find the tension in the cable: F=ma net torke =0 100 ×9.81=981 N T= mg 2) The Force is now increased and the load experiences an acceleration of I ms2 Find the new tension in the cable: Fnet = T-W₂₁ ma= =T-W T = ma+w T = 100 × 1 +100 × 9.81 T= 1081N acceleration is decreasing. 3. terminal Drag: The acceleration is not constant. Da v² If an object travels through a fluid, it will experience drag. factors that affect drag in air: → Surface Area →Speed velocity. acceleration is 9 = IF A=0 TEW: 1) mg >no resultant force 21 mg Fnet-mg-D = 9.81*100=981N 3) 60 √ D = mg Fnet = 0 Practical investigation of the motion in a fuuid: 11 Flil a...
Physics chapter 4: Forces in action revision: Newtons 2nd law: The net force-mass & acceleration (For a constant mass, F = ma AIN Fner=2-1=1N in upwards. direction. IN -2 1 newton = the force required to accelerate a mass of 1 kg by 1 ms² Rin -2 | N=1 kg x 1 ms ²² Forces: weight fner = IN m Img Upthrust: Friction: Normal Forces: 2 Friction (8) Fnet-ON M w = mg always acts vertically downwards. g=9.81m5²² ↑N = normal 12N W IN Force - 2 electrons cannot occupy the same space at the same time. &w=mg - wil always be equal + opposite to Tension: gravity assuming that the surface is level. CA rope pulling an object) mg Free body diagrams: IN 2N → draw to scale → use dot (•) for object. Fnet in norizontal direction = 2-1 = IN to the right. Example Q: A helicopter is lifting a mass of 100 kg at terminal velocity. 1) Find the tension in the cable: F=ma net torke =0 100 ×9.81=981 N T= mg 2) The Force is now increased and the load experiences an acceleration of I ms2 Find the new tension in the cable: Fnet = T-W₂₁ ma= =T-W T = ma+w T = 100 × 1 +100 × 9.81 T= 1081N acceleration is decreasing. 3. terminal Drag: The acceleration is not constant. Da v² If an object travels through a fluid, it will experience drag. factors that affect drag in air: → Surface Area →Speed velocity. acceleration is 9 = IF A=0 TEW: 1) mg >no resultant force 21 mg Fnet-mg-D = 9.81*100=981N 3) 60 √ D = mg Fnet = 0 Practical investigation of the motion in a fuuid: 11 Flil a...
iOS User
Stefan S, iOS User
SuSSan, iOS User
tube with viscous Fluid. 21 mark equal consequtive (10cm) on the tube. 31 drop a bau bearing, and record time it takes to reach each interval. 4) take multiple readings and find the mean to reduce the uncertai- nty. bau bearing 110 cm use a magnet to retrieve img ball bearing. Forces on a slope: •N= normal reaction 90-e 90° to surface. V terminal velocity. masin t Parallel to Plane: Perpendicular to piane: mg cos o The center of gravity = The point at which the entire weight of an object seems to act. The centre of mass = The point where the entire mass of the object seems to act How to determine center of gravity: 1. Pin an object + hang a plumbline. 2. Rotate the object and repeat. 3. Again. 4. intersection of those lines = the center of gravity. Moments: The product of the force and the perpendicular distance from the pivot to the line of action of the Force. unit = Nm (Force & distance) pivot 7 m = F x Clockwise = Clockwise moment M=FX m=6×0.1 F2 Anticlockwise = 0.5 Nm 1. 3. SOH CAH TOA x Pivot 0.2m FT:5N 30 0.2m H wall 30 f 'al m= T=5N 0.2m 30 FOO Sin 30=0=x H 0.2 String = x X-0.2 Sin 30 x₂ 0.1m shelt Perpendicular d From pivot to line of action of force. 2. araw 90°From Slope to pivot T-5N 30° ↑ 0.2m we want this length = X sin Sin 30 = 0 H 0.2 x = 0.2 sin 30° x = 0.1m The principle of moments: An object will be in equilibrium if: 7 11 There is no net torce acting on the object. 21 Sum of anticlockwise moments = Sum of clockwise moments. FX = F₂ X₂ Xi Torave: Force couples and torames: A Force couple consists of 2 caual opposite Forces acting in the same. plane F₁ 30 x ₂ T=Fxd F₁ = F₂ = F Triangle of Forces: F kg m-3 ↓ F₂ N VW N 30 W Density and pressure: ↓ R=m j=d F₂ F₁ = F₂ in magnitude but opposite in direction. otherwise there win be acceleration. True only if the object is in equilibrium. -2 NM = Pa F J N 50% W Pressure in a Huid: >A M 1. Archimedes Principle: The upthrust acting on a body submerged in a liquid is equal to the weight of the liquid that has been displaced. Lmg P= F = mg A A rotation. d m = ev = pvg = pang png A A M = W water M = mass water g : M=pVg M = p Axg moments Practice Q's: F↓ upiniust density area height +gravity. X : P=png dz JFz m=Fxd anticlockwise M₂ F₂ xd₂ Clockwise : T=5N (30 0.2m M₁ = Fxd perp : = 5 × a perp 1 / 0.2 a perpendicular sin 30° = 0 = ∞ 5x 0.1=0.5 Nm ✓ 2. CW = ACW 0.2 sin 30x 2:0km 1. State the principle of moments: -Sum or CIOCKwise moments = surm or anticlockwise moments For objects. in equilibrium. 50 x 0.46 W x 0.14 50*0.46 = W 3 0.14 164 N w = mg m = @= 164 = 60.7 kg 99.81