Moments Mixed exercise 4
|
|
- Wilfred Watkins
- 5 years ago
- Views:
Transcription
1 Moments Mixed exercise 4 1 a The plank is in equilibrium. Let the reaction forces at the supports be R and R D. Considering moments about point D: R ( ) = ( ) (3 1.5) 3.5R = R = R = 105 The support at exerts a force of 105 N on the plank. b The plank is in equilibrium. Resolving vertically: R + R = c R D D = RD = 140 The support at D exerts a force of 140 N on the plank. F When the plank is on the point of tilting, the new reaction force at support, R = 0 N and plank is again in equilibrium. The child stands a distance x m from support D. Considering moments about point D: (3 1.5) The distance DF is 1.03 m. Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 1
2 2 a Since the rod is uniform, the centre of mass is at the mid-point. Taking moments about A: Wx+ 2= R 1+ R 2.5, Wx+ 300= 3.5 R (1) R( ): W+ = R+ R, 2R= W+ W+ R= 2 Sub (2) into (1) gives: 7 W+ Wx+ 300= 2 2 4( Wx+ 300) = 7W + 7 4Wx+ 1200= 7W = 7W 4Wx W ( x) 7 4 = W = 7 4x (2) 3 a b The range of values of x are: x 0 and 0 7 4x > 7 4x> 0 So 0 x<1.75 4x< 7 x< 7 4 x< 1.75 The plank is in equilibrium. Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 2
3 3 a Resolving vertically: 2R+ R= 40g+ 80g 4 a 3R= R= 1176 R= 392 The value of R is 392 N. b Taking moments about A: 80 gx+ (40g 2) = 3R 80 g( x+ 1) = x+ 1= x+ 1= 1.5 The man stands 0.5 m from A. c i Assuming the plank is uniform means the weight of the plank acts at its centre of mass: i.e. halfway along the plank. ii Assuming the plank is a rod means its width can be ignored. iii Assuming the man is a particle means all his weight acts at the point at which he stands. b R( ): 100+R=W+ R=W+50 Taking moments about A: (W +50) 3.5= x W =x W =x 550+7W = 300x R( ) Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 3
4 4 b 52+R=+W 5 a R=+W 52 = 98+W Taking moments about : (98+W) 0.5= (4 x) W = 600 x Doubling, 410+W = x W = x c Solving the simultaneous equations obtained in a and b: W = 790 (550+7W) 8W = = 240 W = = x (3 s.f.) The lever is in equilibrium. Considering moments about point C: F (2 0.25) + 100(1 0.25) = A 1.75F + 75= 425 A F F The force at A is 200 N. A A = 1.75 = 200 Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 4
5 5 b The lever is again in equilibrium. Let x be the distance of the pivot from. Considering moments about the new support position C : (2 x) + 100(1 x) = 1700x 300 x x 400= 1700x+ 250x 400= 1950x The pivot is now 0.21 m from (to the nearest cm). 6 a Let the mass of the plank be M. Since the plank is uniform, its centre of mass is at its mid-point. Taking moments about C: 48g 1.8=Mg g g= 0.2Mg+ 79.2g 86.4=0.2M M = = 7.2 M = 36kg b Let the distance C be x Taking moments about C: 36gx+ 36 g( x 2) = 48 g(4 x) 3x+ 3( x 2) = 4(4 x) 6x 6= 16 4x m Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 5
6 7 a Taking moments about C: R A g 0.5=12g 2 R A 4.5= 24g 15g = 9g R A = 2g =19.6N b The plank is about to tilt about C reaction at A= 0 Taking moments about C: mg g 2= 93g m= = 22.5 m= 5 8 The plank is in equilibrium. Resolving vertically: T+ 4T = 50g+ 25g 5T = 75g T = 15g 4T = 60g Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 6
7 8 Considering moments about : (50g 2) = 60 gx+ (15g 4) 100g = 60gx+ 60g 100g 60g = 60gx 40g 60g The distance from to C is 0.67 m (to the nearest cm). 9 a b Taking moments about A: =R C 4 R C =125N Let the distance AD be x R ( ) 2R= = 700 R=350N Taking moments about A: R 4= x 1400=4R = x 900=500x 1.8m Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 7
8 10 Distance MP = x m The plank is in equilibrium. Taking moments about M: 6g 7 50sin sin = The distance MP is 4.88 m (3s.f.). The ladder is in equilibrium. Let x be the distance AC. Resolving horizontally: R = 200 N Considering moments about A: (80g x cos 55 ) + (20g 5cos 55 ) = sin 55 (784 cos 55 ) 2000sin cos sin cos cos The distance AC is 2.39 m (to the nearest cm). 12 a Centre of mass of beam is 5 m from C. Taking moments about C: 3000g 10 = (4000g 5) + Mgx 30000= Mx M = x Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 8
9 12 b Maximum load is when x = 5 m: M = = 2000 kg 5 Minimum load is when x = 20 m: M = = 500 kg 20 c It is not very accurate to model the beam as a uniform rod. Since the beam may taper at one end, the centre of mass of the beam may not lie in the middle of the beam. Challenge 1 Let x be the distance from A to the centre of mass. The beam is in equilibrium. Taking moments about A: 10g x cos 35 + (2g 7 cos 35 ) = 8 50sin 70 10gx cos 35 = 400 sin 70 14g cos sin cos 35 = cos 35 The centre of mass of the beam is 3.28 m from A (3s.f.). 2 a When force is a minimum, system is in limiting equilibrium. Taking moments about P: FA ( A ) = 1200 PC (1) Finding A : A = 2cos20 = 1sin 20 A = 2cos20 + sin20 Finding PC : PC = PC cos( θ+ 20) ( PC) = PC= tanθ = 0.5 θ = PC = 1.25 cos( ) PC = 1.25 Substituting values for A and PC into equation (1) F 2 cos 20 + sin 20 = A ( ) FA = 2cos20 + sin20 FA = A horizontal force of 69.0 N at A will tip the refrigerator (3s.f.). Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 9
10 Challenge 2 b When force is a minimum, system is in limiting equilibrium. Taking moments about P: F ( P ) = F 1cos 20 = F = cos20 F = A vertical force of 163 N at will tip the refrigerator (3s.f.). Pearson Education Ltd Copying permitted for purchasing institution only. This material is not copyright free. 10
( ) Applications of forces 7D. 1 Suppose that the rod has length 2a. Taking moments about A: acos30 3
Applications of forces 7D Suppose that the rod has length a. Taking moments about A: at 80 acos0 T 80 T 0. 6 N R( ), F T sin0 0 7. N R, T cos0 + R 80 R 80 0 50N In order for the rod to remain in equilibrium,
More informationApplications of forces Mixed exercise 7
Applications of forces Mied eercise 7 a Finding the components of P along each ais: ( ): P cos70 + 0sin7 ( ): P sin70 0cos7 Py tanθ P y sin70 0cos7 tanθ 0.634... cos70 + 0sin7 θ 3.6... The angle θ is 3.3
More informationName: M1 - Moments. Date: Time: Total marks available: Total marks achieved:
Name: M1 - Moments Date: Time: Total marks available: Total marks achieved: Questions Q1. Figure 3 A beam AB has weight W newtons and length 4 m. The beam is held in equilibrium in a horizontal position
More informationb UVW is a right-angled triangle, therefore VW is the diameter of the circle. Centre of circle = Midpoint of VW = (8 2) + ( 2 6) = 100
Circles 6F a U(, 8), V(7, 7) and W(, ) UV = ( x x ) ( y y ) = (7 ) (7 8) = 8 VW = ( 7) ( 7) = 64 UW = ( ) ( 8) = 8 Use Pythagoras' theorem to show UV UW = VW 8 8 = 64 = VW Therefore, UVW is a right-angled
More informationProof by induction ME 8
Proof by induction ME 8 n Let f ( n) 9, where n. f () 9 8, which is divisible by 8. f ( n) is divisible by 8 when n =. Assume that for n =, f ( ) 9 is divisible by 8 for. f ( ) 9 9.9 9(9 ) f ( ) f ( )
More informationExam-style practice: Paper 3, Section A: Statistics
Exam-style practice: Paper 3, Section A: Statistics a Use the cumulative binomial distribution tables, with n = 4 and p =.5. Then PX ( ) P( X ).5867 =.433 (4 s.f.). b In order for the normal approximation
More information9 Mixed Exercise. vector equation is. 4 a
9 Mixed Exercise a AB r i j k j k c OA AB 7 i j 7 k A7,, and B,,8 8 AB 6 A vector equation is 7 r x 7 y z (i j k) j k a x y z a a 7, Pearson Education Ltd 7. Copying permitted for purchasing institution
More informationM201 assessment Moments
Do the questions as a test circle questions you cannot answer Red 1) A light see-saw is 10 m long with the pivot 3 m from the left. a) A 4 kg weight is placed on the left-hand end of the see-saw. Write
More informationQ Scheme Marks AOs Pearson ( ) 2. Notes. Deduces that 21a 168 = 0 and solves to find a = 8 A1* 2.2a
Further Maths Core Pure (AS/Year 1) Unit Test : Matrices Q Scheme Marks AOs Pearson Finds det M = 3 p p+ 4 = p + 4 p+ 6 1 ( )( ) ( )( ) ( ) Completes the square to show p + 4p+ 6= p+ + M1.a Concludes that
More informationAQA Maths M2. Topic Questions from Papers. Moments and Equilibrium
Q Maths M2 Topic Questions from Papers Moments and Equilibrium PhysicsndMathsTutor.com PhysicsndMathsTutor.com 11 uniform beam,, has mass 20 kg and length 7 metres. rope is attached to the beam at. second
More informationApplications of forces 7C
Applications of forces 7C 1 Let the normal reaction be R N, the friction force be F N and the coefficient of friction be µ. Resolve horizontally to find F, vertically to find R and use F µr to find µ:
More informationReview exercise 2. 1 The equation of the line is: = 5 a The gradient of l1 is 3. y y x x. So the gradient of l2 is. The equation of line l2 is: y =
Review exercise The equation of the line is: y y x x y y x x y 8 x+ 6 8 + y 8 x+ 6 y x x + y 0 y ( ) ( x 9) y+ ( x 9) y+ x 9 x y 0 a, b, c Using points A and B: y y x x y y x x y x 0 k 0 y x k ky k x a
More informationElastic strings and springs Mixed exercise 3
Elastic strings and springs Mixed exercise ( )T cosθ mg By Hooke s law 5mgx T 6a a sinθ a + x () () () a 4 5mg Ifcos θ, T from () 5 8 5mg 5mgx so, from () 8 6a a x 4 If cos θ, then sinθ 5 5 a sinθ from
More informationCHAPTER 12 STATIC EQUILIBRIUM AND ELASTICITY. Conditions for static equilibrium Center of gravity (weight) Examples of static equilibrium
CHAPTER 12 STATIC EQUILIBRIUM AND ELASTICITY As previously defined, an object is in equilibrium when it is at rest or moving with constant velocity, i.e., with no net force acting on it. The following
More information1 The diagram shows an object of weight W and an object of weight Z balanced on a uniform metre rule. object of weight W. object of weight Z
1 The diagram shows an object of weight W and an object of weight Z balanced on a uniform metre rule. object of weight W a b 50 cm mark object of weight Z metre rule Which equation relating to W, Z, a
More informationMoments of forces. Rudolf arnheim
T C Y Moments of forces The photograph shows a mobile tower crane, consisting of the main vertical section housing the engine, winding gear and controls and the boom which supports the load and the counterweight.
More informationQ Scheme Marks AOs Pearson. Notes. Deduces that 21a 168 = 0 and solves to find a = 8 A1* 2.2a
Further Maths Core Pure (AS/Year 1) Unit Test : Matrices Q Scheme Marks AOs Pearson Finds det M 3 p p 4 p 4 p 6 1 Completes the square to show p 4 p 6 p M1.a Concludes that (p + ) + > 0 for all values
More informationChapter 12. Static Equilibrium and Elasticity
Chapter 12 Static Equilibrium and Elasticity Static Equilibrium Equilibrium implies that the object moves with both constant velocity and constant angular velocity relative to an observer in an inertial
More informationSection 2: Static Equilibrium II- Balancing Torques
Section 2: Static Equilibrium II- Balancing Torques Last Section: If (ie. Forces up = Forces down and Forces left = Forces right), then the object will have no translatory motion. In other words, the object
More informationEdexcel GCE Mechanics M1 Advanced/Advanced Subsidiary
Centre No. Candidate No. Paper Reference(s) 6677/01 Edexcel GCE Mechanics M1 Advanced/Advanced Subsidiary Wednesday 16 May 2012 Morning Time: 1 hour 30 minutes Materials required for examination Mathematical
More informationChapter Objectives. Copyright 2011 Pearson Education South Asia Pte Ltd
Chapter Objectives To develop the equations of equilibrium for a rigid body. To introduce the concept of the free-body diagram for a rigid body. To show how to solve rigid-body equilibrium problems using
More informationReview Exercise 2. 1 a Chemical A 5x+ Chemical B 2x+ 2y12 [ x+ Chemical C [ 4 12]
Review Exercise a Chemical A 5x+ y 0 Chemical B x+ y [ x+ y 6] b Chemical C 6 [ ] x+ y x+ y x, y 0 c T = x+ y d ( x, y) = (, ) T = Pearson Education Ltd 08. Copying permitted for purchasing institution
More informationCircles, Mixed Exercise 6
Circles, Mixed Exercise 6 a QR is the diameter of the circle so the centre, C, is the midpoint of QR ( 5) 0 Midpoint = +, + = (, 6) C(, 6) b Radius = of diameter = of QR = of ( x x ) + ( y y ) = of ( 5
More informationConstant acceleration, Mixed Exercise 9
Constant acceleration, Mixed Exercise 9 a 45 000 45 km h = m s 3600 =.5 m s 3 min = 80 s b s= ( a+ bh ) = (60 + 80).5 = 5 a The distance from A to B is 5 m. b s= ( a+ bh ) 5 570 = (3 + 3 + T ) 5 ( T +
More informationSolutionbank M1 Edexcel AS and A Level Modular Mathematics
Page of Solutionbank M Exercise A, Question A particle P of mass 0. kg is moving along a straight horizontal line with constant speed m s. Another particle Q of mass 0.8 kg is moving in the same direction
More informationTorque and Static Equilibrium
Torque and Static Equilibrium Rigid Bodies Rigid body: An extended object in which the distance between any two points in the object is constant in time. Examples: sphere, disk Effect of external forces
More informationTaylor Series 6B. lim s x. 1 a We can evaluate the limit directly since there are no singularities: b Again, there are no singularities, so:
Taylor Series 6B a We can evaluate the it directly since there are no singularities: 7+ 7+ 7 5 5 5 b Again, there are no singularities, so: + + c Here we should divide through by in the numerator and denominator
More informationChapter 12 Static Equilibrium; Elasticity and Fracture
2009 Pearson Education, Inc. This work is protected by United States copyright laws and is provided solely for the use of instructors in teaching their courses and assessing student learning. Dissemination
More informationUnit 4 Statics. Static Equilibrium Translational Forces Torque
Unit 4 Statics Static Equilibrium Translational Forces Torque 1 Dynamics vs Statics Dynamics: is the study of forces and motion. We study why objects move. Statics: is the study of forces and NO motion.
More informationis the study of and. We study objects. is the study of and. We study objects.
Static Equilibrium Translational Forces Torque Unit 4 Statics Dynamics vs Statics is the study of and. We study objects. is the study of and. We study objects. Recall Newton s First Law All objects remain
More informationFr h mg rh h. h 2( m)( m) ( (0.800 kg)(9.80 m/s )
5. We consider the wheel as it leaves the lower floor. The floor no longer exerts a force on the wheel, and the only forces acting are the force F applied horizontally at the axle, the force of gravity
More information( y) ( ) ( ) ( ) ( ) ( ) Trigonometric ratios, Mixed Exercise 9. 2 b. Using the sine rule. a Using area of ABC = sin x sin80. So 10 = 24sinθ.
Trigonometric ratios, Mixed Exercise 9 b a Using area of ABC acsin B 0cm 6 8 sinθ cm So 0 4sinθ So sinθ 0 4 θ 4.6 or 3 s.f. (.) As θ is obtuse, ABC 3 s.f b Using the cosine rule b a + c ac cos B AC 8 +
More informationTorque. Physics 6A. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB
Physics 6A Torque is what causes angular acceleration (just like a force causes linear acceleration) Torque is what causes angular acceleration (just like a force causes linear acceleration) For a torque
More information( ) ( ) or ( ) ( ) Review Exercise 1. 3 a 80 Use. 1 a. bc = b c 8 = 2 = 4. b 8. Use = 16 = First find 8 = 1+ = 21 8 = =
Review Eercise a Use m m a a, so a a a Use c c 6 5 ( a ) 5 a First find Use a 5 m n m n m a m ( a ) or ( a) 5 5 65 m n m a n m a m a a n m or m n (Use a a a ) cancelling y 6 ecause n n ( 5) ( 5)( 5) (
More information( ) Trigonometric identities and equations, Mixed exercise 10
Trigonometric identities and equations, Mixed exercise 0 a is in the third quadrant, so cos is ve. The angle made with the horizontal is. So cos cos a cos 0 0 b sin sin ( 80 + 4) sin 4 b is in the fourth
More informationConsider two students pushing with equal force on opposite sides of a desk. Looking top-down on the desk:
1 Bodies in Equilibrium Recall Newton's First Law: if there is no unbalanced force on a body (i.e. if F Net = 0), the body is in equilibrium. That is, if a body is in equilibrium, then all the forces on
More informationCHAPTER 8: ROTATIONAL OF RIGID BODY PHYSICS. 1. Define Torque
7 1. Define Torque 2. State the conditions for equilibrium of rigid body (Hint: 2 conditions) 3. Define angular displacement 4. Define average angular velocity 5. Define instantaneous angular velocity
More information2008 FXA THREE FORCES IN EQUILIBRIUM 1. Candidates should be able to : TRIANGLE OF FORCES RULE
THREE ORCES IN EQUILIBRIUM 1 Candidates should be able to : TRIANGLE O ORCES RULE Draw and use a triangle of forces to represent the equilibrium of three forces acting at a point in an object. State that
More informationEdexcel GCE Mechanics M1 Advanced/Advanced Subsidiary
Centre No. Candidate No. Paper Reference(s) 6677/01 Edexcel GCE Mechanics M1 Advanced/Advanced Subsidiary Monday 13 May 2013 Afternoon Time: 1 hour 30 minutes Materials required for examination Mathematical
More informationFinds the value of θ: θ = ( ). Accept awrt θ = 77.5 ( ). A1 ft 1.1b
1a States that a = 4. 6 + a = 0 may be seen. B1 1.1b 4th States that b = 5. 4 + 9 + b = 0 may be seen. B1 1.1b () Understand Newton s first law and the concept of equilibrium. 1b States that R = i 9j (N).
More informationPhysicsAndMathsTutor.com
1. A beam AB is supported by two vertical ropes, which are attached to the beam at points P and Q, where AP = 0.3 m and BQ = 0.3 m. The beam is modelled as a uniform rod, of length 2 m and mass 20 kg.
More informationPhysicsAndMathsTutor.com
1. A uniform rod AB, of mass 20 kg and length 4 m, rests with one end A on rough horizontal ground. The rod is held in limiting equilibrium at an angle α to the horizontal, where tan 3 α =, by a force
More informationPhysics 8 Wednesday, October 28, 2015
Physics 8 Wednesday, October 8, 015 HW7 (due this Friday will be quite easy in comparison with HW6, to make up for your having a lot to read this week. For today, you read Chapter 3 (analyzes cables, trusses,
More informationRotational N.2 nd Law
Lecture 19 Chapter 12 Rotational N.2 nd Law Torque Newton 2 nd Law again!? That s it. He crossed the line! Course website: http://faculty.uml.edu/andriy_danylov/teaching/physicsi IN THIS CHAPTER, you will
More informationChapter 8. Centripetal Force and The Law of Gravity
Chapter 8 Centripetal Force and The Law of Gravity Centripetal Acceleration An object traveling in a circle, even though it moves with a constant speed, will have an acceleration The centripetal acceleration
More information40 N cos 35 =49N. T 1 = 28 T y
16 (a) Analyzing vertical forces where string 1 and string 2 meet, we find T 1 = 40 N cos 35 =49N (b) Looking at the horizontal forces at that point leads to T 2 = T 1 sin 35 = (49 N) sin 35 =28N (c) We
More informationP12 Torque Notes.notebook. March 26, Torques
Torques The size of a torque depends on two things: 1. The size of the force being applied (a larger force will have a greater effect) 2. The distance away from the pivot point (the further away from this
More informationMark Scheme (Results) June GCE Mechanics M3 (6679) Paper 1
Mark (Results) June 011 GCE Mechanics M (6679) Paper 1 Edexcel is one of the leading examining and awarding bodies in the UK and throughout the world. We provide a wide range of qualifications including
More informationElastic collisions in two dimensions 5B
Elastic collisions in two dimensions 5B a First collision: e=0.5 cos α = cos30 () sin α = 0.5 sin30 () Squaring and adding equations () and () gies: cos α+ sin α = 4cos 30 + sin 30 (cos α+ sin α)= 4 3
More informationStatic Equilibrium; Torque
Static Equilibrium; Torque The Conditions for Equilibrium An object with forces acting on it, but that is not moving, is said to be in equilibrium. The first condition for equilibrium is that the net force
More informationQ Scheme Marks AOs. 1a States or uses I = F t M1 1.2 TBC. Notes
Q Scheme Marks AOs Pearson 1a States or uses I = F t M1 1.2 TBC I = 5 0.4 = 2 N s Answer must include units. 1b 1c Starts with F = m a and v = u + at Substitutes to get Ft = m(v u) Cue ball begins at rest
More informationA level Exam-style practice
A level Exam-style practice 1 a e 0 b Using conservation of momentum for the system : 6.5 40 10v 15 10v 15 3 v 10 v 1.5 m s 1 c Kinetic energy lost = initial kinetic energy final kinetic energy 1 1 6.5
More informationHATZIC SECONDARY SCHOOL
HATZIC SECONDARY SCHOOL PROVINCIAL EXAMINATION ASSIGNMENT STATIC EQUILIBRIUM MULTIPLE CHOICE / 33 OPEN ENDED / 80 TOTAL / 113 NAME: 1. State the condition for translational equilibrium. A. ΣF = 0 B. ΣF
More informationEngineering Mechanics Statics
Mechanical Systems Engineering _ 2016 Engineering Mechanics Statics 7. Equilibrium of a Rigid Body Dr. Rami Zakaria Conditions for Rigid-Body Equilibrium Forces on a particle Forces on a rigid body The
More informationA-Level Mathematics. MM2B Mechanics 2B Final Mark scheme June Version/Stage: v1.0
A-Level Mathematics MM2B Mechanics 2B Final Mark scheme 6360 June 2017 Version/Stage: v1.0 Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant questions,
More informationSolutionbank Edexcel AS and A Level Modular Mathematics
Page of Exercise A, Question The curve C, with equation y = x ln x, x > 0, has a stationary point P. Find, in terms of e, the coordinates of P. (7) y = x ln x, x > 0 Differentiate as a product: = x + x
More informationThe normal distribution Mixed exercise 3
The normal distribution Mixed exercise 3 ~ N(78, 4 ) a Using the normal CD function, P( 85).459....4 (4 d.p.) b Using the normal CD function, P( 8).6946... The probability that three men, selected at random,
More informationWhen the applied force is not perpendicular to the crowbar, for example, the lever arm is found by drawing the perpendicular line from the fulcrum to
When the applied force is not perpendicular to the crowbar, for example, the lever arm is found by drawing the perpendicular line from the fulcrum to the line of action of the force. We call torques that
More informationA Level. A Level Physics. Mechanics: Equilibrium And Moments (Answers) AQA, Edexcel, OCR. Name: Total Marks: /30
Visit http://www.mathsmadeeasy.co.uk/ for more fantastic resources. AQA, Edexcel, OCR A Level A Level Physics Mechanics: Equilibrium And Moments (Answers) Name: Total Marks: /30 Maths Made Easy Complete
More informationChapter 12 Static Equilibrium
Chapter Static Equilibrium. Analysis Model: Rigid Body in Equilibrium. More on the Center of Gravity. Examples of Rigid Objects in Static Equilibrium CHAPTER : STATIC EQUILIBRIUM AND ELASTICITY.) The Conditions
More informationPure Mathematics Year 1 (AS) Unit Test 1: Algebra and Functions
Pure Mathematics Year (AS) Unit Test : Algebra and Functions Simplify 6 4, giving your answer in the form p 8 q, where p and q are positive rational numbers. f( x) x ( k 8) x (8k ) a Find the discriminant
More informationMark scheme Pure Mathematics Year 1 (AS) Unit Test 2: Coordinate geometry in the (x, y) plane
Mark scheme Pure Mathematics Year 1 (AS) Unit Test : Coordinate in the (x, y) plane Q Scheme Marks AOs Pearson 1a Use of the gradient formula to begin attempt to find k. k 1 ( ) or 1 (k 4) ( k 1) (i.e.
More informationHow Can Motion be Described? and Explained?
How Can Motion be Described? and Explained? Lesson 14: Torque and the Stability of Structures Stable Structures Explain why structures should be stable. What are the conditions for a structure to be stable?
More informationElastic collisions in two dimensions 5C
Elastic collisions in two dimensions 5C 1 No change in component of velocity perpendicular to line of centres. So component of velocity for A=6sin10 Since B is stationary before impact, it will be moving
More informationSection 2: Static Equilibrium II- Balancing Torques
Section 2: Static Equilibrium II- Balancing Torques Last Section: If (ie. Forces up = Forces down and Forces left = Forces right), then the object will have no translatory motion. In other words, the object
More informationEdexcel GCE Mechanics M1 Advanced/Advanced Subsidiary
Centre No. Candidate No. Paper Reference 6 6 7 7 0 1 Paper Reference(s) 6677/01 Edexcel GCE Mechanics M1 Advanced/Advanced Subsidiary Friday 20 January 2012 Afternoon Time: 1 hour 30 minutes Surname Signature
More informationPHYSICS - CLUTCH CH 13: ROTATIONAL EQUILIBRIUM.
!! www.clutchprep.com EXAMPLE: POSITION OF SECOND KID ON SEESAW EXAMPLE: A 4 m-long seesaw 50 kg in mass and of uniform mass distribution is pivoted on a fulcrum at its middle, as shown. Two kids sit on
More informationRotational N.2 nd Law
Lecture 0 Chapter 1 Physics I Rotational N. nd Law Torque Course website: http://faculty.uml.edu/andriy_danylov/teaching/physicsi IN THIS CHAPTER, you will continue discussing rotational dynamics Today
More informationThe gradient of the radius from the centre of the circle ( 1, 6) to (2, 3) is: ( 6)
Circles 6E a (x + ) + (y + 6) = r, (, ) Substitute x = and y = into the equation (x + ) + (y + 6) = r + + + 6 = r ( ) ( ) 9 + 8 = r r = 90 = 0 b The line has equation x + y = 0 y = x + y = x + The gradient
More informationApplications of Statics, Including Problem-Solving Strategies
Applications of Statics, Including Problem-Solving Strategies Bởi: OpenStaxCollege Statics can be applied to a variety of situations, ranging from raising a drawbridge to bad posture and back strain. We
More information4.0 m s 2. 2 A submarine descends vertically at constant velocity. The three forces acting on the submarine are viscous drag, upthrust and weight.
1 1 wooden block of mass 0.60 kg is on a rough horizontal surface. force of 12 N is applied to the block and it accelerates at 4.0 m s 2. wooden block 4.0 m s 2 12 N hat is the magnitude of the frictional
More informationKinematics. Exam-style assessment
1 Exam-style assessment Kinematics 1. golfer hits a ball from a point O on a horizontal surface. The initial velocity of the ball is 35 m s -1 at an angle of 40 to the horizontal. (a) Calculate the maximum
More informationStatic Equilibrium. Lecture 22. Chapter 12. Physics I Department of Physics and Applied Physics
Lecture 22 Chapter 12 Physics I 12.02.2013 Static Equilibrium Course website: http://faculty.uml.edu/andriy_danylov/teaching/physicsi Lecture Capture: http://echo360.uml.edu/danylov2013/physics1fall.html
More informationTEST REPORT. Question file: P Copyright:
Date: February-12-16 Time: 2:00:28 PM TEST REPORT Question file: P12-2006 Copyright: Test Date: 21/10/2010 Test Name: EquilibriumPractice Test Form: 0 Test Version: 0 Test Points: 138.00 Test File: EquilibriumPractice
More informationMEI Mechanics 2. A Model for Friction. Section 1: Friction
Notes and Examples These notes contain subsections on model for friction Modelling with friction MEI Mechanics Model for Friction Section 1: Friction Look at the discussion point at the foot of page 1.
More informationStatics. Phys101 Lectures 19,20. Key points: The Conditions for static equilibrium Solving statics problems Stress and strain. Ref: 9-1,2,3,4,5.
Phys101 Lectures 19,20 Statics Key points: The Conditions for static equilibrium Solving statics problems Stress and strain Ref: 9-1,2,3,4,5. Page 1 The Conditions for Static Equilibrium An object in static
More information( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) Trigonometric ratios 9E. b Using the line of symmetry through A. 1 a. cos 48 = 14.6 So y = 29.
Trigonometric ratios 9E a b Using the line of symmetry through A y cos.6 So y 9. cos 9. s.f. Using sin x sin 6..7.sin 6 sin x.7.sin 6 x sin.7 7.6 x 7.7 s.f. So y 0 6+ 7.7 6. y 6. s.f. b a Using sin sin
More informationGravitational potential energy
Gravitational potential energ m1 Consider a rigid bod of arbitrar shape. We want to obtain a value for its gravitational potential energ. O r1 1 x The gravitational potential energ of an assembl of N point-like
More informationDraft Version 1 Mark scheme Further Maths Core Pure (AS/Year 1) Unit Test 1: Complex numbers 1
1 w z k k States or implies that 4 i TBC Uses the definition of argument to write 4 k π tan 1 k 4 Makes an attempt to solve for k, for example 4 + k = k is seen. M1.a Finds k = 6 (4 marks) Pearson Education
More informationTrigonometry and modelling 7E
Trigonometry and modelling 7E sinq +cosq º sinq cosa + cosq sina Comparing sin : cos Comparing cos : sin Divide the equations: sin tan cos Square and add the equations: cos sin (cos sin ) since cos sin
More informationTrigonometric Functions 6C
Trigonometric Functions 6C a b c d e sin 3 q æ ö ø 4 tan 6 q 4 æ ö tanq ø cos q æ ö ø 3 cosec 3 q - sin q sin q cos q sin q (using sin q + cos q ) So - sin q sin q æ ö ø 6 4cot 6 q sec q cot q secq cos
More information( ) Momentum and impulse Mixed exercise 1. 1 a. Using conservation of momentum: ( )
Momentum and impulse Mixed exercise 1 1 a Using conseration of momentum: ( ) 6mu 4mu= 4m 1 u= After the collision the direction of Q is reersed and its speed is 1 u b Impulse = change in momentum I = (3m
More information1a States correct answer: 5.3 (m s 1 ) B1 2.2a 4th Understand the difference between a scalar and a vector. Notes
1a States correct answer: 5.3 (m s 1 ) B1.a 4th Understand the difference between a scalar and a vector. 1b States correct answer: 4.8 (m s 1 ) B1.a 4th Understand the difference between a scalar and a
More informationPhysicsAndMathsTutor.com
1. [In this question, the unit vectors i and j are horizontal vectors due east and north respectively.] At time t = 0, a football player kicks a ball from the point A with position vector (2i + j) m on
More informationMark Scheme (Results) January 2011
Mark (Results) January 2011 GCE GCE Mechanics (6677) Paper 1 Edexcel Limited. Registered in England and Wales No. 4496750 Registered Office: One90 High Holborn, London WC1V 7BH Edexcel is one of the leading
More informationApplications of Statics, Including Problem-Solving Strategies
OpenStax-CNX module: m42173 1 Applications of Statics, Including Problem-Solving Strategies OpenStax College This work is produced by OpenStax-CNX and licensed under the Creative Commons Attribution License
More informationMathematics Assessment Unit M1
ADVANCED SUBSIDIARY (AS) General Certificate of Education January 2011 Mathematics Assessment Unit M1 assessing Module M1: Mechanics 1 [AMM11] WEDNESDAY 19 JANUARY, AFTERNOON TIME 1 hour 30 minutes. INSTRUCTIONS
More informationTorques and Static Equilibrium
Torques and Static Equilibrium INTRODUCTION Archimedes, Greek mathematician, physicist, engineer, inventor and astronomer, was widely regarded as the leading scientist of the ancient world. He made a study
More informationMark scheme Mechanics Year 1 (AS) Unit Test 7: Kinematics 1 (constant acceleration)
1a Figure 1 General shape of the graph is correct. i.e. horizontal line, followed by negative gradient, followed by a positive gradient. Vertical axis labelled correctly. Horizontal axis labelled correctly.
More informationConnected Bodies 1. Two 10 kg bodies are attached to a spring balance as shown in figure. The reading of the balance will be 10 kg 10 kg 1) 0 kg-wt ) 10 kg-wt 3) Zero 4) 5 kg-wt. In the given arrangement,
More informationChapter 9- Static Equilibrium
Chapter 9- Static Equilibrium Changes in Office-hours The following changes will take place until the end of the semester Office-hours: - Monday, 12:00-13:00h - Wednesday, 14:00-15:00h - Friday, 13:00-14:00h
More informationEpisode 203: Turning effects
Episode 203: Turning effects Many students will recall the principle of moments from earlier work. The application of moments to situations beyond the simple see-saw examples met earlier in the school
More informationSprings Old Exam Questions
Springs Old Exam Questions Q1. A spring has a stiffness of 15 Nm 1. Calculate the extension of the spring when a weight of 8.0 N is suspended on it. Give your answer in metres. extension of spring... m
More informationFind the value of λ. (Total 9 marks)
1. A particle of mass 0.5 kg is attached to one end of a light elastic spring of natural length 0.9 m and modulus of elasticity λ newtons. The other end of the spring is attached to a fixed point O 3 on
More informationCreated by T. Madas. Candidates may use any calculator allowed by the regulations of this examination.
IYGB GCE Mathematics MMS Advanced Level Practice Paper Q Difficulty Rating: 3.400/0.6993 Time: 3 hours Candidates may use any calculator allowed by the regulations of this examination. Information for
More informationPhysics 211 Week 10. Statics: Walking the Plank (Solution)
Statics: Walking the Plank (Solution) A uniform horizontal beam 8 m long is attached by a frictionless pivot to a wall. A cable making an angle of 37 o, attached to the beam 5 m from the pivot point, supports
More informationApplication of Forces. Chapter Eight. Torque. Force vs. Torque. Torque, cont. Direction of Torque 4/7/2015
Raymond A. Serway Chris Vuille Chapter Eight Rotational Equilibrium and Rotational Dynamics Application of Forces The point of application of a force is important This was ignored in treating objects as
More informationAdvanced/Advanced Subsidiary. You must have: Mathematical Formulae and Statistical Tables (Pink)
Write your name here Surname Other names Pearson Edexcel GCE Centre Number Mechanics M1 Advanced/Advanced Subsidiary Candidate Number Wednesday 8 June 2016 Morning Time: 1 hour 30 minutes You must have:
More informationPaper Reference R. Mechanics M5 Advanced/Advanced Subsidiary. Wednesday 18 June 2014 Afternoon Time: 1 hour 30 minutes
Centre No. Candidate No. Surname Paper Reference(s) 6681/01R Edexcel GCE Mechanics M5 Advanced/Advanced Subsidiary Wednesday 18 June 2014 Afternoon Time: 1 hour 30 minutes Materials required for examination
More informationθ Beam Pivot F r Figure 1. Figure 2. STATICS (Force Vectors, Tension & Torque) MBL-32 (Ver. 3/20/2006) Name: Lab Partner: Lab Partner:
Please Circle Your Lab day: M T W T F Name: Lab Partner: Lab Partner: Project #1: Kinesthetic experiences with force vectors and torque. Project #2: How does torque depend on the lever arm? Project #1:
More information