Sample/practice exam chapter 3, questions statics PDF

Preview

Summary

Revision based one chapter book here for students taking statics in engineering...

Description

9/3/2015

Eighth Edition

CHAPTER

3

VECTOR MECHANICS FOR ENGINEERS:

STATICS Ferdinand P. Beer E. Russell Johnston, Jr. Lecture Notes: J. Walt Oler

Rigid Bodies: Equivalent Systems of Forces

Texas Tech University

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

Eighth Edition

Vector Mechanics for Engineers: Statics Contents Introduction

Moment of a Force About a Given Axis

External and Internal Forces

Sample Problem 3.5

Principle of Transmissibility: Equivalent Forces

Moment of a Couple

Vector Products of Two Vectors

Couples Can Be Represented By Vectors

Moment of a Force About a Point

Resolution of a Force Into a Force at O and a Couple

Varigon’s Theorem Rectangular Components of the Moment of a Force

Addition of Couples

Sample Problem 3.6

Sample Problem 3.1

System of Forces: Reduction to a Force and a Couple

Scalar Product of Two Vectors

Further Reduction of a System of Forces

Scalar Product of Two Vectors: Applications

Sample Problem 3.8

Sample Problem 3.10

Mixed Triple Product of Three Vectors

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3-2

1

9/3/2015

Eighth Edition

Vector Mechanics for Engineers: Statics Introduction • Treatment of a body as a single particle is not always possible. In general, the size of the body and the specific points of application of the forces must be considered. • Most bodies in elementary mechanics are assumed to be rigid, i.e., the actual deformations are small and do not affect the conditions of equilibrium or motion of the body. • Current chapter describes the effect of forces exerted on a rigid body and how to replace a given system of forces with a simpler equivalent system. • moment of a force about a point • moment of a force about an axis • moment due to a couple • Any system of forces acting on a rigid body can be replaced by an equivalent system consisting of one force acting at a given point and one couple. © 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3-3

Eighth Edition

Vector Mechanics for Engineers: Statics External and Internal Forces • Forces acting on rigid bodies are divided into two groups: - External forces - Internal forces

• External forces are shown in a free-body diagram.

• If unopposed, each external force can impart a motion of translation or rotation, or both. © 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3-4

2

9/3/2015

Eighth Edition

Vector Mechanics for Engineers: Statics Principle of Transmissibility: Equivalent Forces • Principle of Transmissibility Conditions of equilibrium or motion are not affected by transmitting a force along its line of action. NOTE: F and F’ are equivalent forces.

• Moving the point of application of the force F to the rear bumper does not affect the motion or the other forces acting on the truck.

• Principle of transmissibility may not always apply in determining internal forces and deformations.

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3-5

Eighth Edition

Vector Mechanics for Engineers: Statics Vector Product of Two Vectors • Concept of the moment of a force about a point is more easily understood through applications of the vector product or cross product.

• Vector product of two vectors P and Q is defined as the vector V which satisfies the following conditions: 1. Line of action of V is perpendicular to plane containing P and Q. 2. Magnitude of V is V  PQ sin 3. Direction of V is obtained from the right-hand rule. • Vector products: - are not commutative, Q  P  P  Q  P  Q1  Q2   P  Q1  P  Q2 - are distributive, - are not associative,  P  Q  S  P  Q  S © 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3-6

3

9/3/2015

Eighth Edition

Vector Mechanics for Engineers: Statics Vector Products: Rectangular Components • Vector products of Cartesian unit vectors,         i i  0 j i  k k i  j         i j  k j j  0 k  j  i         i  k   j jk  i kk  0 • Vector products in terms of rectangular coordinates        V  Pxi  Py j  Pzk  Q xi  Q y j  Q z k    Py Qz  Pz Q y i   Pz Qx  Px Qz  j   Px Q y  Py Qx k

 

 









 i

 j

 k

 Px Qx

Py Qy

Pz Qz

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3-7

Eighth Edition

Vector Mechanics for Engineers: Statics Moment of a Force About a Point • A force vector is defined by its magnitude and direction. Its effect on the rigid body also depends on it point of application. • The moment of F about O is defined as M O  rF • The moment vector MO is perpendicular to the plane containing O and the force F.

• Magnitude of MO measures the tendency of the force to cause rotation of the body about an axis along MO. M O  rF sin   Fd The sense of the moment may be determined by the right-hand rule. • Any force F’ that has the same magnitude and direction as F, is equivalent if it also has the same line of action and therefore, produces the same moment. © 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3-8

4

9/3/2015

Eighth Edition

Vector Mechanics for Engineers: Statics Moment of a Force About a Point • Two-dimensional structures have length and breadth but negligible depth and are subjected to forces contained in the plane of the structure. • The plane of the structure contains the point O and the force F. MO, the moment of the force about O is perpendicular to the plane. • If the force tends to rotate the structure clockwise, the sense of the moment vector is out of the plane of the structure and the magnitude of the moment is positive. • If the force tends to rotate the structure counterclockwise, the sense of the moment vector is into the plane of the structure and the magnitude of the moment is negative. © 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3-9

Eighth Edition

Vector Mechanics for Engineers: Statics Varignon’s Theorem • The moment about a give point O of the resultant of several concurrent forces is equal to the sum of the moments of the various moments about the same point O.        r  F1  F2    r  F1  r  F2   • Varigon’s Theorem makes it possible to replace the direct determination of the moment of a force F by the moments of two or more component forces of F.

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3 - 10

5

9/3/2015

Eighth Edition

Vector Mechanics for Engineers: Statics Rectangular Components of the Moment of a Force The moment of F about O,        M O  r  F , r  xi  yj  zk     F  F x i  F y j  Fz k

    MO  M x i  M y j  M z k  i

 j

 x y Fx Fy

 k z Fz

     yFz  zFy  i   zFx  xFz  j   xFy  yFx  k

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3 - 11

Eighth Edition

Vector Mechanics for Engineers: Statics Rectangular Components of the Moment of a Force The moment of F about B,    M B  rA/ B  F

   rA /B  rA  rB

     xA  xB  i   yA  yB  j   zA  zB  k     F  F xi  Fy j  Fz k

 i  M B   xA  xB  Fx

 j  yA  yB  Fy

 k  z A  zB 

Fz

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3 - 12

6

9/3/2015

Eighth Edition

Vector Mechanics for Engineers: Statics Rectangular Components of the Moment of a Force For two-dimensional structures,   M O  xFy  yFz k





MO  M Z  xF y  yFz

  MO   xA  xB  Fy   yA  yB Fz k





MO M Z   x A  xB  Fy   y A  y B Fz

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3 - 13

Eighth Edition

Vector Mechanics for Engineers: Statics Sample Problem 3.1 A 100-lb vertical force is applied to the end of a lever which is attached to a shaft at O.

Determine: a) moment about O, b) horizontal force at A which creates the same moment, c) smallest force at A which produces the same moment, d) location for a 240-lb vertical force to produce the same moment, e) whether any of the forces from b, c, and d is equivalent to the original force.

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3 - 14

7

9/3/2015

Eighth Edition

Vector Mechanics for Engineers: Statics Sample Problem 3.1 a) Moment about O is equal to the product of the force and the perpendicular distance between the line of action of the force and O. Since the force tends to rotate the lever clockwise, the moment vector is into the plane of the paper.

M O  Fd d   24 in. cos 60  12 in. M O  100 lb12 in.

M O  1200 lb  in

3 - 15

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

Eighth Edition

Vector Mechanics for Engineers: Statics Sample Problem 3.1 c) Horizontal force at A that produces the same moment,

d  24 in. sin 60   20.8 in. M O  Fd 1200 lb  in.  F  20.8 in. F 

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

1200 lb  in. 20.8 in.

F  57.7 lb

3 - 16

8

9/3/2015

Eighth Edition

Vector Mechanics for Engineers: Statics Sample Problem 3.1 c) The smallest force A to produce the same moment occurs when the perpendicular distance is a maximum or when F is perpendicular to OA.

M O  Fd 1200 lb  in.  F 24 in. F 

1200 lb  in. 24 in.

F  50 lb

3 - 17

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

Eighth Edition

Vector Mechanics for Engineers: Statics Sample Problem 3.1 d) To determine the point of application of a 240 lb force to produce the same moment,

M O  Fd 1200 lb  in.  240 lbd 1200 lb  in.  5 in. 240 lb OB cos60   5 in. d 

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

OB  10 in.

3 - 18

9

9/3/2015

Eighth Edition

Vector Mechanics for Engineers: Statics Sample Problem 3.1 e) Although each of the forces in parts b), c), and d) produces the same moment as the 100 lb force, none are of the same magnitude and sense, or on the same line of action. None of the forces is equivalent to the 100 lb force.

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3 - 19

Eighth Edition

Vector Mechanics for Engineers: Statics Sample Problem 3.4 SOLUTION: The moment MA of the force F exerted by the wire is obtained by evaluating the vector product,    M A  rC A  F

The rectangular plate is supported by the brackets at A and B and by a wire CD. Knowing that the tension in the wire is 200 N, determine the moment about A of the force exerted by the wire at C. © 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3 - 20

10

9/3/2015

Eighth Edition

Vector Mechanics for Engineers: Statics Sample Problem 3.4 SOLUTION:    M A  rC A  F

     rC A  rC  rA  0.3 m i  0.08 m  j    r F  F   200 N C D rC D     0.3 m i  0.24 m  j  0.32 m k  200 N 0.5 m     120 N  i  96 N j  128 Nk    i j k  M A  0.3 0 0.08

 120 96  128     M A   7.68 N  m  i  28.8 N  m  j  28.8 N  m k © 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3 - 21

Eighth Edition

Vector Mechanics for Engineers: Statics Scalar Product of Two Vectors • The scalar product or dot product between two vectors P and Q is defined as   P  Q  PQ cos scalar result • Scalar products: - are commutative,

    P Q  Q  P        - are distributive, P  Q1  Q2   P  Q1  P  Q2    - are not associative,  P  Q  S  undefined

• Scalar products with Cartesian unit components,         P  Q   Pxi  Py j  P zk   Q xi  Q y j  Qz k 

            i  i  1 j  j  1 k  k  1 i  j  0 j  k  0 k i  0   P  Q  PxQ x  Py Qy  Pz Qz   P  P  Px2  Py2  Pz2  P 2 © 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3 - 22

11

9/3/2015

Eighth Edition

Vector Mechanics for Engineers: Statics Scalar Product of Two Vectors: Applications • Angle between two vectors:   P Q  PQ cos   P xQ x  P yQ y  PzQ z

cos  

PxQ x  Py Qy  Pz Qz PQ

• Projection of a vector on a given axis:

POL  P cos  projection of P along OL   P  Q  PQ cos    PQ  P cos  POL Q • For an axis defined by a unit vector:   POL  P  

 Px cos x  Py cos y  Pz cos z 3 - 23

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

Eighth Edition

Vector Mechanics for Engineers: Statics Mixed Triple Product of Three Vectors • Mixed triple product of three vectors,    S  P  Q  scalar result • The six mixed triple products formed from S, P, and Q have equal magnitudes but not the same sign,          S  P  Q   P  Q  S   Q  S  P            S  Q  P   P  S  Q  Q  P  S  • Evaluating the mixed triple product,    S  P  Q   S x PyQ z  PzQ y  S y  PzQ x  PxQ z 







 S z Px Qy  Py Qx Sx  Px

Sy

Sz

Py

Pz

Q x Qy

Qz

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.



3 - 24

12

9/3/2015

Eighth Edition

Vector Mechanics for Engineers: Statics Moment of a Force About a Given Axis • Moment MO of a force F applied at the point A about a point O,    MO  r  F • Scalar moment MOL about an axis OL is the projection of the moment vector MO onto the axis,      M OL    M O    r  F  • Moments of F about the coordinate axes,

M x  yF z  zF y M y  zF x  xF z M z  xFy  yFx

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3 - 25

Eighth Edition

Vector Mechanics for Engineers: Statics Moment of a Force About a Given Axis • Moment of a force about an arbitrary axis,   M BL    M B       rA B  F     rA B  rA  rB • The result is independent of the point B along the given axis.

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3 - 26

13

9/3/2015

Eighth Edition

Vector Mechanics for Engineers: Statics Sample Problem 3.5 A cube is acted on by a force P as shown. Determine the moment of P a) about A b) about the edge AB and c) about the diagonal AG of the cube. d) Determine the perpendicular distance between AG and FC.

3 - 27

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

Eighth Edition

Vector Mechanics for Engineers: Statics Sample Problem 3.5 • Moment of P about A,    M A  rF A  P      rF A  ai  a j  ai  j       P  P  2 i  2 j   P 2 i  j       M A  a i  j  P 2 i  j      M A  aP 2 i  j  k  • Moment of P about AB,   M AB  i  M A      i  aP 2 i  j  k 

M AB  aP 2

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3 - 28

14

9/3/2015

Eighth Edition

Vector Mechanics for Engineers: Statics Sample Problem 3.5 • Moment of P about the diagonal AG,   M AG    M A      rA G ai  aj  ak 1    i  j  k     rA G a 3 3  aP    i  j  k  MA  2 1    aP    i  j  k  2 i  j  k  M AG  3 aP 1 1 1  6

M AG  

aP 6

3 - 29

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

Eighth Edition

Vector Mechanics for Engineers: Statics Sample Problem 3.5 • Perpendicular distance between AG and FC,

  P   1    P 0  1 1  P   j  k i  j  k  2 3 6 0 Therefore, P is perpendicular to AG.

M AG 

aP  Pd 6 d

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

a 6

3 - 30

15

9/3/2015

Eighth Edition

Vector Mechanics for Engineers: Statics Moment of a Couple • Two forces F and -F having the same magnitude, parallel lines of action, and opposite sense are said to form a couple. • Moment of the couple,      M  rA  F  rB   F       rA  rB   F   r  F M  rF sin  Fd • The moment vector of the couple is independent of the choice of the origin of the coordinate axes, i.e., it is a free vector that can be applied at any point with the same effect.

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3 - 31

Eighth Edition

Vector Mechanics for Engineers: Statics Moment of a Couple Two couples will have equal moments if • F1d1  F2 d 2 • the two couples lie in parallel planes, and • the two couples have the same sense or the tendency to cause rotation in the same direction.

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3 - 32

16

9/3/2015

Eighth Edition

Vector Mechanics for Engineers: Statics Addition of Couples • Consider two intersecting planes P1 and P2 with each containing a couple    M1  r  F1 in plane P1    M 2  r  F2 in plane P2 • Resultants of the vectors also form a couple       M  r  R  r   F1  F2  • By Varigon’s theorem      M  r  F1  r  F2    M1  M 2 • Sum of two couples is also a couple that is equal to the vector sum of the two couples © 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3 - 33

Eighth Edition

Vector Mechanics for Engineers: Statics Couples Can Be Represented by Vectors

• A couple can be represented by a vector with magnitude and direction equal to the moment of the couple. • Couple vectors obey the law of addition of vectors. • Couple vectors are free vectors, i.e., the point of application is not significant. • Couple vectors may be resolved into component vectors.

© 2007 The McGraw-Hill Companies, Inc. All rights reserved.

3 - 34

17

9/3/2015

Eighth Edition

Vector Mechanics for Engineers: Statics Resolution of a Force Into a Force at O and a Couple

• Force vector F can not be simply moved to O without modifying its action on th...