[Yu Yue education] theoretical mechanics reference materials of Shanghai Jiaotong University

education
- Theoretical mechanics - Chapter materials, examination materials - Shanghai Jiaotong University 【】
Mathematical basis
1、【 Single topic selection 】
A、
B、
C、
D、
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2、【 Single topic selection 】<img src="//img.inotgo.com/imagesLocal/202202/15/202202150236105246_62.jpg> A、
B、‍
C、
D、‍
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3、【 Single topic selection 】<img src="//img.inotgo.com/imagesLocal/202202/15/202202150236105246_4.jpg> A、 vector
B、 Scalar
C、 matrix
D、 Vector matrix
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4、【 Single topic selection 】<img src="//img.inotgo.com/imagesLocal/202202/15/202202150236105246_79.jpg> A、
B、
C、
D、
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5、【 Single topic selection 】<img src="//img.inotgo.com/imagesLocal/202202/15/202202150236105246_88.jpg> A、5
B、
C、4
D、3
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statics ： Force and couple
1、【 Single topic selection 】 The converging force system can be simplified to a （ ）
A、 force
B、 Couple
C、 Moment
D、 Force spiral
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2、【 Single topic selection 】 The couple system can be simplified into a （ ）
A、 force
B、 Couple
C、 Moment
D、 Couple or moment
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3、【 Multiple choice 】 The three elements of a couple acting on a rigid body are （ ）
A、 size
B、 Direction of rotation
C、 Action surface
D、 Action point
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statics ： Simplification of force system
1、【 Single topic selection 】 There are several simplest results of the simplification of a parallel force system （ ）
A、1 Kind of
B、2 Kind of
C、3 Kind of
D、 4 Kind of
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2、【 Single topic selection 】 A general spatial force system , The principal vector is not a zero vector , The principal moment of a point is not a zero vector , The simplest result of this force system is （ ）
A、 force ‍
B、 Force or couple ‍
C、 Force spiral ‍
D、 Force or force spiral
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3、【 Single topic selection 】 The following figure shows the position of the center of the parallel force system (x,y) Coordinate for （ ）
A、（3,2）
B、（3,4）
C、（2,3）
D、（4,3）
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4、【 Completion 】 A spatial force system , As shown in the figure ,F1=400N,F2=300N, F3=500N, The principal vector is _____N（ Data retention 2 Decimal place ）<img src="//img.inotgo.com/imagesLocal/202202/15/202202150236105246_77.jpg> A、
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5、【 Completion 】 A spatial force system , As shown in the figure ,F1=400N,F2=300N, F3=500N, Yes O The point principal moment is _____N.m（ Data retention 2 Decimal place ）<img src="//img.inotgo.com/imagesLocal/202202/15/202202150236105246_77.jpg> A、
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statics ： Balance of force system
1、【 Single topic selection 】 A spatial parallel force system has at most several independent equilibrium equations ？
A、1
B、2
C、3
D、 Insufficient conditions to determine
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2、【 Single topic selection 】 If two plane intersecting force systems form a plane force system , Then the force system has at most several independent equilibrium equations ？
A、1
B、2
C、3
D、4
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3、【 Single topic selection 】 The structure and its stress are shown in the figure 1 Shown , Ruoli F With the rod 1 And rods 2 parallel , Regardless of the dead weight of the member , Which compression bars ？
A、 rod 1
B、 rod 2
C、 rod 3
D、 Insufficient conditions to determine
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4、【 Single topic selection 】 As shown in the figure , Straight bar AB And bent bar BC With hinges B Connect , A、C They are fixed hinge bearings . On the pole AB Act on a couple , Regardless of the weight of the member , Find the fixed hinge support A Binding force at .
A、
B、
C、
D、
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5、【 Single topic selection 】 A composite beam ABC The support and load of are shown in the figure , The dead weight of beam and support rod is not counted . It is known that F＝1kN,M=0.5kNm, Find the fixed end A Constraint couple .
A、3kNm
B、3.5kNm
C、4kNm
D、4.5kNm
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Statics test
1、【 Single topic selection 】 A spatial parallel force system has at most several independent equilibrium equations ？（ ）
A、4
B、3
C、2
D、1
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2、【 Single topic selection 】 The principal vector of a force system is not zero , Point to point O The principal moment of is not zero , Then the force system can be simplified into a （ ）
A、 Together
B、 Resultant couple
C、 Resultant force or force spiral
D、 Force spiral
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3、【 Single topic selection 】 The active force of a rigid body 、 Ideal binding force 、 Balance under sliding friction , Then the static friction force meets （ ）
A、
B、
C、
D、
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4、【 Single topic selection 】 The self weight of the components of the structure shown in the figure is not counted ,B For smooth hinges . It is known that q=3kN/m,F＝4kN,M=6kNm,l =2m, Movable hinge support C The binding force at is （ ） <img src="//img.inotgo.com/imagesLocal/202202/15/202202150236105246_22.jpg> A、1.7321
B、2
C、3.4641
D、4
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5、【 Single topic selection 】 The self weight of the components of the structure shown in the figure is not counted ,B For smooth hinges . It is known that q=3kN/m,F＝4kN,M=6kNm,l =2m, Fixed end A The size of the constraint couple at is （ ）<img src="//img.inotgo.com/imagesLocal/202202/15/202202150236105246_22.jpg> A、8
B、8.5
C、9
D、9.5
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6、【 Single topic selection 】 Long for L The quality is m Homogeneous rod A Put the end on the horizontal ground ,B The end is lifted horizontally with a thin rope , The included angle between the rod and the horizontal plane is , Pictured 4 Shown . When trying to maintain balance , Minimum static sliding friction coefficient between ground and rod
A、
B、
C、
D、
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7、【 Single topic selection 】 Long for L The quality is m Homogeneous rod A Put the end on the horizontal ground ,B The end is lifted horizontally with a thin rope , The included angle between the rod and the horizontal plane is , Pictured 4 Shown . When trying to maintain balance , At this time, the tension of the rope （ ）<img src="//img.inotgo.com/imagesLocal/202202/15/202202150236105246_9.jpg> A、
B、
C、
D、
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Planar kinematics of rigid body ： The configuration of a rigid body
1、【 Single topic selection 】 The directional cosine matrix is a （ ）.
A、 Symmetric matrix
B、 Antisymmetric matrix
C、 Orthogonal matrix
D、 Unit matrix
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2、【 Single topic selection 】 The derivative of a vector over time is another vector , The direction of these two vectors （ ）
A、 Must be perpendicular to each other
B、 It must not be perpendicular to each other
C、 Not necessarily perpendicular to each other
D、 In the same direction
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3、【 Single topic selection 】 In general , Of the nine elements of the directional cosine matrix , There are several independent elements ？( )
A、2 individual
B、3 individual
C、 4 individual
D、5 individual
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4、【 Single topic selection 】 The car goes straight or turns on a level road , What is the movement of the car body ? ( )
A、 General motion in space
B、 Plane motion
C、 Circular motion
D、 Undetermined
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5、【 Single topic selection 】 The car turned on a level road , What is the motion of the wheel ? ( )
A、 General motion in space
B、 Plane motion
C、 Circular motion
D、 Undetermined
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6、【 Single topic selection 】 As shown in the figure , rod AB Put it in a place with a radius of R Movement in a fixed semicircular groove ,A、B The two points are always in contact with the circular groove , be AB The movement of the rod is （ ）.<img src="//img.inotgo.com/imagesLocal/202202/15/202202150236105246_49.jpg> A、 Fixed shaft rotation , The rotating shaft is O
B、 Fixed shaft rotation , The shaft is not O
C、 Not fixed axis rotation
D、 Undetermined
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Kinematics test
1、【 Single topic selection 】 The directional cosine matrix is a
A、 Symmetric matrix
B、 Antisymmetric matrix
C、 Orthogonal matrix
D、 Unit matrix
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2、【 Single topic selection 】 A plane moving rigid body makes a translational motion at a certain instant , Then the angular acceleration at this time is which of the following ？
A、 It must not be zero
B、 It must be zero
C、 Not necessarily zero
D、 Unable to judge
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3、【 Single topic selection 】<img src="//img.inotgo.com/imagesLocal/202202/15/202202150236105246_40.jpg> A、 A straight line
B、 parabola
C、 The circumference
D、 Undetermined
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4、【 Single topic selection 】
A、0°
B、30°
C、60°
D、90°
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5、【 Multiple choice 】 After the velocity analysis of the moving point is completed , Which components of the acceleration of the moving point can certainly be calculated directly ？
A、 Coriolis acceleration
B、 Relative acceleration
C、 Rotation involves centripetal acceleration
D、 Rotation involves tangential acceleration
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6、【 Completion 】
A、
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7、【 Completion 】<img src="//img.inotgo.com/imagesLocal/202202/15/202202150236105246_31.jpg> A、
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8、【 Completion 】<img src="//img.inotgo.com/imagesLocal/202202/15/202202150236105246_63.jpg> A、
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9、【 Completion 】
A、
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10、【 Completion 】
A、
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11、【 Completion 】
A、
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12、【 Completion 】
A、
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Kinematics test 2
1、【 Completion 】 The straight rod moves as shown in the figure ,h=1m. Position and speed shown v=2m/s, The angular velocity of the position lever shown in the figure = rad/s.
A、
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2、【 Completion 】 The radius is r A round wheel with a radius of R Pure rolling on a fixed circular arc track , angular velocity w=2rad/s, The angular acceleration is 0.r=0.5m,R=1.5m. Then the acceleration of the wheel center is m/s^2<img src="//img.inotgo.com/imagesLocal/202202/15/202202150236105246_64.jpg> A、
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The twelfth speak Computer aided analysis of rigid system kinematics ： Theoretical basis
1、【 Single topic selection 】 For a mechanical system , Which of the following quantities remains constant
A、 freedom δ
B、 Number of configuration coordinates n
C、 Number of constraint equations s
D、 All of these are
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2、【 Single topic selection 】 The mechanism shown below , Its degree of freedom is
A、0
B、1
C、2
D、3
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3、【 Single topic selection 】 As shown in the figure below, the mechanism , Generalized coordinates q=（l1,θ,y3）, Its Jacobian is
A、
B、
C、
D、
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4、【 Single topic selection 】 The mechanism shown below , The right term of the acceleration constraint equation is
A、
B、
C、
D、
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5、【 Single topic selection 】 As shown in the figure below, the mechanism , In the figure d=0.32m. connecting rod B2 The length of is l2=0.32m. The piston cylinder is consolidated on the ground , dip angle q=45°. piston B1 The driving law of is l1=l0+0.08t Push point P, When q=45° Time slider B3 The acceleration of is <img src="//img.inotgo.com/imagesLocal/202202/15/202202150236105246_1.jpg> A、0.01
B、-0.01
C、0.02
D、-0.02
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About 14 Fundamentals of vector dynamics ： inertia 、 The theorem of momentum
1、【 Single topic selection 】 The quality is m The radius is R A homogeneous disk , Dig a hole with a radius of r=R/2 A round hole in the , As shown in the figure . Then the disc is opposite to the center of the circle O Moment of inertia of the shaft JO by （ ）<img src="//img.inotgo.com/imagesLocal/202202/15/202202150236105246_45.jpg> A、
B、
C、
D、
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2、【 Judgment questions 】 The moment of inertia of a plane moving rigid body to the center of mass is the smallest ; The moment of inertia of the fixed axis rotating rigid body to the rotating axis is the smallest .
A、 correct
B、 error
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3、【 Judgment questions 】 For any rigid system , The total momentum of the rigid system is only related to the total mass of the rigid system and the velocity at the center of mass of the rigid system .
A、 correct
B、 error
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4、【 Judgment questions 】 The acceleration of the center of mass of a rigid body depends entirely on the size and direction of the principal vector of the external force system , It has nothing to do with the internal force of the particle system , It is also independent of the action position of external force .
A、 correct
B、 error
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5、【 Completion 】OA The rod is wound around O The shaft rotates counterclockwise , A homogeneous disk is formed along OA The rod makes pure rolling . The mass of the disc is known m =20kg, radius R =10cm. When in the position shown ,OA The inclination of the rod is 30°, Its angular speed of rotation is ω1=1rad/s, The disk is relative to OA The angular speed at which the rod rotates ω2=4rad/s,OB=40cm, Then the momentum of the disk at this time p The size is Ns.
A、
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Fundamentals of vector dynamics ： The theorem of moment of momentum
1、【 Single topic selection 】 A rope that crosses a pulley , Tie a heavy object at one end , A weight equal to that of a monkey at the other end , Start at rest at constant speed v Climb up , As shown in the figure . If the mass and friction of rope and pulley are not taken into account , Then the speed of the weight （ ）<img src="//img.inotgo.com/imagesLocal/202202/15/202202150236105246_34.jpg> A、 be equal to v, Direction down
B、 be equal to v, Direction up
C、 It's not equal to v
D、 The weight doesn't move
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2、【 Single topic selection 】 Homogeneous elliptical gauge AB The quality of is 2m, crank OC The quality of is m, slider A,B All are of m.OC=AC=BC=l, The gauge and crank are homogeneous rods , The crank moves at an equal angular speed ω Around the O Shaft rotation . Point to point when the system is in the position shown in the figure O The moment of momentum is （ ）<img src="//img.inotgo.com/imagesLocal/202202/15/202202150236105246_14.jpg> A、
B、
C、
D、
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3、【 Single topic selection 】 The radius is r The homogeneous disk is in the vertical plane around the horizontal axis A Swing as shown in the figure . Set the center of the disc O as for A The distance to b. Try to ask b Why is it worth it , The period of oscillation is the minimum ？（ ）
A、
B、
C、
D、
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4、【 Single topic selection 】 Long for l, The quality is m Homogeneous rod AB and BC Use a hinge B Connect , And use hinges A Fix , In a balanced position . Today C A horizontal force acts on the end F, be AB Rod and BC The angular accelerations of the rod are （ ）.<img src="//img.inotgo.com/imagesLocal/202202/15/202202150236105246_46.jpg> A、
B、
C、
D、
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5、【 Completion 】 A homogeneous rod OA With the homogeneous disk in the center of the disk A At the hinge , When in the position shown ,OA The rod revolves around the fixed shaft O The angular velocity of rotation is ω=1rad/s, The disc is relative to the rod OA The angular velocity of is also ω. set up OA The mass of the rod and disc is m=10kg, The radius of the disk is R=2m, Rod length l=3R. At this time, the system fixes the shaft O The magnitude of the moment of momentum is kg*m^2/s.（ Just give the size , No sign ）<img src="//img.inotgo.com/imagesLocal/202202/15/202202150236105246_12.jpg> A、
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