Transcript
Absolute Dependent Motion Analysis of Two Particles • Motion of one particle will depend on depend on the corresponding motion of another particle • Dependency occurs when particles are interconnected by the inextensible cords which are wrapped around pulleys • For example, the movement of block A block A downward along the inclined plane will cause a corresponding movement of block B up the other incline • Specify the locations of the blocks using position coordinate s A and sB
Absolute Dependent Motion Analysis of Two Particles • Note each of the coordinate axes is (1) referenced from a fixed point fixed point (O) or fixed datum line, (2) measured along each inclined plane in the direction of motion of block A and block B and (3) has a positive sense from C to A and D to B • If total cord length is l T, the position coordinate are related by the equation s A l CD s B l T
Absolute Dependent Motion Analysis of Two Particles • Note each of the coordinate axes is (1) referenced from a fixed point fixed point (O) or fixed datum line, (2) measured along each inclined plane in the direction of motion of block A and block B and (3) has a positive sense from C to A and D to B • If total cord length is l T, the position coordinate are related by the equation s A l CD s B l T
Absolute Dependent Motion Analysis of Two Particles • Here l CD is the length passing over arc CD • Taking time derivative of this expression, realizing that l CD and l T remain constant , while s A and sB measure the lengths of the changing segments of the cord ds A ds B 0 or v B v A dt dt • The negative sign indicates that when block A has a velocity downward in the direction of position s A, it causes a corresponding upward velocity of block B; B moving in the negative sB direction
Absolute Dependent Motion Analysis of Two Particles • Time differentiation of the velocities yields the relation between accelerations aB = - a A • For example involving dependent motion of two blocks • Position of block A block A is specified by s A, and the position of the end of the cord which block B is suspended is defined by sB
Absolute Dependent Motion Analysis of Two Particles • Choose coordinate axes which are (1) referenced from fixed points and datums, (2) measured in the direction of motion of each block, (3) positive to the right (s A) and positive downward (sB) • During the motion, the red colored segments of the cord remain constant • If l represents the total length of the cord minus these segments, then the position coordinates can be related by
2 s B h s A l
Absolute Dependent Motion Analysis of Two Particles Since l and h are constant during the motion, the two time derivatives yields 2v B v A 2a B a A • When B moves downward (+sB), A moves to left (-s A) with two times the motion • This example can also be worked by defining the position of block B from the center of the bottom pulley ( a fixed point)
Absolute Dependent Motion Analysis of Two Particles 2(h s B ) h s A l
Time differentiation yields 2v B v A
2a B a A
Absolute Dependent Motion Analysis of Two Particles PROCEDURE FOR ANALYSIS Position-Coordinate Equation • Establish position coordinates which have their origin located at a fixed point or datum • The coordinates are directed along the path of motion and extend to a point having the same motion as each of the particles • It is not necessary that the origin be the same for each of the coordinates; however, it is important that each coordinate axis selected be directed along the path of motion of the particle
Absolute Dependent Motion Analysis of Two Particles • Using geometry or trigonometry, relate the coordinates to the total length of the cord, l T , or to that portion of cord, l , which excludes the segments that do not change length as the particles move – such as arc segments wrapped over pulleys • For problem involving a system of two or more cords wrapped over pulleys, then the position of a point on one cord must be related to the position of a point on another cord using the above procedure • Separate equations must be written for a fixed length of each cord of the system.
Absolute Dependent Motion Analysis of Two Particles Time Derivatives • Two successive time derivatives of the positioncoordinates equations yield the required velocity and acceleration equations which relate motions of the particles • The signs of the terms in these equations will be consistent with those that specify the positive and negative sense of the position coordinates
EXAMPLE 12.21
Determine the speed of block A if block B has an upward speed of 2 m/s.
EXAMPLE 12.21
Position Coordinate System. There is one cord in this system having segments which are changing length. Position coordinates s A and sB will be used since each is measured from a fixed point (C or D) and extends along each block’s path of motion. In particular, sB is directed to point E since motion of B and E is the same. The red colored segments of the cords remain at a constant length and do not have to be considered as the block move.
EXAMPLE 12.21
The remaining length of the cord, l , is also considered and is related to the changing position coordinates s A and sB by the equation s A 3 s B l
Time Derivative. yields
Taking the time derivative v A 3vB 0
so that when v B = -2m/s (upward) v A = 6m/s ↓
EXAMPLE 12.22
Determine the speed of block A if block B has an upward speed of 2m/s.
EXAMPLE 12.22
Position Coordinate Equation. Positions of A and B are defined using coordinates s A and sB. Since the system has two cords which change length, it is necessary to use a third coordinate sC in order to relate s A to sB. Length of the cords can be expressed in terms of s A and sC , and the length of the other cord can be expressed in terms of sB and sC . The red colored segments are not considered in this analysis.
EXAMPLE 12.22
For the remaining cord length, s A 2 sC l 1
s B ( s B sC ) l 2
Eliminating sC yields, s A 4 s B 2l 2 l 1 Time Derivative.
The time derivative gives v A 4vB 0
so that v B = -2m/s (upward) v B 8m / s 8m / s
EXAMPLE 12.23
Determine the speed with which block B rises if the end of the cord at A is pulled down with a speed of 2m/s.
EXAMPLE 12.23
Position-Coordinate Equation. The position of A is defined by s A, and the position of block B is specified by sB since point E on the pulley will have the same motion as the block. Both coordinates are measured from a horizontal datum passing through the fixed pin at pulley D. Since the system consists of two cords, the coordinates s A and sB cannot be related directly. By establishing a third position coordinate, sC , and the length of the other cord in terms of s A, sB and sC .
EXAMPLE 12.23
Excluding the red colored segments of the cords, the remaining constant cord lengths l 1 and l 2 (along the hook and link dimensions) can be expressed as sC s B l 1
s A sC s B sC s B l 2 Eliminating sC yields sC 4 s B l 2 2l 1
EXAMPLE 12.23
Time Derivative.
The time derivative give v A 4v B 0
when v A = 2m/s (downward) v B 0.5m / s 0.5m / s
EXAMPLE 12.24
A man at A is hoisting a safe S by walking to the right with a constant velocity v A = 0.5m/s. Determine the velocity and acceleration of the safe when it reaches the elevation at E . The rope is 30m long and passes over a small pulley at D.
EXAMPLE 12.24 View Free Body Diagram
Position Coordinate System. Rope segment DA changes both direction and magnitude. However, the ends of the rope, which define the position of S and A, are specified by means of the x and y coordinates measured from a fixed point and directed along the paths of motion of the ends of the rope. The x and y coordinates may be related since the rope has a fixed length l = 30m, which at all times is equal to the length of the segment DA plus CD.
EXAMPLE 12.24 Using Pythagorean Theorem and the equation for l CD to get the relationship between y and x: l DA
152 x 2 (eq.1)
l CD 15 y (eq.2)
l l DA l CD (eq.3) 30
15 x 2 15 y 2
y 225 x 2 15 (eq.4)
©2007 Pearson Education South Asia Pte Ltd
EXAMPLE 12.24
Time Derivative. Taking time derivative, using the chain rule where, v S = dy/dt and v A = dx/dt dx dy 1 2 x vS dt 2 225 x 2 dt
x 225 x
2
v A
(2)
At y = 10 m, x = 20 m, v A = 0.5 m/s, v S = 400mm/s ↑
EXAMPLE 12.24
The acceleration is determined by taking the time derivative of eqn (2),
x(dx / dt ) dx 1 aS 2 xv v A 2 3 / 2 A dt 225 x 2 dt (225 x ) d 2 y
2 dv A 1 225v A x 3/ 2 2 2 dt 225 x 225 x
At x = 20 m, with v A = 0.5 m/s, aS 3.6mm / s 2
Relative Motion Analysis of Two Particles Using Translating Axes • There are many cases where the path of the motion for a particle is complicated, so that it may be feasible to analyze the motions in parts by using two or more frames of reference • For example, motion of an particle located at the tip of an airplane propeller while the plane is in flight, is more easily described if one observes first the motion of the airplane from a fixed reference and then superimposes (vectorially) the circular motion of the particle measured from a reference attached to the airplane
Relative Motion Analysis of Two Particles Using Translating Axes Position. • Consider particle A and B, which moves along the arbitrary paths aa and bb, respectively • The absolute position of each particle r A and r B, is measured from the common origin O of the fixed x , y, z reference frame
Relative Motion Analysis of Two Particles Using Translating Axes • Origin of the second frame of reference x’, y’ and z’ is attached to and moves with particle A • Axes of this frame only permitted to translate relative to fixed frame • Relative position of “B with respect to A” is designated by a relative-position vector r B/A • Using vector addition
r B r A r B / A
Relative Motion Analysis of Two Particles Using Translating Axes Velocity. • By time derivatives,
v B v A v B / A
• Here v B d r B / dt and v A d r A / dt refer to absolute velocities, since they are observed from the fixed frame • Relative velocity v B / A d r B / A / dt is observed from the translating frame
Relative Motion Analysis of Two Particles Using Translating Axes • Since the x’, y’ and z’ axes translate, the components of r B/A will not change direction and therefore time derivative o this vector components will only have to account for the change in the vector magnitude • Velocity of B is equal to the velocity of A plus (vectorially) the relative velocity of “B relative to A” as measured by the translating observer fixed in the x’, y’ and z’ reference
Relative Motion Analysis of Two Particles Using Translating Axes Acceleration. • The time derivative yields a similar relationship between the absolute and relative accelerations of the particles A and B
a B a A a B / A
• Here aB/A is the acceleration of B as seen by the observer located at A and translating with the x’, y’ and z’ reference frame
Relative Motion Analysis of Two Particles Using Translating Axes PROCEDURE FOR ANALYSIS • When applying the relative position equations, r B = r A + r B/A, it is necessary to specify the location of the fixed x, y , z and translating x’, y’ and z’ • Usually, the origin A of the translating axes is located at a point having a known position r A • A graphical representation of the vector addition can be shown, and both the known and unknown quantities labeled on this sketch
Relative Motion Analysis of Two Particles Using Translating Axes • Since vector addition forms a triangle, there can be at most two unknowns, represented by the magnitudes and/or directions of the vector quantities • These unknown can be solved for either graphically, using trigonometry, or resolving each of the three vectors r A, r B and r B/A into rectangular or Cartesian components, thereby generating a set of scalar equations
Relative Motion Analysis of Two Particles Using Translating Axes • The relative motion equations vB = v A + vB/A and aB = a A + aB/A are applied in the same manner as explained above, except in this case, origin O of the fixed axes x, y, z axes does not have to be specified
EXAMPLE 12.25
A train, traveling at a constant speed of 90km/h, crosses over a road. If automobile A is traveling t 67.5km/h along the road, determine the magnitude and direction of relative velocity of the train with respect to the automobile
EXAMPLE 12.25
Vector Analysis. The relative velocity is measured from the translating x’, y’ axes attached to the automobile. Since vT and v A are known in both magnitude and direction, the unknowns become the x and y components of vT/A. Using the x, y axes and a Cartesian vector analysis,
vT v A vT / A ~ ~ ~ 90 i (67.5 cos 45 i 67.5 sin 45 j ) vT / A ~ ~ vT / A {42.3i 47.7 j )km / h
EXAMPLE 12.25
The magnitude of vT/A is 2 2 vT / A (42.3 47.7 ) 63.8km / h
The direction of vT/A defined from the x axis is
vT / A y 47.7 tan vT / A x 42.3 48.40
EXAMPLE 12.26
Plane A is flying along a straight-line path, while plane B is flying along a circular path having a radius of curvature of ρB = 400 km. Determine the velocity and acceleration of B as measured by the pilot of A.
EXAMPLE 12.26
Velocity. The x, y axes are located at an arbitrary fixed point. Since the motion relative to plane A is to be determined, the translating frame of reference x’. y’ is attached to it. Applying the relative-velocity equation in scalar form since the velocity vectors of both plane are parallel at the instant shown, ( )
v B v A v B / A 600 700 v B / A v B / A 100km / h 100km / h
EXAMPLE 12.26
Acceleration. Plane B has both tangential and normal components of acceleration, since it is flying along a curved path. Magnitude of normal acceleration, 2 vB a B n 900km / h 2 Applying the relative-acceleration equation, a B a A a B / A ~ ~ ~ 900 i 100 j 50 j a B / A ~ ~ a B / A 900 i 150 j km / h 2
EXAMPLE 12.26
From the figure shown, the magnitude and direction of a B / A 1 150 2 a B / A 912km / h tan 9.46 900
EXAMPLE 12.27
At the instant, car A and B are traveling with the speed of 18 m/s and 12 m/s respectively. Also at this instant, A has a decrease in speed of 2 m/s 2, and B has an increase in speed of 3 m/s2. Determine the velocity and acceleration of B with respect to A.
EXAMPLE 12.27 View Free Body Diagram
Velocity. The fixed x, y axes are established at a point on the ground and the translating x’, y’ axes are attached to car A. Using Cartesian vector analysis, v B v A v B / A ~ i 18 sin 60 j v B / A 12 j~ 18 cos 60 ~ ~ ~ v B / A 9 i 3.588 j m / s
Thus, v B / A 92 3.5882 9.69m / s
EXAMPLE 12.27
Its direction is
v B / A y 3.588 tan v B / A x 9 21.7
Acceleration. component is
The magnitude of the normal
a B n
2 vB
1.440m / s
2
