mechanics - university of manchester · meriam jl and kraige lg, engineering mechanics (volume 1...

G F Lane-Serff 1 15-Sep-09

Mechanics Introduction Aim To introduce the basic concepts in mechanics, especially as they apply to civil engineering. Mechanics Mathematical models describing the effects of forces and motion on objects (structures, machines, etc). In constructing mathematical models we make assumptions and approximations. It is important that you are clear about the assumptions being made and the limitations of the approximations. Outline

A Statics B Kinematics C Dynamics D Energy E Rotation F Oscillations

In this unit we will concentrate on the behaviour of particles and rigid bodies in two dimensions (typically height and one horizontal dimension). We will use SI units throughout but it is useful to have some general knowledge of other systems of units. Staff Dr G F Lane-Serff: sections A-C and unit coordinator. Extn 64602, room P/B20, [email protected] Dr D D Apsley: sections D-F Extn 63732, room P/B15, [email protected] Assessment Coursework (laboratory work and problems):

Forces and moments 3% Structures 3%

Momentum 4% Energy 5% Rotation and oscillation 5% Exam: Four out of six questions 80% (Two from three on sections A-C, two from three on sections D-F.) It is important that you develop the habit of working in a clear and logical way, setting out your calculations so that they are easy to follow and check. Marks will be deducted for inadequate working (even if you get the correct answer). Books Meriam JL and Kraige LG, Engineering Mechanics (volume 1 Statics, volume 2 Dynamics) 531 Beer FP and Johnston ER, Mechanics for Engineers (volume 1 Statics, volume 2 Dynamics) 620.1 Both have lots of example questions: there’s no substitute for plenty of practice! These and many other useful books (e.g. Young HD, Fundamentals of mechanics and heat) can be found in the Joule Library, classmarks: 620.1 mechanics: applied; 531 mechanics: physics.


A Statics

A1 Forces Force: action on a body Newton’s First Law

A body remains at rest or moves with constant velocity unless acted upon by a net force.

Forces are vectors, having magnitude (size) and direction; written in various ways, eg F∼ or F or F

→ .

F To completely describe a force acting on a body we need to know its magnitude, direction and point of application (where the force is applied). Internal Forces: Forces within the body. Important in considering strength (failure) elasticity and plasticity (bending and deformation) but not studied in this lecture course. External forces: Forces on the body under consideration. Rigid bodies: Bodies that stay the same shape (don’t deform) under the action of forces. Principle of Transmissibility (Line of Action)

= For a rigid body, a force can be treated as being applied at any point along its line of action (this doesn’t have to be within the body). Adding forces Forces applied at the same point F1 R = F1 + F2 R parallelogram rule F2 Forces applied at different points (on a rigid body)

Use principle of transmissibility to move forces along their lines of action to the point

where they meet.


Adding parallel (or nearly parallel) forces. First add two equal and opposite forces

(acting along a common line)

F1 -F

R1

R2 F2 F Then find the resultant force (this can be moved back along its line of action if desired). R1 R2 R2 R = R1 + R2 R1 Components A force can be decomposed into two components along any two non-parallel directions by constructing the appropriate parallelogram. It is often convenient to use rectangular components. F F = Fx + Fy or F = Fxi + Fyj Fy i unit vector in x-direction θ j unit vector in y-direction (k unit vector in z-direction) Fx Fx = F cosθ tanθ = Fy/Fx Fy = F sinθ θ = tan-1(Fy/Fx) F = Fx

2 + Fy2

The resultant of two forces can be found be summing the components separately: y R = F1 + F2 Rx = F1x + F2x Ry = F1y + F2y F1 R x e.g. F1 = (2,3), F2 = (4,-2) F2 R = (6,1). Provided the axes are perpendicular, you can make your own choice of axes to simplify the calculations.


A2 Moments and couples Moment of a force about an axis is the tendency of the force to rotate an object about that axis. M

The force F is applied at right angles to the spanner handle. The magnitude of the moment about the centre of the bolt is:

M = d F d F M If the force is not perpendicular to the line joining the point of application to the F axis, we need to find the perpendicular r distance from the line of action to the axis. α d

M = d F or M = r sinα F

The usual sign convention follows the “right hand rule,” with a positive moment for anticlockwise motion. The moment can be represented as a vector, with the direction of the vector being the axis of rotation. For the example above M = d F k, where k is the unit vector in the z-direction. For 2D problems the rotation axis is always perpendicular to the plane we are working in, and so the sign is all that is needed to define the sense of the rotation. For 3D problems the moment vector is given by

M = r × F or r ∧ F (Not F × r which equals -r × F.) As with forces, where we needed to state the point of application, a complete description of a moment requires its magnitude, direction and position of the axis (this is simply a point in 2D problems). Units of moment are N m (force × distance). Varignon’s Theorem The moment of a force about a point (axis) is equal to the sum of the moments of the components of the force about the same point. (The components need not be rectangular components.)

R = F1 + F2 r × R = r × (F1 + F2) = (r × F1) + (r × F2)

Moment about A d R = d1 F1 + d2 F2

d1

d2

d

F1

F2

R r A


Couples Consider a pair of equal and opposite forces acting on points separated by a perpendicular distance d. (It is not possible to calculate a single resultant force from such a pair: try it!)

Total moment about A is 0 + d F = d F

Total moment about B is d F + 0 = d F

Total moment about C is -x F + (x + d) F = d F

The moment of a couple is the same about all points. A couple is a “free vector.”

Force-couple systems Any force acting on a body tends to push/pull in the direction of the force and tends to rotate about any axis not on the line of action of the force. We can replace any force at one point by a force and couple applied at a different point. This can be useful when finding the reaction at a support.

1) Add equal and opposite forces at B. 2) Then these two forces can be treated as a couple M = -d F (clockwise) So the force applied at A is equivalent to a force and couple applied at B.

d A

B

C

x

F -F

B

A F

d

B

A F

F -F

B

A

F M


A3 Tutorial: Forces and moments Resultant of several forces at a point. Moments.


A4 Equilibrium and free body diagrams Equilibrium For a body to be in equilibrium, the net effect of the forces on the body must neither tend to make the body move nor rotate. Thus the resultant force R and resultant couple N must both be zero.

R = ∑F = 0 N = ∑M = 0 In two dimensions, with forces in the x- and y-directions and all moments are about an axis perpendicular to the plane, this can be written as

∑Fx = 0 ∑Fy = 0 ∑MA = 0, where the MA are all the moments about any point A (which may be on or off the body). Other versions of these equilibrium equations are

∑Fx = 0 ∑MA = 0 ∑MB = 0, where the points A and B must not lie on a line perpendicular to the x-direction, and

∑MA = 0 ∑MB = 0 ∑MC = 0, where the three points A, B and C must not lie on a straight line. In tackling a given problem, it is worth considering which choice of equations, coordinate system and points will give the simplest set of equations to solve. Free Body Diagrams 1) Define: In considering a body in equilibrium it is first necessary to define the body under

consideration. This may be a composite body consisting of several elements. 2) Isolate: Draw a diagram that represents the complete external boundary of the body. 3) Forces: Represent all the forces acting on the body. Supports and connections Support or connection Forces/couples supported pin

forces, two components (no couple)

rollers

normal force only

smooth contact

normal force only

rough contact

normal and friction

cable

tension (in direction of cable at body)

fixed

forces and couple

mass under gravity

weight W = mg (down)

Rx

Ry

N

N

N F

T

Fx Fy M

m W


Examples Mass at mid-point on beam (length L) Free body diagram

x-component forces y-component forces moments about mid-point (or use A or B) FAx = 0 FAy + FBy – W = 0 -½L FAy + ½L FBy = 0 Final result FAx = 0, FAy = FBy = ½ W = ½ mg Simple structure with a cable x-component forces y-component forces moments about A FAx + T = 0 FAy + FBy = 0 2 FBy – 2 sin 60° T= 0 Final result FAx = -10 N, FAy = -FBy = -10 sin 60° = -8.66 N Structure with cable and mass moments about A x-component forces y-component forces 2T – 4W = 0 FAx – T sin 60° = 0 FAy + T cos 60° - W = 0 Final result T = 2W = 2mg FAx = T sin 60° = 1.73 mg FAy = W - T cos 60° = 0

FAx

FAy FBy

W = mg m A B

T=10N

A B

2m

C T=10N

2m

FAx

FAy

60°

FBy

T

mass m

2m

cable

60°

FAx

FAy W=mg 4m

2m


A5 Structures Statically determinate structures These are structures where the equilibrium equations are sufficient to calculate all the forces. We will concentrate on statically determined structures in this course, but you should be aware that not all structures are statically determinate. Statically indeterminate

With the beam pinned at both ends, the x-component of the forces gives

FAx = -FBx

and there is no way to calculate the undetermined compression/tension in the beam.

Inadequate constraints

In this case there is still an unknown tension in the beam but also there is no way of counteracting the moment about A.

(Plane) Trusses Trusses are made up of simple two-force members (Equal and opposite forces.) Forces on the members: Tension or Compression We will ignore (for now) any bending or deformation of the members and we will usually consider them to have negligible mass (weight). Members under tension could be replaced by cables. The basic unit for plane trusses is the triangle. Method of Joints We consider the forces acting on the connecting pins (these are in the opposite direction to the forces acting on the members). The resultant force on each pin must be zero.

m A B FBx

FBy

FAx

FAy

m A B

T

T

C

C

T

C

C forces on the connecting pin


Example of a plane truss 1) Draw a free body diagram Calculate the external forces

FBx = 0

FAy = FBy = L/2 2) Use method of joints Work round the truss joint by joint, finding equilibrium equations (in e.g. x- and y-directions) for each joint. By using a sensible order, the number of simultaneous equations to be solved can be minimized or eliminated.

Forces on the pins (joints)

Add the forces to your diagram as you calculate them.

Equations at joint B gives compression L/2 in

vertical member, no force in horizontal member.

At joint A, the y-direction equation implies a compressive load in the diagonal member,

which in turn implies tension in the horizontal member (from the x-equation).

Most of the rest of the forces can be

calculated from the joint above B. The force in the final member can be calculated from the joint where the load is applied, and checked with the joint vertically above this.

2m

2m 2m L

B A

L FAy

FBx

FBy

L/2

L/2

L/2

L

C

0

L/√2

L/2

L/2

L/2

L

C

0

L/2 C

L/2

T

L/2

L/√2

C

L/√2

L/2 L/2 T

L/√2

L/√2

L/2

L/2

L/2

L

C

0

L/2 C

L/2

T

L/2

L/√2

C

L/√2

L/2 L/2 T

L/√2

L/2

L/2

T


A6 Tutorial: Statics Structures


B Kinematics

B1 Velocity and acceleration: rectilinear motion Rectilinear motion: motion in a straight line (1D) Position of object at time t is s(t), at time t+∆t the object is at s+∆s.

Velocity, v = lim

∆t→0 ∆s∆t =

dsdt = s. acceleration, a =

lim∆t→0

∆v∆t =

dvdt = v. =

d2sdt2 = s..

Can also make use of a = ddt (v) =

dsdt

dds (v) = v

dvds

⌡⌠s1

s2

ds = ⌡⌠t1

t2

vdt

Distance travelled is equal to the area under a graph of velocity versus time.

example

s = 5t3 + 2t + 7 v = dsdt = 15t2 + 2 a =

dvdt = 30t

Constant acceleration If a is constant (and v = v0 and s = s0 at t=0) then v = v0 + a t s = s0 + v0 t + ½a t2 v2 = v0

2 + 2 a (s - s0) Example: a = -5 m s-2, v0 = 30 m s-1, s0 = 0 m.

Mathematically, time to stop, t1 0 = 30 – 5 t1, ⇒ t1 = 6 seconds s = 0 + 30 × 6 – ½ × 5 × 62 = 90 m

s=0 s s+∆s

∆s

t t+∆t

s

t

∆s ∆t

t

v

∆v ∆t

t

v

t1 t2

v (m s –1)

6 t (seconds)

30

Graphically, area = 90 m


The graphical representation of distance versus time (where the slope of the graph is the velocity) is useful when considering problems involving two bodies. Example A train leaves Station A travelling at 15 m s-1. One minute later a train leaves Station B travelling at 20 m s-1. The two lines meet at Junction C, 3 km from A and 1.5 km from B. Which train gets to the junction first and by how long? You may wish (or need) to do some of the calculations mathematically but the graphical representation is at least a very useful tool in clarifying the problem. It is sometimes also useful to plot a velocity-time graph (the integral of which is a distance-time graph). E.g. acceleration 2 m s-2 for 6 s, then constant velocity.

0 km

3 km

1.5 km Station B

Junction C distance

time (seconds) 50 100 200 150

15 m s-1 20 m s-1

65 s

0

5

10

15

0 2 4 6 8 10 time (s)

velo

city

(m s

-1)

0

20

40

60

80

100

0 2 4 6 8 10 time (s)

dist

ance

(m)


B2 Plane curvilinear motion Motion in two dimensions

Position is now a vector r(t), with position moving to r+∆r at time t+∆t.

Velocity, v = lim

∆t→0 ∆r∆t =

drdt = r. ,

acceleration, a = lim

∆t→0 ∆v∆t =

dvdt = v. =

d2rdt2 = r..

In rectangular coordinates:

r = (x,y) = xi + yj

v = (vx,vy) = x. i + y. j magnitude (speed): v = vx2 + vy

2

a = (ax,ay) = x..i + y..j tanθ = vy/vx Projectiles Constant acceleration (due to gravity) ax = 0, ay = -g (where g≈9.81 m s-2). (Neglect air resistance.) Initial conditions: x = y = 0 at t = 0 (or r = 0 at t = 0).

vx = vx0 = v0 cos θ (constant horizontal velocity)

vy = vy0 – gt = v0 sin θ – gt (constant downward acceleration) Initial conditions: x0 = y0 = 0 x = vx0t y = vy0t – ½gt2 Where does it hit the ground?

y = 0, ⇒ vy0t = ½gt2, and so t = 2vy0/g (t = 0 is also a solution).

x = vx0t = vx02vy0

g = 2v0

2

g sin θ cos θ = v0

2

g sin 2θ

y

∆r

r

x

y

x

θ

v0


Example Firing a projectile over an obstacle, with initial speed 100 m s-1. First find the angle

From solving “where does it hit the ground?” we have: x = v0

2

g sin 2θ, and here x = 150 m.

Putting in v0 = 100 m s-1 and g = 9.81 m s-2 gives sin2θ = 0.147, so 2θ = 8.5° or 171.5° and so,

θ = 4.25° or 85.75° (it is important to notice that there are usually two solutions).

Will it clear the obstacle? θ = 4.25° vx0 = v0 cos4.25° = 99.7 m s-1, vy0 = v0 sin4.25° = 7.4 m s-1., At x = 100 m, t = x/vx0 = 1.003 s. Thus y = 7.4×1.003 – ½×9.81×1.0032 = 2.49 m. So the projectile will not clear the obstacle at this angle. θ = 85.75° vx0 = v0 cos85.75° = 7.4 m s-1, vy0 = v0 sin85.75° = 99.7 m s-1., At x = 100 m, t = x/vx0 = 13.514 s. Thus y = 99.7×13.514 – ½×9.81×13.5142 = 452 m. So the projectile will clear the obstacle at this angle by a considerable height!

y

x

θ

v0

target

100 m

3 m

50 m


B3 Tutorial: kinematics A car accelerates from 0 to 144 kmh (40 m s-1) in 8 seconds (with constant acceleration). It then continues travelling at this speed. There is a police car 90 m from the position where the first car started. Immediately the police car is passed, it begins to accelerate to 180 kmh (50 m s-1) in 10 seconds. With the aid of graphical representations, find out when and where the police car overtakes the first car?


C Dynamics

C1 Force and acceleration: Newton’s Laws of Motion Newton’s Laws

(Galileo’s Principle of Inertia) For bodies of fixed mass:

F = ma

m is mass (kg), a is acceleration (m s-2), F is the force (Newtons N).

So 1 N = 1 kg m s-2 [We are using a system of units such that the constant of proportionality is one.] More generally,

F = ddt(mv)

Momentum is a vector, p = mv. These opposing forces act on the different interacting bodies. This includes contact forces and body forces. Contact forces

Force F by A on B Force –F by B on A

Body forces, e.g. gravity

Gravitational force F of A on B Gravitational force –F of B on A

If A has mass m, B has mass M and r is the distance between the centres of mass, the magnitude of the force is

F = GmM

r2

G = 6.67×10-11 N m2 kg-2 (universal constant)

On the surface of the earth r is approximately constant,

F =

GM

r2 m = g m, g ≈ 9.81 m s-2

Newton’s First Law A body remains at rest or continues to move at constant velocity unless acted upon by a net force.

Newton’s Second Law The rate of change of momentum of a body is proportional to the force acting on it.

Newton’s Third Law Action and reaction are equal (in magnitude) and opposite (in direction).

F -F

A B

F -F

A

B


Examples 1) If a car has a mass of 500 kg, what is the force required to accelerate it from 0 to 30 m s-1 in 5 s (at a constant rate)? a = 30/5 = 6 m s-2, F = ma = 500×6 = 3000 N (or 3 kN) 2) The engines of a ship provide a maximum thrust T (N) which depends on the ship’s speed V (m s-1)

T = 105(1 - 0.01V)

The drag on the ship as it moves through the water is

D = 200V2 If the mass of the ship is 5000 tonnes, calculate the initial acceleration if the ship starts with full engine power from rest and the maximum speed of the ship. Draw up a table of acceleration as a function of velocity and sketch velocity as a function of time for the ship as it accelerates from rest to 95% of maximum speed.

Force F = 105(1 - 0.01V) - 200V2 = ma

a = F/m = … = 2×10-2(1 – 0.01V) – 4×10-5V2

At rest (V=0) a=2×10-2 m s-2 Maximum speed when a=0, solving the quadratic gives Vmax=20 m s-1 (the other, unrealistic, solution is –25 m s-1) Draw up table of acceleration as a function of velocity, then assume that during acceleration we can approximate the time taken to accelerate from v to v+∆v as ∆t≈∆v/a. (In the table below ∆v = 1 m s-1.) v (m s-1) a (m s-2) ∆t (s) t = Σ ∆t (s)

0 0.02 50 0 1 0.01976 50.60729 50 2 0.01944 51.44033 100.60733 0.01904 52.52101 152.04764 0.01856 53.87931 204.56865 0.018 55.55556 258.44796 0.01736 57.60369 314.00357 0.01664 60.09615 371.60728 0.01584 63.13131 431.70339 0.01496 66.84492 494.8346

10 0.014 71.42857 561.679611 0.01296 77.16049 633.108112 0.01184 84.45946 710.268613 0.01064 93.98496 794.728114 0.00936 106.8376 888.713115 0.008 125 995.550716 0.00656 152.439 1120.551 17 0.00504 198.4127 1272.99 18 0.00344 290.6977 1471.402 19 0.00176 568.1818 1762.1

[It is also possible to solve the equation for v as a function of t analytically by integrating the equation for dv/dt (=a), after some rearrangement (actually integrating dt/dv).]

05

10152025

0 1000 2000 3000 4000time (s)

v (m

s-1

)

analytic approx


C2 Conservation of momentum Conservation of momentum follows from Newton’s Second and Third Laws. Consider two interacting bodies: The rate of change of momentum of the first body is

ddt(m1v1) =

ddt(p1) = -F

and of the second body

ddt(m2v2) =

ddt(p2) = F

Adding these gives ddt(p1) +

ddt(p2) = -F + F = 0, or

ddt(p1 + p2) = 0, so (p1 + p2) is constant.

Principle of conservation of momentum: The total momentum of an isolated system of two (or more) interacting bodies doesn’t change. Examples 1) Two masses (m1=100 kg and m2=50 kg) are joined together and moving at 10 m s-1. A spring is released which separates the two masses such that the final speed of the second mass is 20 m s-1. What is the final speed of the first mass? Initial momentum (100 + 50)10 = 1500 kg m s-1 Final momentum 100 v1 + 50×20 = 100 v1 + 1000, Initial momentum = Final momentum, so the final momentum of the first body is

100 v1 = 500 kg m s-1, And thus v1 = 5 m s-1 2) A small mass (0.1 kg) travelling at 250 m s-1 strikes a stationary larger mass (500 kg) suspended on a cable. After the impact, the small mass remains imbedded in the larger mass. What is the final speed of the combined masses?

Initial momentum Final momentum 0.1×250 = 25 kg m s-1 500.1×v = 25 kg m s-1 ∴ v = 25/500.1 = 0.050 m s-1

F -F

m1

m2

v1

v2

10 m s-1 20 m s-1

m1 m2

v1


Force and rate of change of momentum 3) A car of mass 700 kg is travelling eastward at 20 m s-1. The road bends 45° and the car slows to 16 m s-1, now travelling north-east. Assuming that the change in speed and direction of the car is achieved by a constant applied force F acting over 3 seconds, find F. Use rectangular components with the initial velocity in the x-direction. Initial velocity, vi = (20, 0), momentum pi = (20×700, 0) = (14000, 0) Final velocity, vf = (16×cos45°, 16×sin45°) = (11.31, 11.31), momentum pf = (11.31×700, 11.31×700) = (7917, 7917) Change in momentum (final - initial)

pf - pi = (7917, 7917) - (14000, 0) = (-6083, 7917) If this change happened in a time of 3 seconds, then the rate of change of momentum (which is equal to the force) is given by,

F = (pf - pi)/∆t = (-6083, 7917)/3 = (-2028, 2639) N Note the total change of momentum (pf - pi) = F×∆t is given by force multiplied by time.

20 m s-1

16 m s-1

45°

F


C3 Impacts and impulse In the first two example problems in the previous lecture we didn’t calculate the forces involved (to do so we would need information about the length of time of the interactions). From the final problem we saw how the total change in momentum of a body was given by force×time (when the force acting on the body is constant). Impacts (collisions)

When two bodies collide the interaction between the bodies involves deformation of the bodies and a complicated force-time history (with rapidly changing and very large forces).

We will begin with cases where the motion is in a straight line, so that we only have to consider one component of the velocities and momentum. For angled collisions both the x- and y-components of the total momentum must be conserved (use separate equations for the x and y momentum). Impulse The total change of momentum that a body experiences in such an interaction is known as an impulse.

I = (p' - p) impulse = (final momentum – initial momentum) For a steady force I = F×∆t, in general we would need to integrate the varying force over the duration of the interaction.

I = ⌡⌠F dt.

Coefficient of restitution In an impact between two bodies, the elasticity of the collision can be described by the coefficient of restitution, e, given by

e = speed of separationspeed of approach .

In terms of the velocities this is:

e = |v1' - v2'||v1 - v2|

.

For motion in two-dimensions, note that you find the difference in velocities first, then the magnitude of this difference. E.g. if v1 = (3, -1) m/s and v2 = (4, 2) m/s, then |v1 - v2| = |(-1, -3)| = √10 = 3.16 m/s. The coefficient of restitution is also the ratio of restorative forces (as the bodies resume their shape) to deformative forces (as the bodies deform) and e is always between 0 and 1. Perfectly inelastic or plastic impacts have e=0 (the bodies stick together after impact). Perfectly elastic impacts have e=1 (the bodies fly apart at the same relative speed as they approached).

v1 v2

v1' v2'

m1 m2


Example Find the final velocities after a collision between two bodies travelling along a straight line, with:

m1 =1 kg, m2 = 10 kg, v1 = 15 m s-1 and v2 = 4 m s-1; for the three cases e=0, e=0.5 and e=1. Total initial momentum

m1v1 + m2v2 = 1×15 + 10×4 = 15 + 40 = 55 kg m s-1 a) e=0 (perfectly plastic) In this case the bodies stick together after collision with v1' = v2'. Total momentum

m1v1' + m2v2' = 11× v1' = 55 kg m s-1, so v1' = 5 m s-1 = v2'. b) e=0.5

|v1' - v2'| = 0.5×|v1 - v2| = 0.5×(15 – 4) = 5.5 m s-1. We expect v2'>v1', so this implies that

v2' - v1' = 5.5 m s-1. (So, v1' = v2' - 5.5.)

From the final momentum we have

m1v1' + m2v2' = 1v1' + 10v2' = 55

v2' - 5.5 + 10v2' = 55

11v2' = 60.5

v2' = 60.5/11 = 5.5 m s-1. And so v1' = 0 (i.e. the smaller mass is stationary after the collision). c) e=1 (perfectly elastic)

|v1' - v2'| = 1.0×|v1 - v2| = 1.0×(15 – 4) = 11 m s-1. We expect v2'>v1', so this implies that

v2' - v1' = 11 m s-1. (So, v1' = v2' - 11.)

From the final momentum we have

m1v1' + m2v2' = 1v1' + 10v2' = 55

v2' - 11 + 10v2' = 55

11v2' = 66

v2' = 66/11 = 6 m s-1. And so v1' = -5 m s-1 (i.e. the smaller mass bounces back in the opposite direction after the collision).


C4 Tutorial: Momentum Plastic/elastic collisions

mechanics - university of manchester · meriam jl and kraige lg, engineering mechanics (volume 1...

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