Example 13.1

The 50-kg crate rests on a horizontal plane for which the coefficient of kinetic friction is μk = 0.3. If the crate is subjected to a 400-N towing force, determine the velocity of the crate in 3 s starting from rest.

Free-Body Diagram. The weight of the crate is W = mg = 50 (9.81) = 490.5 N. The frictional force has a magnitude F = μkNC and acts to the left, since it opposes the motion of the crate. The acceleration a is assumed to act horizontally, in the positive xdirection. There are 2 unknowns, namely NC and a.

Example 13.1

Equations of Motion.

2/19.5

5.290

030sin4005.490;

503.030cos400;

sma

NN

NmaF

aNmaF

C

Cyy

Cxx

=

=

=+−=↑+

=−=→

∑∑

+

o

o

Solving for the two equations yields

Example 13.1

Kinematics. Acceleration is constant, since the applied force P is constant. Initial velocity is zero, the velocity of the crate in 3 s is

→=+=+=

sm

tavv c

/6.15)3(19.50

0

Example 13.1

A 10-kg projectile is fired vertically upward from the ground, with an initial velocity of 50 m/s. Determine the max height to which it will travel if

(a) atmospheric resistance is neglected.

(b) atmospheric resistance is measured as FD = (0.01v2) N, where v is the speed at any instant, measured in m/s

Example 13.2

Part (a) Free-Body Diagram. The projectile’s weight is W = mg = 10(9.81) = 98.1 N. Assuming the unknown acceleration a acts upward in the positive z direction.

Equations of Motion.2/81.9,101.98; smaamaF zz −==−=↑+ ∑

The results indicates that the projectile is subjected to a constant downward acceleration of 9.81 m/s2

Example 13.2

Kinematics. Initially, z0 = 0 and v0 = 50 m/s. At max height, z = h and v = 0. Since acceleration is constant, then

( )

mhh

zzavv c

127)0)(81.9(2)50(0

)(22

020

2

=−−+=

−+=↑+

Example 13.2

Part (b) Free-Body Diagram. Since the force FD = (0.01v2) N tends to retard the upward motion of the projectile, it acts downward as shown

Equations of Motion.

22

2

/81.901.0

,101.9801.0;

smva

avmaF zz

−−=

=−−=↑+ ∑

Example 13.2

Kinematics. Here the acceleration is not constant since FD depends on the velocity. Since a = f(v), we relate a to position using

( ) dvvdzvdvvdza =−−=↑+ )81.9001.0(; 2

Integrating with initially z0 = 0, v0 = 50 m/s(positive upward), and at z = h, v = 0

mh

vvdvvdz

h

114

)9810ln(50081.9001.0

0

5020

50 20

=

+−=+

−= ∫∫

Example 13.2

The baggage truck A has a weight of 3600 N and tows a 2200 N cart B and a 1300 N cart C. For a short time the driving frictional force developed at the wheels is FA = (160t) N where t is in seconds. If the truck starts from rest, determine its speed in 2 seconds. What is the horizontal force acting on the coupling between the truck and cart B at this instant?

Example 13.3

Free-Body Diagram. As shown, it is the frictional driving force that gives both the truck and carts an acceleration, we have to consider all 3 vehicles.

Example 13.3

Equations of Motion. Only motion in the horizontal direction has to be considered.

ta

atmaF xx

221.081.9

130022003600160;

=

⎟⎠⎞

⎜⎝⎛ ++

==← ∑+

Kinematics. Since the acceleration is a function of time, the velocity of the truck is obtained using a = dv/dt with the initial condition that v0 = 0 at t = 0,

Example 13.3

smtvdttdvv

/442.01105.0;)221.0(2

022

00=== ∫∫

Free-Body Diagram. Considering the FBD of the truck, we can “expose” the coupling force T as external to the FBD.

Example 13.3

Equations of Motion. When t = 2 s, then

[ ]

NT

TmaF xx

8.157

)2(221.081.9

3600)2(160;

=

⎟⎠⎞

⎜⎝⎛=−=← ∑

+

Example 13.3

A smooth 2-kg collar C is attached to a spring having a stiffness k = 3 N/m and an unstretched length of 0.75 m. If the collar is released from rest at A, determine its acceleration and normal force of the rod on the collar at the instant y = 1 m.

Example 13.4

Free-Body Diagram. The FBD of the collar when it is located at the arbitrary position y is as shown. The weight is W = 2(9.81) = 19.62 N. The collar is assumed to be accelerating so that “a” acts downward in the positive y direction. There are four unknown, namely, NC, Fs, a and θ.

Example 13.4

aFmaF

FNmaF

syy

sCxx

2sin62.19;

0cos;

=−=↓+

=+−=→

∑∑

+

θ

θ

Equations of Motion.

The magnitude of the spring force is a function of the stretch s of the spring; i.e. Fs = ks. The unstretched length is AB = 0.75m. Therefore

)75.0)75.0((3 22 −+=−= yABCBs

Example 13.4

Since k = 3 N/m,

)75.0)75.0((3 22 −+== yksFs

The angle θ is related to y by trigonometry

75.0tan y

=θ

For y = 1 m, θ = 53.1 and Fs = 1.50 N. Therefore, NC = 0.900 N and a = 9.21 m/s2

Example 13.4

Equations of Motion: Translation

Example 17.5The car has a mass of 2Mg and a center of mass at G. Determine the car’s acceleration if the “driving” wheels in the back are always slipping, whereas the front wheel freely rotate. Neglect the mass of the wheels. The coefficient of kinetic friction between the wheels and the road is μk = 0.25.

SolutionSection IFBD• The rear wheel frictional

force pushes the car forward, and since slipping occurs, FB= 0.25NB

• Frictional forces acting on the front wheels = 0m, since these wheels have negligible mass

• Car (point G) is assumed to accelerate to the left, in the negative x direction

SolutionSection IEquations of Motion

Solving,kNNkNNsma

mNmNmNM

NNN

aF

akgNamF

BAG

BBA

G

BA

yGy

GB

xGx

7.1288.6/59.1

0)75.0()3.0(25.0)25.1(;0

0)81.9(2000

;)(

)2000(25.0;)(

2 ==←=

=+−−=∑

=−+

=∑↑+

−=−=∑→+

SolutionSection IIFBD• Apply moment equation at point A,

unknown NA will be eliminated from the equation

• Use the kinetic diagram o visualize the moment at A

SolutionSection IIEquations of Motion

Solve and proceed with Section I equations

)3.0()2000()25.1()81.9(2000)2(;

makgmNmNM

GB

AA

=−∑=Μ∑

Example 17.6The motorcycle has a mass of 125kg and a center of mass at G1, while the rider has a mass of 75kg and a center of mass at G2. Determine the minimum coefficient of static friction between the wheels and the pavement in order for the rider to do a “wheely” ie, lift the front wheel off the ground as shown. What acceleration is necessary to do this? Neglect the mass of wheels and assume that the front wheel is free to roll.

SolutionFBD and Kinetic Diagrams• Consider both the motorcycle and the rider as the system to be analyzed• Determine center of mass, however, consider the separate weight and mass

of each of its components parts • Both parts move with same acceleration• Assume that the front wheel is about to leave the ground so that normal

reaction NA ≈ 0• Three unknowns NB, FB and aG

FBD and Kinetic Diagrams

SolutionEquations of Motion

)6.0)(125()9.0)(75()8.0)(25.1226()4.0)(75.735(

;)(025.122675.735

;)(

)12575(;)(

makgmakgmNmN

MNNN

aF

akgkgFamF

G

G

BkB

B

yGy

GB

xGx

−−=−−

∑=Μ∑=−−

=∑↑+

+==∑→+

SolutionEquations of Motion• Solving

• For minimum coefficient of static friction912.01962/1790/)(

17901962

/95.8

min

2

===

==

→=

NNNF

NFNNsma

BBs

B

B

G

μ

Example 17.7A uniform 50-kg crate rest on a horizontal surface for which the coefficient of kinetic friction is 0.2. determine the acceleration if a force of P = 600N is applied to the crate.

SolutionFBD • Force P can cause the crate to either slide or to tip over• Assume that the crate slides so that F = μkNC = 0.2NC

• Resultant force NC acts at O, a distance x where 0 < x ≤ 0.5m from the crate’s center line

• Three unknowns NC, x and aG

SolutionEquations of Motion

0)5.0()()3.0(600;0

05.490

;)(

)50(2.0600;)(

=−+−=Μ∑

=−

=∑↑+

=−=∑→+

mNxNmN

NN

aF

akgNNamF

CC

G

C

yGy

GC

xGx

• Solving

• x < 0.5m, the crates slides as assumed• If x > 0.5m, problem would be reworked with the assumption that the

crates tips • In that case, NC acts at corner point A and F ≤ 0.2NC

mxNNsma

C

G

467.10490

/0.10 2

==

→=

Example 17.8The 100kg beam is supported by two rods having negligible mass. Determine the force created in each rod if at the instant θ = 30° and ω = 6rad/s.

SolutionFBD • Beam moves with curvilinear translation since points B, D and the center of

mass all move along circular paths, each path having the same radius of 0.5m

• Using normal and tangential coordinates, the FBD for the beam is shown• Because of translation, G has the same motion as the pin at B, which is

connected to both the rod and the beam

View Free Body Diagram

• By studying the angular motion of AB, the tangential component of accelerations acts downward to the left due to the clockwise direction of α

• Normal component of acceleration is always directed towards the center of curvature (towards point A for rod AB)

• Since angular velocity of AB is 6 rad/s, (aG)n = ω2r = (6 rad/s)2(0.5m) = 18m/s2

FBD

• Three unknowns TB, TD and (aG)t

Equations of Motion

0)3.0)(30sin()4.0)(30cos(

;0)(10030sin981

;)()/18(10030cos981

;)(2

=+−

=Μ∑=

=∑=−+

=∑

mTmT

akg

aFsmkgNTT

amF

DB

G

tG

tGt

DB

nGn

oo

o

o

• Solving

2/90.4)(

32.1

sma

kNTT

tG

DB

=

==