Homework 2 Solutions - Purdue University

Introduction to Dynamics (N. Zabaras)

Homework 2 Solutions:Newton’s Second Law

Prof. Nicholas Zabaras

Warwick Centre for Predictive Modelling

University of Warwick

Coventry CV4 7AL

United Kingdom

Email: [email protected]

URL: http://www.zabaras.com/

March 5, 2016

mailto:[email protected]

http://www.zabaras.com/


Problem 1

2

Weight: W =5 lb Spring constant: k =10 lb/ft

unstretched length of 3ft, Find the normal force N &

and acceleration a at the position shown.

2

: cos cos ( )( )snnW v

F ma W N F tg

t

W mg

N

sF

nmatma

t

90

2

3

2 322 2

2

2

tan 2 63.435 ( 2)

1 ( )1 ( 2)

11.18 ( )1

x

dyx slope of the curve at x

dx

dy

dxft radius of curvature

d y

dx

n

𝑦 = 8 −1

2𝑥2,

𝑑𝑦

𝑑𝑥= −x

Newton’s law in the normal

direction to the curve that A travels

Note the spring force Fs is NOT in the normal

direction. normal

direction

spring

direction

normal

component of

acceleration

t

Tangent to

the curve

Normal to

the curve

y direction

x direction


Problem 1

3

2 2

1

2 6 6.3246

10( 3) 33.246( )

6tan ( ) 71.565 90 45

2

s

OA

F kx OA lb

t

2

2

22 2

: cos cos ( )( )

cos 63.435 cos ( )( )

24.3551( )

5: sin sin

32.2

32.2( sin 63.435 33.246sin 45 )

5

~ 180 /

t t

t

s

s

s

n

t n

n

t

W vF ma W N F t

g

W vN W F t

g

N

F ma W F t a

a W

a a a ft s

W mg

90

tN

sF

nma tma

X=stretch of the

spring beyond its

initial length of 3 ft

𝑦 = 8 −1

2𝑥2 = 6,

𝑑𝑦

𝑑𝑥= −2

𝑦 = 6See Fig. on the

left explaining the

angles

Newton’s law in the tangential

direction to the curve that A travels

Magnitude of the acceleration

vector


Problem 2

The 12-lb block B starts from rest

and slides on the 30-lb wedge A,

which is supported by a horizontal

surface.

Neglecting friction, determine

(a) the acceleration of the wedge,

and

(b) the acceleration of the block

relative to the wedge.

SOLUTION:

• The block is constrained to slide

down the wedge. Therefore, their

motions are dependent. Express the

acceleration of block as the

acceleration of wedge plus the

acceleration of the block relative to

the wedge.

• Write the equations of motion for the

wedge and block.

• Solve for the accelerations.

4


Problem 2

SOLUTION:

• The block is constrained to slide down the wedge.

Therefore, their motions are dependent.

ABAB aaa

• Write equations of motion for wedge and block.

x

y

:AAx amF

AA

AA

agWN

amN

1

1

5.0

30sin

:30cos ABABxBx aamamF

30sin30cos

30cos30sin

gaa

aagWW

AAB

ABABB

:30sin AByBy amamF

30sin30cos1 ABB agWWN5


Problem 2

AA agWN 15.0

• Solve for the accelerations.

30sinlb12lb302

30coslb12sft2.32

30sin2

30cos

30sin30cos2

30sin30cos

2

1

A

BA

BA

ABBAA

ABB

a

WW

gWa

agWWagW

agWWN

2sft07.5Aa

30sinsft2.3230cossft07.5

30sin30cos

22AB

AAB

a

gaa

2sft5.20ABa

6


Problem 3

The bob of a 2-m pendulum

describes an arc of a circle in a

vertical plane. If the tension in the

cord is 2.5 times the weight of the

bob for the position shown, find the

velocity and acceleration of the bob

in that position.

SOLUTION:

• Resolve the equation of motion for

the bob into tangential and normal

components.

• Solve the component equations for

the normal and tangential

accelerations.

• Solve for the velocity in terms of the

normal acceleration.

7


Problem 3

SOLUTION:

• Resolve the equation of motion for the bob into

tangential and normal components.

• Solve the component equations for the normal

and tangential accelerations.

:tt maF

30sin

30sin

ga

mamg

t

t

2sm9.4ta

:nn maF

30cos5.2

30cos5.2

ga

mamgmg

n

n

2sm03.16na

• Solve for velocity in terms of normal acceleration.

22

sm03.16m2 nn avv

a

sm66.5v

8


Problem 4

Determine the rated speed of a

highway curve of radius = 400 ft

banked through an angle = 18o.

The rated speed of a banked

highway curve is the speed at

which a car should travel if no

lateral friction force is to be exerted

at its wheels.

SOLUTION:

• The car travels in a horizontal

circular path with a normal

component of acceleration directed

toward the center of the path.The

forces acting on the car are its

weight and a normal reaction from

the road surface.


the car into vertical and normal

components.

• Solve for the vehicle speed.

9


Problem 4

SOLUTION:

• The car travels in a horizontal

circular path with a normal

component of acceleration directed

toward the center of the path.The

forces acting on the car are its

weight and a normal reaction from

the road surface.


the car into vertical and normal

components.

:0 yF

cos

0cos

WR

WR

:nn maF

2

sincos

sin

v

g

WW

ag

WR n

• Solve for the vehicle speed.

18tanft400sft2.32

tan

2

2 gv

hmi1.44sft7.64 v10


x

ymg

N k N

40 9.81 cos 20 0

369

N

N N

0;yF

x xF ma sink N mg ma

cos 0N mg

2

0.25(369) 40 9.81 sin 20 40

5.66 /

x

x

a

a m s

a = ?

Problem 5

11


Example 13-5

y yF ma 981 2 100 AT a

Block A

y yF ma 196.2 20 BT a

Block B

2 A Bs s l 2 0A B 2 0A Ba a

o Ba t

Problem 6

12

The 100-kg block A is released

from rest. If the masses of the

pulleys and the cord are

neglected, determine the velocity

of the 20-kg block B in 2 sec.

2 2: 3.27 / , 6.54 / , 327A BFinally a m s a m s T N

0 ( 6.54)2 13.1 / ( )o Ba t m s upwards


VA = ?

2Sc + SA = L

0 = 2ac + aA aA = -2ac

ca

0: 200*9.81cos30 0 1699

: 0.5 2 sin 30 200

y

x cc

F N N N

F ma N T mg a

Problem 7

13

What is VA when A reaches

the ground B? We start with VA=VC=0

2Sc + SA = L

0 = 2aC+ aA aA = -2aC

Note the directions of positive acceleration we have

selected: aA >0 when A goes down, aC is positive

when C goes down. Obviously we expect that for

aA>0, aC<0.

Note the x-direction

selected (consistent

with the notation for

sC and aC)


mgAma

2

2 4.62( / )

(this is derived from ds dv )2

A A

AA

V a s m s

Va v a s

125*9.81 125

1226.25 125 (2)

y A A

A

F ma T a

T a

2 2

(2) (1)

849.5 2(1226.25 125 ) 981 200

622 200 250

200 500

0.888( / ) 1.777( / ) 1004

A C

C A

C C

C A

a a

a a

a a

a m s a m s T N

Problem 7

14

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