a 8 dec 2003 solution

December 2003 Applications of Engineering Mechanics in Medicine, GED – University of Puerto Rico, Mayaguez 1

Education is not received. It is achieved---Anonymous

ENGINEERING BIOMECHANICS: STATICS¹ <Josue Aracon, Lizbeth Agront, Laritza Lara, Dania Vazquez>²

1

1 This review article was prepared on December 8,

2003. Course Instructor: Dr. Megh R. Goyal,

Professor in Agricultural and Biomedical

Engineering, General Engineering Department, PO

Box 5984, Mayaguez, Puerto Rico 00681-5984. For

details contact:

[email protected] or visit at:

http://www.ece.uprm.edu/~m_goyal/home.htm

² The authors are in the alphabetical order.


DIMENSION AND UNITS

A woman goes to Buenos Aires, Argentina, when she starts to feel sick and goes to the

hospital. She arrives, the nurse asks for her weight in kilograms, but she answered that her

weight was 125 lb. The nurse asks here to do the conversion to kg. How much is her mass

in kg?

SOLUTION

Conversion rate:

2.2 lb = 1 kg

125 lb * 1 kg = 275 kg

2.2 lb

125 lb = 56.8 kg ←


DIMENSIONS AND UNITS

The human body contains three types of fluids. These include the extracellular, intracellular

and transcellular fluids. It is known that these fluids form 60% of the body weight for an

average 70 kg male. What will be the eight in lb.?

[Goyal 2003, A-1]

SOLUTION

(70kg)(60%) = 42 kg

42 kg → lb = ?

42 kg (1 lb/ 0.4536 kg) = 92.6 lb

92.6 lb←



Suppose your hair grows at a rate of 1/32 in. per day. Find the rate at which it grows in

nanometers per day.

[Serway and Jewett 2000, 32]

SOLUTION

( 1 /32 ) ( 2.54 x 10 EE -2 meters ) ( 1 nanometers )

____________________________________________________________

( 1 inch ) ( 1 x 10 EE -9 meters )

= 7937500 nanometers

7937500 nanometers←



If an Olympic sprinter runs 100 meters in seconds, his average

velocity is 10 m/s. What is his average velocity in mi/hr?

[Bedford and Fowler 1998, 10]

SOLUTION

10 m/s = 10 m/s (1ft / 0.3048) (1 mi / 5280 ft) (3600s / 1hr

Vave = 22.4 mi / hr

Vave = 22.4 mi / hr←


STATICS OF PARTICLES

SOLUTION

FBD

A

B

Forces at A

Ax = (3 N) (cos 0o) = 3 N

Ay = (3 N) (sin 0o) = 0 N

Forces at B

Bx = (2 N) (cos 25o) = 1.81 N

By = (2 N) (sin 250) = 0.84 N

Resultant

R = 2 2(3 1.81) (.84)+ +

R = 4.88 N

or R = 4.88 N �

Direction

Θ = 1tan (.84 / 4.81)−

Θ = 9.9 o

or Θ = 9.9

o �

A man is walking on the street. Each foot moves with a force of 3 N to

the right and 2 N upwardly, at a 25º angle from the street level. Calculate

the resultant force from the foot?



A hospital patient is recovering from an accident she endured recently. Her doctor

suggested she work out in order to get a good circulation in her arms. Knowing that

�=25¬, what is the angle formed with the line CB and the x-axis? Determine the

tension in cable AC and in cable BC.

[Beer and Johnston 2004, 41]

SOLUTION

FDB: TBC

y

TAC 5º

TAC

x 25 º 5 lb 5 lb TBC

Law of Sines:

TAC / sin 115º = TBC / sin 5º = 5 lb / sin 60º

TAC = (5 lb* sin 115º) / (sin 60º) = 5.23 lb

TBC = (5 lb* sin 5º) / (sin 60º) = 0.503 lb

TAC = 5.23 lb ←

TBC = 0.503 lb ←



The helicopter view shows two peoples pulling on a stubborn mule. .Find a single

force that is equivalent to the force shown and the force that a third person would

have to exert on the mule to make the resultant force equal to zero. The force are

measured in units of Newtons.


SOLUTION

FBD

X = 120 cos 60 + 80 cos 75

= 80.7

Y = 120 sen 60 + 80 sen 75

= 181.3

Resultant

( ( 80.7) ^ 2 + (181.3) ^ 2 ) ) ^ .5 = 313


STATIC OF PARTICLES

While steadily pushing the machine up an incline, a person

exerts a 180-N force P as shown. Determine the components

of P, which are parallel and perpendicular to the incline.

[J.L.Meriam and L.G.Kraige 2000, 31]

SOLUTION

Pn = P sin 65°

= 180 sin 65°

= 163.1 N

Pt = P cos 65°

= 180 cos 65°

= 76.1 N

Pn = 163.1 N←

P = 76.1 N←


EQUIVALENT SYSTEM OF FORCES

A hand is applying a 13.2 N force at point P. And P is applied to the lever

which controls the auger of a snow blower. Determine the moment of the

hand at point P about A when α is equal to 30º.


SOLUTION

First note

Px = Psinα = (13.2 N) sin 30o = 6.60 N

Py = Pcosα = (13.2 N) cos 30o = 11.4315 N

Notting that the direction of the moment of each

force component about A is counterclockwise.

MA = x B/A Py + y B/A Px

= (0.086 m) (11.4315 N) + (.0122 m ) (6.60 N)

= 1.78831 N . m

MA = 1.788 N. m ←


EQUIVALENT SYSTEMS OF FORCES

A person was hospitalized for bad blood circulation in the legs. To

help the circulation in the legs, doctors recommended the he should

sleep in a position bed. His legs will be lifted at the end of the position

bed. The bottom of the position bed (ABCD) is supported by cables at

corners C and D. The tension in each of the cables is 360 lb.

Determine the moment about A of the force exerted by (a) the cable at

D, (b) the cable at C.


SOLUTION

a) MA = r E/A * TDE

r E/A = (92in) j

where TDE = λDE * TDE

= {((24 in) i + (132 in) j – (120 in) k ) / √(24)² +

(132)² + (120)²}* {(360 lb)}

= 48 i + 264 j – 240 k

MA = i j k

0 92 0 lb-in

48 264 -240

= -(22080) i – (4416) k lb-in

b) MA = r G/A * TCG

where r G/A = (108 in) i + (92 in) j

TCG = λCG * TCG

= {((-24 in) i + (132 in) j + (120 in) k ) / √(24)²

+ (132)² + (120)²}* {(360 lb)}

= -48 i + 264 j – 240 k

MA = i j k

108 92 0 lb-in

-48 264 -240

= -22080 i + 25920 j + 32928 k lb-in


EQUIVALENT SYSTEMS OF FORCES

SOLUTION

d = sin 50° x 13”

= 9.96”

= 8 N

ΣMA = Fd sin θ

120° = 8 (9.96) sin 90°

= 79.68

8-lb

A person locks his elbow so that the angle ABC is maintained at 130° it

rotates the shoulder joint at A so that the arm remains in the vertical

plane shown. The shoulder joint remains fixed. Determine the moment

about point A of the weight of the 8-lb sphere when the angle θ is 0°. [J.L.Meriam and L.G.Kraige 2000, 101]


EQUIVALENT SYSTEM OF FORCES

A person bending forward to lift a load “with his back” rather than “with

his knees” can be injured by large forces exerted on the muscle and

vertebrae. The spine pivots mainly at the fifth lumbar vertebra, with the

principal supporting force provided by erector spinals muscle in the

back. To see the magnitude of the force involved and to understand why

back problems are so common among humans, consider the model for a

person bending forward to lift 200 N weight 350 N, pivoted at the base

of the spine. The erector spinals muscle, attached at a point two thirds of

the way up the spine, maintains the positions of the back The angle

between the spine and this muscle 12º. Find the tension in the back

muscle.


SOLUTION

∑ Fx = Rx – 200 – T sen 12

Rx = T sen 12

∑ Fy = Ry –200 – 350 + T sen 12

Ry = T sen 12 + 200 + 350

Ty * 2/3 d – 350 * ½ d –200 = 0

Ty = 562.5

T = 2.71 KN


EQUILIBRIUM OF RIGID BODIES

A modified peavey is used to lift a 0.2 m diameter log of mass 36 kg.

Knowing that θ= 45º and that the force exerted at C by the worker is

perpendicular to the handle of the peavey, determine the force exerted by

the arms of the worker.


SOLUTION

FBD

The weight is:

W = 36*9.8 = 353.16

From Geometry:

ß = tan-1

(1.1 m / (1.1 m + .2 m) = 40.236o

α = 45o – ß = 45

o – 40.236

o = 4.7636

o

Law of Sin:

W = C = A

sin ß sin α sin 35o

or

353.16 = C = A

sin ß sin α sin 35o

a) C = 45.404 N

or C = 45.404 N at 45o ←

b) A = 386.60 N

or A = 386.60 N at 85.2o ←


EQILIBRIUM OF RIGID BODIES

A 60 N weight is held in the hand with the forearm making a 90¬ angle with

the upper arm. The biceps muscle exerts a force Fm that acts 3.4 cm from the

pivot point O at the elbow joint. Neglecting the weight of the arm and hand,

(a) Find the magnitude of Fm if the distance from the weight to the pivot

point is 30 cm, and (b) Find the force exerted on the elbow joint by the upper

arm.

[Tiple, 1999, 353]

SOLUTION

FBD:

Fm

O

mg

Fua

The torque about the elbow exerted by the weight must be balanced by the torque

exerted by the force Fm. The force Fua exerted by the upper arm at O is found by

setting the net force on the hand and forearm equal to zero. We choose upward to be

the positive direction.

a) �rrrr= 0 about point O gives Fm: Fua * (0) + Fm * (3.4 cm) – (60 N) (30cm) =0

Fm = {(30 cm) (60 N)} / (3.4 cm) = 529 N

529 N ←

b) �F = 0 gives Fua: -Fua + Fm – 60 N = 0

Fua = Fm – 60 N = 469 N

469 N ←



Figure shows a claw hammer as it being used to pull a nail out of horizontal board. If a force of

magnitude 150 N is exerted horizontally as shown, find the force exerted by the hammer claws

on the nail and the force exerted by the surface on the point of contact with the hammer head.

Assume that the force the hammer head exerts on the nails is parallel to the nail.


SOLUTION

FDB

SOLUTION

-.3 ( 150 ) + F ( .5 ) = 0

F = 779.4

R y – F cos 30 = 0

R y = F cos 30

R y = 674

.3 R x + 150 N = F sen 30

R x = 799



SOLUTION

J1 = r1f1 sin θ1

J1 0.25 x 100 sin 75°

= 24.1 N ● M

J2 = -r2 F2 sin θ2

J2 = -0.2 x 80 sin 65°

= - 14.5 N ● M

∑ J = J1 + J2

= 24.1 – 14.5

= 9.6 N ● M

∑ J = 9.6 N ● M ←

The wrench is used to loosen the bolt. Determine the moment of each

force about the bolt’s axis passing through point O.



CENTROIDS AND CENTER OF GRAVITY

David is doing an exercise of lifting body with hands. His center

gravity is found over point P. The range is 0.9 m of his feet and 0.6 m

of his hands. His weight is of 54 kg. What is the force exerted by his

hands on the floor?

[Tipler 1999, 364]

SOLUTION

FBD

Center of Gravity

Find Weight:

W = 54 kg * 9.8 m/s^2 = 526.2 N

or W = 526.2 N ←

Find Moment at A:

MA = 0 = -526.2 N * .9 m + B*1.5

= 315.72 N-m

or B = 315.72 N-m ←



A gravity board for locating the center of gravity of a person consists of

a horizontal board supported by a fulcrum at one end and by a scale at

the other end. A physics student lies horizontally on the board with the

top of his head above the fulcrum point as shown in the figure. The scale

is 2 m from the fulcrum. The student has a mass of 70 kg. When he is

on the board, the scale advances 250 N. Where is the center of gravity of

the student?

[Tipler 1999, 366]

SOLUTION

Apply �rrrr= 0 about the fulcrum:

70 gxcg = (2 m) * (250 N-m)

xcg = 72.8 cm from the fulcrum

72.8 cm ←



A person has a mass of 85 kg that rests on the fulcrum and a scale, the scale

read 320 N. The scales are separated by a distance of 2.00 m. Where is the

center of gravity of the person?


SOLUTION

∑ y = 0 about the fulcrum

85 Kg = 2 x 320 N . m

= 75.3 cm from the fulcrum



SOLUTION

130°

F1 = F3 – WL2 / L1

= (15 kg) (9.80 m/s^2) (35 cm) – (2.0 kg)

(9.80 m/s^2) (15 cm) / 2.5 cm

= 1.9 x 10^3 N

and

F2 = F1 + F3 – W

= 1.9 x 10^3 N + (15 kg) (9.80 m/s^2)

=2.0 x 10^3 N

F1 points upward and F2 points downward

A 15 kg block is being lifted by the pulley system shown. The upper arm

is vertical, whereas the forearm makes an angle of 30° with the

horizontal. What are the forces on the forearm from (a) the triceps

muscle and (b) the upper-arm bone (the humerus)? The forearm and

hand together have a mass of 2.0 kg with a center of mass 15 cm

(measured along the arm) from the point where the forearm and upper-

arm bones are in contact. The triceps muscle pulls vertically upward at a

point 2.5 cm behind that contact point.

[Halliday and Resnick and Walker 2000, 290]


TRUSSES

The compound-lever pruning can be adjusted by placing pin A

at various ratchet positions on blade is ACE. Knowing that 1.3

KN vertical forces required to complete the pruning of a small

branch, determine the magnitude P of the force that the hand

must applied to the handles shears are adjusted as shown.


SOLUTION

FBD of AC

Find Moment at C

MC = 0 = (32 mm) * 1.5 KN – (28 mm) * Ay – (10 mm) * Ax

10Ax + 28Ay = 18 KN

Ax = 1.42466 KN

Ax = 1.42466 KN ←

Ay = 1.20248 KN

Ay = 1.20248 KN ←

FBD at AD

Find Moment at D

MD = 0 = (15 mm) (1.20248 KN) – (5 mm) (1.42466 KN) – (70 mm)P

P = 0.1566 KN

P = 0.1566 KN ←


TRUSSES

A physical therapist has recommended the patient to work out a few

days a week to rebuilt his strength. He was told to use a free weight

training machine. He needs to lift 10 kg. The weight is supported by a

rope-and-pulley arrangement as shown in the figure. What is the tension

in the rope, if the tension in the rope is the same on each of a simple

pulley, when the man pulls on the bar?


SOLUTION

FBD:

T T + ↑ � Fy = 0: 2T – (10 kg) (9.81 m/s) = 0

T = (1/2) (98.1 N)

T = 49.05 N

10 kg T = 49.05 N ←


TRUSSES

SOLUTION

2ft 2ft

Tan θ = 1.5 / 2 A B D

= 0.013°

ΣFx = Fx – 250 cos 0.013°

Fx = 250 cos 0.013°

= 249 N C

ΣFy = F1 + Fy – 250 sin 0.013°

F + Fy = 0.056 N

ΣJ = - F1(4) + 250 (2) sin 0.013° B

4F1 = 500 sin 0.013°

F1 = 0.028 N

1 ft 5 in Fy 250 sin 0.013° – 0.028

Fy 248.9 N

F = - 249 i + 0.028 j←

5 ft in 5 in

A football player works out with a “squat thrust”

machine. To rotate the bar A B d, he must exert a

vertical force at A such that the axial force in member

BC is 250 lb. When bar A B D is on the verge of

rotating, what force does he exert at A and what forces

are exerted on member C D E at D?



FORCES IN BEAMS AND CABLES

A man as injured in the leg. He is getting physical therapy.

The leg is stretched and has a counter weight of 2 kg and is

at 12º angle. Determine the tension in the leg of the man?

SOLUTION

FBD

Find T:

Fy = 0 = -(2*9.8) + T

T = 19.6 N

T = 19.6 N ←

Find A:

Fx = 0 = Ax*cos 12o

Ax = 0

Ax = 0 N ←

Fy = 0 = 19.6 – Ay* sin 12 o

Ay = -94.27 N

Ay = -94.27 N ←


FORCES IN BEANS AND CABLES

Two climbers on an icy (frictionless) slope, tied together by a 30 m

rope, are in the predicament shown in the figure. At t = 0, the speed of

each is zero, but the top climber, Paul (mass 52 kg), has taken one step

too many and his friend Jay (mass 74 kg) has dropped his pick. (a)

Find the tension in the rope as Paul falls and his speed just before he

hits the ground. (b) If Paul unhooks his rope after hitting the ground,

find Jay’s speed as he hits the ground.

[Tipler 1999, 110]

SOLUTION

FBD:

a) �Fx = m1*a : T- m1*g*sin 40º = m1*a

�F = m2*a : m2*g –T = m2 *a

Add the two equations and solve for a:

a = g{( m2 – m1 sin 40 º)/ (m1 + m2)}

Use expression for a to find T:

T = g{ m1* m2 / (m1 + m2)(1 + sin 40º)}

a = (9.81) {(52 – 74 sin 40º) / 126} m/s² = 0.345 m/s²

T = (9.81) (52*74/126) (1 + sin 40º) = 492 N

v = √(2as) = √(2*0.345*20) m/s = 3.71 m/s

b) 1. Find the distance Jay slides

s1 = (25 m)/ sin 40º = 33.9 m

2. Find the acceleration of Jay.

a1 = g sin 40º = 6.31 m/s²

3. Find v1 = √(2 a1* s1).

v1 = √(2*33.986.31) = 20.7 m/s



A man cleaning his apartment pulls a vacuum cleaner with a force

of magnitude F = 50 N. The forces make an angle of 30.0 with the

horizontal as shown. The vacuum done by the 50.0 N force on the

vacuum cleaner.


SOLUTION

F B D

Using the definition of work, we have:

W = ( 50.0 N ) ( cos 30.0) ( 3.00m)

= 130 N.m

130 J ←



The lever ABC is pin-supported at A and connect to a short link BD as

shown. If the weight of the members is negligible, determine the tension

of the pin on the lever at A.

[R.C.Hebbler, 205]

SOLUTION C 400 N

θ 0.5 m ΣFx = 0; FA cos 60.3° - F cos 45° + 400 N

ΣFy = 0; FA sin 60.3°

0.2 m FA = 1075 N ←

F = 1320 N ←

0.2m


DRY FRICTION

SOLUTION

Find Friction:

WF = 0.8 * (125 * sin 8o)

= 13.9 #

or WF = 13.9 # ←

Find Work:

W = (.5*125*3^2) + (125*1) + 13.9

= 701.4 # - ft

or W = 701.4 # - ft ←

A boy is pushing his grandma on a wheelchair up

a ramp. Grandma weighs 110 lb. How much work

is done by the boy?


DRY FRICTION

Two children that were hurt in an avalanche accident are pulled on

a sled over snow-covered ground to be taken to the hospital. The

sled, which is initially at rest, is pulled by a rope that makes an

angle of 40¬ with the horizontal. The children have a combined

mass of 45 kg and the sled has a mass of 5 kg. The coefficients of

static and kinetic friction are ±s=0.2 and ±k=0.15. Find the

frictional force exerted by the ground on the sled and the

acceleration of the children and sled, starting from rest, if then

tension in the rope is (a) 100 N and (b) 140 N.

[Tipler 1999, 118]

SOLUTION

FBD:

T

Fn

40º

f

mg

a) 1. fs,max = µs * Fn

2. �Fx = Tx – f = max

�Fy = Fn + Ty – mg = may

3. Tx = T cos 40º = (100 N) (0.766) = 76.6 N

Ty = T sin 40º = (100 N) (0.643) = 64.3 N

4. Fn + Ty – mg = 0

Fn = mg – Ty = (50 kg) (9.81 m/s²) – 64.3 N = 426 N

5. fs,max = µs * Fn = 0.2 (426 N) = 85.2 N

6. Tx – f = max = 0

F = Tx = 76.6 N

b) 1. Tx = (140 N) (cos 40º) = 107 N

Ty = (140 N0 ( sin 40º) = 90 N

2. �Fy = Fn + Ty – mg = may

Fn + Ty – mg = may = 0

Fn = mg – Ty = 490 N – 90 N = 400 N

3. fs,max = µs * Fn = 0.2(400 N) = 80 N


(CONTINUED)

4. fk = µk * fn = 0.15 (400 N) = 60 N

5. �Fx = Tx – fk = max

ax = (Tx - fk) / m = (107 – 60) N/ 50 kg = 0.940 m/s²


DRY FRICTION

A child mass 60 pounds an external force of 20 pounds takes a ride on

slide of height h = 2.00 m . The child makes an angle of 80º with the

slide. Determine the friction of the child with a slide.

SOLUTION

∑Fy = o

N – 60 + 20 cos 80 =

N = 56.53

Friction = M * N

Friction = ( .2 ) * ( 56.53 )

Friction = 11.30


DRY FRICTION

A rock climber with mass m = 55 kg rests during a “chimney climb”,

pressing only with her shoulders and feet against the walls of a fissure of

width w = 1.0 m. her center of mass is a horizontal distance d = 0.20 m

from the wall against which her shoulders are pressed. The coefficient of

static friction between her shoes and the wall is µ1 = 1.1, and between

her shoulders and the wall it is µ2 = 0.70. The rest, the climber wants to

minimize her horizontal push on the walls. The minimum occurs when

her feet and her shoulders are both on the verge of sliding.


SOLUTION

Fnet, y = 0

F1 = f2 – mg = 0

F1 = µ1N and f2 = µ2N

N = mg / µ1 + µ2

= (55 kg) (9.8 m/s^2) / 1.1 = 0.70

= 300N

Fnet = 300 N←


MOMENT OF INERTIA

An Olympic competitor of ice skating is spinning in the

same point. A physics is watching the competition and is

intrigue to know the moment of inertia of the competitor.

Her mass is of 125 lb and the radio is 3.6 ft. What is the

competitors moment of inertia?

SOLUTION

The equation is

I = 2/5 * m * R^2

I = 2/5 * 125 *3.6 ^ 2

I= 648 lb – ft

I= 648 lb – ft ←


MOMENTS OF INERTIA

The small wheel of a wheelchair, 1m in diameter consists of a thin rim

having a mass of 8 kg and six spokes each having a mass of 1.2 kg.

Determine the moment of inertia of the small wheel for the rotation about its

axis.

[Tipler 1999, 286]

SOLUTION

I = MR² ; I= (2/3) MR²

Add both of them:

I = {(8 * 0.5²) + (6 * 1.2 * 0.5²/3)} kg m² = 2.6 kg m²

2.6 kg m² ←


MOMENT OF INERTIA

400 mm

100 mm

600

mm

400

mm

Compute the moments of inertia of the bond cross sectional area shown in figure. About

the x and y centroids axes.

[Seywar and Jewett 2000, 245]

SOLUTION

Composite Parts. The cross section can be considered as three composite rectangular areas A, B,

D shown in figure. For the calculation the centroids of each of these rectangles is located in the

figure.

Rectangle A

Ix = Ix + Ad^2y = 1/12 (100)(300)^3 + (100)(300)(200)^2

= 1.425(10^9)mm^4

or 1.425 �

Iy = Ad^2x= 1/129300)(100)^3 + (100)(300)(250)^2

= 1.90(10^9) mm ^4

Rectangle B

Ix = 1/12(600)(100)^3

= .05(10)^9mm^4

Iy = 1/12 (100)(600)^3

= 1.80(10^9)mm^4

Rectangle D

Ix = Ix + Ad^2y= 1/12 (100)(300)^3 + (100)(300)(200)^2

= 1.425(10^9)mm^4

Iy = Iy + Ad^2x= 1/12(300)(100)^3 + (100)(300)(250)^2

= 1.90(10^9)mm^4

A

B

D


(CONTINUED)

The moments of inertia for the entires cross section are thus

Ix = 1.425(10^9) + .05 (10^9) + 1.425(10^9)

= 2.90(10^9)mm^4

Iy = 1.90(10^9) + 1.80 (10^9) + 1.90(10^9)

= 5.60(10^9)mm^4


MOMENT OF INERTIA

To throw an 80 kg opponent with a basic judo hip throw, you intend to

pull his uniform with a force F and a moment arm d1 = 0.30 m from a

pivot point (rotation axis) on your right hip. You wish to rotate him

about the pivot point with an angular acceleration α of -6.0 rad / s^2-

that is, with an angular acceleration that is clockwise in the figure.

Assume that his rotational inertia I relative to the pivot point is 15 kg · m^2.


SOLUTION

-d1F = Iα

F = - Iα / d1

= -(15 kg ● m^2) (-6.0 rad / s^2) / 0.30 m

= 300 N

F = 300 N←

a 8 dec 2003 solution

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