Problem 3-26

If blocks D and F each have weight W1, determine the weight of block E if the sag is s. Neglectthe size of the pulleys.

Given:

W1 5N:=

s 0.3m:=

a 0.4m:=

Solution:

Sum forces in the y direction

2s

s2 a2+

⋅ W1 W− 0=

W 2s

s2 a2+

⋅ W1⋅:= W 6 N=

Problem 3-27

The block of mass M is supported by two springs having the stiffness shown. Determine theunstretched length of each spring.

Units Used:

kN 103N:=

Given:

M 30kg:=

l1 0.6m:=

l2 0.4m:=

l3 0.5m:=

kAC 1.5kNm

:=

kAB 1.2kNm

:=

g 9.81m

s2:=

Solution:

Initial guesses: FAC 20N:= FAB 30N:=

Given

+→ ΣFx = 0;l2 FAB⋅

l22 l3

2+

l1 FAC⋅

l12 l3

2+

− 0=

+↑ ΣFy = 0;l3 FAB⋅

l22 l3

2+

l3 FAC⋅

l12 l3

2+

+ M g⋅− 0=

FAC

FAB

⎛⎜⎜⎝

⎞⎟⎟⎠

Find FAC FAB,( ):=FAC

FAB

⎛⎜⎜⎝

⎞⎟⎟⎠

183.88

226.13⎛⎜⎝

⎞⎟⎠

N=

Then guess LAB 0.1m:= LAC 0.1m:=

Given FAC kAC l12 l3

2+ LAC−⎛

⎝⎞⎠⋅=

FAB kAB l22 l3

2+ LAB−⎛

⎝⎞⎠⋅=

LAB

LAC

⎛⎜⎜⎝

⎞⎟⎟⎠

Find LAB LAC,( ):=LAB

LAC

⎛⎜⎜⎝

⎞⎟⎟⎠

0.452

0.658⎛⎜⎝

⎞⎟⎠

m=

Problem 3-28

Three blocks are supported using the cords and two pulleys. If they have weights of WA = WC =W, WB = kW,, determine the angle θ for equilibrium.

Given:

k 0.25:=

Solution:

+→ ΣFx = 0; W cos φ( )⋅ k W⋅ cos θ( )⋅− 0=

+↑ ΣFy = 0; W sin φ( )⋅ k W⋅ sin θ( )⋅+ W− 0=

cos φ( ) k cos θ( )⋅=

sin φ( ) 1 k sin θ( )⋅−=

1 k2 cos θ( )2 1 k sin θ( )⋅−( )2+= 1 k2

+ 2 k⋅ sin θ( )⋅−=

θ asink2⎛⎜⎝⎞⎟⎠

:= θ 7.18 deg=

Problem 3-29

A continuous cable of total length l is wrapped around the small pulleys at A, B, C, and D. Ifeach spring is stretched a distance b, determine the mass M of each block. Neglect the weight ofthe pulleys and cords. The springs are unstretched when d = l/2.

Given:

l 4m:=

k 500Nm

:=

b 300mm:=

g 9.81m

s2:=

Solution:

Fs k b⋅:= Fs 150.00 N=

Guesses

T 1N:= θ 10deg:= M 1kg:=

Given

2 T⋅ sin θ( )⋅ Fs− 0=

2− T⋅ cos θ( )⋅ M g⋅+ 0=

bl4

sin θ( )⋅+l4

=

b

l/4l/4

T

θ

M

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

Find T θ, M,( ):=

T 107.14 N= θ 44.43 deg= M 15.60 kg=

Problem 3-30

Prove Lami's theorem, which states that if three concurrent forces are in equilibrium, each isproportional to the sine of the angle of the other two; that is, P/sin α = Q/sin β = R/sin γ.

Solution:

Sine law:

Rsin 180deg γ−( )

Qsin 180deg β−( )

=P

sin 180deg α−( )=

However, in general sin 180deg φ−( ) sin φ( )= , hence

Rsin γ( )

Qsin β( )

=P

sin α( )= Q.E.D.

Problem 3-31

A vertical force P is applied to the ends of cord AB of length a and spring AC. If the spring has anunstretched length δ, determine the angle θ for equilibrium.

Given:

P 10N:=

δ 0.2m:=

k 150Nm

:=

a 0.2m:=

b 0.2m:=

Guesses

θ 10deg:= φ 10deg:=

T 1N:= F 1N:=

x 0.1m:=

Given

T− cos θ( )⋅ F cos φ( )⋅+ 0=

T sin θ( )⋅ F sin φ( )⋅+ P− 0=

F k x δ−( )⋅=

a sin θ( )⋅ x sin φ( )⋅=

a cos θ( )⋅ x cos φ( )⋅+ a b+=

θ

φ

T

F

x

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

Find θ φ, T, F, x,( ):=T

F

⎛⎜⎝

⎞⎟⎠

10.296

9.377⎛⎜⎝

⎞⎟⎠

N= θ 35.0 deg=

Problem 3-32

Determine the unstretched length δ of spring AC if a force P causes the angle θ for equilibrium.Cord AB has length a.

Given:

P 80N:=

θ 60deg:=

k 500Nm

:=

a 0.2m:=

b 0.2m:=

Guesses

δ 0.1m:= φ 10deg:=

T 1N:= F 1N:=

x 0.1m:=

Given

T− cos θ( )⋅ F cos φ( )⋅+ 0=

T sin θ( )⋅ F sin φ( )⋅+ P− 0=

F k x δ−( )⋅=

a sin θ( )⋅ x sin φ( )⋅=

a cos θ( )⋅ x cos φ( )⋅+ a b+=

δ

φ

T

F

x

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

Find δ φ, T, F, x,( ):=T

F

⎛⎜⎝

⎞⎟⎠

69.282

40⎛⎜⎝

⎞⎟⎠

N= δ 0.266 m=

Problem 3-33

The flowerpot of mass M is suspended from three wires and supported by the hooks at B and C.Determine the tension in AB and AC for equilibrium.

Given:

M 20kg:=

l1 3.5m:=

l2 2m:=

l3 4m:=

l4 0.5m:=

g 9.81m

s2:=

Solution:

Initial guesses:

TAB 1N:= TAC 1N:=

θ 10deg:= φ 10deg:=

Given

TAC− cos φ( )⋅ TAB cos θ( )⋅+ 0=

TAC sin φ( )⋅ TAB sin θ( )⋅+ M g⋅− 0=

l1 cos φ( )⋅ l2 cos θ( )⋅+ l3=

l1 sin φ( )⋅ l2 sin θ( )⋅ l4+=

TAB

TAC

θ

φ

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

Find TAB TAC, θ, φ,( ):=θ

φ

⎛⎜⎝⎞⎟⎠

53.13

36.87⎛⎜⎝

⎞⎟⎠

deg=TAB

TAC

⎛⎜⎜⎝

⎞⎟⎟⎠

156.96

117.72⎛⎜⎝

⎞⎟⎠

N=

Problem 3-34

A car is to be towed using the rope arrangement shown. The towing force required is P. Determinethe minimum length l of rope AB so that the tension in either rope AB or AC does not exceed T.Hint: Use the equilibrium condition at point A to determine the required angle θ for attachment, thendetermine l using trigonometry applied to triangle ABC.

Given:

P 3000N:=

T 3750N:=

φ 30deg:=

d 1.2m:=

Solution:

The initial guesses

TAB T:=

TAC T:=

θ 30deg:=

l 0.6m:=

Case 1: Assume TAC T:=

Given+→ Σ Fx = 0; TAC cos φ( )⋅ TAB cos θ( )⋅− 0=

+↑ P TAC sin φ( )⋅− TAB sin θ( )⋅− 0=Σ Fy = 0;

lsin φ( )

dsin 180deg θ− φ−( )

=

TAB

θ

l1

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

Find TAB θ, l,( ):= TAB 3.437 kN= θ 19.11 deg= l1 0.794 m=

Case 2: Assume TAB T:=

Given

+→ Σ Fx = 0; TAC cos φ( )⋅ TAB cos θ( )⋅− 0=

+↑ P TAC sin φ( )⋅− TAB sin θ( )⋅− 0=Σ Fy = 0;

lsin φ( )

dsin 180deg θ− φ−( )

=

TAC

θ

l2

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

Find TAC θ, l,( ):= TAC 4.204 kN= θ 13.85 deg= l2 0.866 m=

l min l1 l2,( ):= l 0.794 m=

Problem 3-35

Determine the mass of each of the two cylinders if they cause a sag of distance d whensuspended from the rings at A and B. Note that s = 0 when the cylinders are removed.

Given:

d 0.5m:=

l1 1.5m:=

l2 2m:=

l3 1m:=

k 100Nm

:=

g 9.81m

s2:=

Solution:

TAC k l1 d+( )2 l22

+ l12 l2

2+−⎡

⎣⎤⎦⋅:=

TAC 32.84 N=

θ atanl1 d+

l2

⎛⎜⎝

⎞⎟⎠

:=

θ 45 deg=

MTAC sin θ( )⋅

g:=

M 2.37 kg=

Problem 3-36

The sling BAC is used to lift the load W with constant velocity. Determine the force in the sling andplot its value T (ordinate) as a function of its orientation θ , where 0 θ≤ 90°≤ .

Solution:

W 2 T⋅ cos θ( )⋅− 0=

T12

Wcos θ( )⋅:=

θ

Problem 3-37

The lamp fixture has weight W and is suspended from two springs, each having unstretchedlength L and stiffness k. Determine the angle θ for equilibrium.

Units Used:

kN 103N:=

Given:

W 45N:=

L 1.2m:=

k 75Nm

:=

a 1.2m:=

Solution :

The initial guesses: T 1N:= θ 10deg:=

Given

Spring T ka

cos θ( )L−⎛⎜

⎝⎞⎟⎠

⋅=

+↑Σ Fy = 0; 2T sin θ( )⋅ W− 0=

T

θ

⎛⎜⎝

⎞⎟⎠

Find T θ,( ):= T 33.0 N= θ 43.0 deg=

Problem 3-38

The uniform tank of weight W is suspended by means of a cable, of length l, which is attachedto the sides of the tank and passes over the small pulley located at O. If the cable can beattached at either points A and B, or C and D, determine which attachment produces the leastamount of tension in the cable.What is this tension?

Given:

W 1000N:=

l 1.8m:=

a 0.3m:=

b 0.6m:=

c b:=

d 2a:=

Solution:

Free Body Diagram : Byobservation, the force F has tosupport the entire weight of thetank. Thus, F = W. The tension incable is the same throughout thecable.

Equations of Equilibrium

ΣFy = 0; W 2 T⋅ sin θ( )⋅− 0=

Attached to CD θ1 acos2al

⎛⎜⎝

⎞⎟⎠

:= θ1 70.53 deg=

Attached to AB θ2 acos2 b⋅

l⎛⎜⎝

⎞⎟⎠

:= θ2 48.19 deg=

We choose the largest angle (which will produce the smallest force)

θ max θ1 θ2,( ):= θ 70.53 deg=

T12

Wsin θ( )⋅:= T 530.3 N=

Problem 3-39

A sphere of mass ms rests on the smooth parabolic surface. Determine the normal force it exertson the surface and the mass mB of block B needed to hold it in the equilibrium position shown.

Given:

ms 4kg:=

a 0.4m:=

b 0.4m:=

θ 60deg:=

g 9.81m

s2:=

Solution :

ka

b2:=

Geometry : The angle θ1 which the surface make with the horizontal is to be determined first.

tan θ1( ) dydx

= 2 k⋅ x⋅= evaluated at x a:= θ1 atan 2 k⋅ a⋅( ):= θ1 63.435 deg=

Free Body Diagram : The tension in the cord is the same throughout the cord and is equto the weight of block B, mBg.

The initial guesses: mB 200kg:= FN 200N:=

Given

+→ Σ Fx = 0; mB g⋅ cos θ( )⋅ FN sin θ1( )⋅− 0=

+↑Σ Fy = 0; mB g⋅ sin θ( )⋅ FN cos θ1( )⋅+ ms g⋅− 0=

mB

FN

⎛⎜⎜⎝

⎞⎟⎟⎠

Find mB FN,( ):= FN 19.655 N=

mB 3.584 kg=

Problem 3-40

The pipe of mass M is supported at A by a system of five cords. Determine the force in eachcord for equilibrium.

Given:

M 30kg:= c 3:=

d 4:=g 9.81m

s2:=

θ 60deg:=

Solution:

Initial guesses:

TAB 1N:= TAE 1N:=

TBC 1N:= TBD 1N:=

Given

TAB sin θ( )⋅ M g⋅− 0=

TAE TAB cos θ( )⋅− 0=

TBDc

c2 d2+

⋅ TAB sin θ( )⋅− 0=

TBDd

c2 d2+

⋅ TAB cos θ( )⋅+ TBC− 0=

TAB

TAE

TBC

TBD

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

Find TAB TAE, TBC, TBD,( ):=

TAB

TAE

TBC

TBD

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

339.8

169.9

562.3

490.5

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

N=

Problem 3-41

The joint of a space frame is subjected to four forces. Strut OA lies in the x-y plane and strut OBlies in the y-z plane. Determine the forces acting in each of the three struts required forequilibrium.Units Used:

kN 103N:=

Given:

F 2kN:=

θ1 45deg:=

θ2 40deg:=

Solution:

ΣFx = 0; R− sin θ1( )⋅ 0=

R 0:=

ΣFz = 0; P sin θ2( )⋅ F− 0=

PF

sin θ2( ):=

P 3.11 kN=

ΣFy = 0; Q P cos θ2( )⋅− 0=

Q P cos θ2( )⋅:=

Q 2.38 kN=

Problem 3-42

Determine the magnitudes of F1, F2, and F3 for equilibrium of the particle.

Units Used:

kN 103N:=

Given:

F4 800N:=

α 60deg:=

β 30deg:=

γ 30deg:=

c 3:=

d 4:=

Solution :

The initial guesses: F1 100N:= F2 100N:= F3 100N:=

Given

F1

cos α( )

0

sin α( )

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅F2

c2 d2+

c

d−

0

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅+ F3

cos γ( )−

sin γ( )−

0

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅+ F4

0

sin β( )

cos β( )−

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅+ 0=

F1

F2

F3

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

Find F1 F2, F3,( ):=

F1

F2

F3

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

800

147

564

⎛⎜⎜⎝

⎞⎟⎟⎠

N=

Problem 3-43

Determine the magnitudes of F1, F2, and F3 for equilibrium of the particle.

Units Used:

kN 1000N:=

Given:

F4 8.5kN:=

F5 2.8kN:=

α 15deg:=

β 30deg:=

c 7:=

d 24:=

Solution:

Initial Guesses: F1 1kN:= F2 1kN:= F3 1kN:=

Given

F1

cos β( )−

0

sin β( )

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅F2

c2 d2+

c−

d−

0

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅+ F3

1

0

0

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅+ F4

sin α( )−

cos α( )

0

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅+ F5

0

0

1−

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅+ 0=

F1

F2

F3

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

Find F1 F2, F3,( ):=

F1

F2

F3

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

5.60

8.55

9.44

⎛⎜⎜⎝

⎞⎟⎟⎠

kN=

Problem 3-44

Determine the magnitudes of F1, F2 and F3 for equilibrium of the particle F = {-9i - 8j -5k}.

Units Used:

kN 103N:=

Given:

F

9−

8−

5−

⎛⎜⎜⎝

⎞⎟⎟⎠

kN:=

a 4m:=

b 2m:=

c 4m:=

θ1 30deg:=

θ2 60deg:=

θ3 135deg:=

θ4 60deg:=

θ5 60deg:=

Solution:

Initial guesses: F1 8kN:= F2 3kN:= F3 12kN:=

Given

F1

cos θ2( ) cos θ1( )⋅

cos θ2( )− sin θ1( )⋅

sin θ2( )

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

⋅ F2

cos θ3( )cos θ5( )cos θ4( )

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

⋅+F3

a2 b2+ c2

+

a

c

b−

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

⋅+ F+ 0=

F1

F2

F3

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

Find F1 F2, F3,( ):=

F1

F2

F3

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

8.26

3.84

12.21

⎛⎜⎜⎝

⎞⎟⎟⎠

kN=

Problem 3.45

The three cables are used to support the lamp of weight W. Determine the force developed ineach cable for equilibrium.

Units Used:

kN 103N:=

Given:

W 800N:= b 4m:=

a 4m:= c 2m:=

Solution:

Initial Guesses:

FAB 1N:= FAC 1N:= FAD 1N:=

Given

FAB

0

1

0

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅ FAC

1

0

0

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅+FAD

a2 b2+ c2

+

c−

b−

a

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

⋅+ W

0

0

1−

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅+ 0=

FAB

FAC

FAD

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

Find FAB FAC, FAD,( ):=

FAB

FAC

FAD

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

800

400

1200

⎛⎜⎜⎝

⎞⎟⎟⎠

N=

Problem 3-46

Determine the force in each cable needed to support the load W.

Given:

a 0.8m:=

b 0.6m:=

c 0.2m:=

d 0.2m:=

e 0.6m:=

W 500N:=

Solution:

Initial guesses:

FCD 600N:= FCA 195N:= FCB 195N:=

Given

FCA

c2 e2+

c

e−

0

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅FCB

d2 e2+

d−

e−

0

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅+FCD

a2 b2+

0

b

a

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅+ W

0

0

1−

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅+ 0=

FCD

FCA

FCB

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

Find FCD FCA, FCB,( ):=

FCD

FCA

FCB

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

625.0

197.6

197.6

⎛⎜⎜⎝

⎞⎟⎟⎠

N=

Problem 3-47

Determine the stretch in each of the two springs required to hold the crate of mass mc in theequilibrium position shown. Each spring has an unstretched length δ and a stiffness k.

Given:

mc 20kg:=

δ 2m:=

k 300Nm

:=

a 4m:=

b 6m:=

c 12m:=

Solution:

Initial Guesses

FOA 1N:=

FOB 1N:=

FOC 1N:=

Given

FOA

0

1−

0

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅ FOB

1−

0

0

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅+FOC

a2 b2+ c2

+

b

a

c

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

⋅+ mc g⋅

0

0

1−

⎛⎜⎜⎝

⎞⎟⎟⎠

⋅+ 0=

FOA

FOB

FOC

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

Find FOA FOB, FOC,( ):=

FOA

FOB

FOC

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

65.378

98.066

228.822

⎛⎜⎜⎝

⎞⎟⎟⎠

N=

δOAFOA

k:= δOA 218 mm=

δOBFOB

k:= δOB 327 mm=

Problem 3-48

If the bucket and its contents have total weight W, determine the force in the supporting cables DA,DB, and DC.

Given:

W 200N:=

a 3m:=

b 4.5m:=

c 2.5m:=

d 3m:=

e 1.5m:=

f 1.5m:=

Solution:

The initial guesses:

FDA 10N:= FDB 20N:= FDC 30N:=

Given

Σ Fx = 0; b e−

b e−( )2 f 2+ a2

+

FDAe

e2 d2+ c f−( )2

+

FDC− 0=

Σ Fy = 0; f−

b e−( )2 f 2+ a2

+

FDAc f−

e2 d2+ c f−( )2

+

FDC+ FDB− 0=

Σ Fz= 0; a

b e−( )2 f 2+ a2

+

FDAd

e2 d2+ c f−( )2

+

FDC+ W− 0=

FDA

FDB

FDC

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

Find FDA FDB, FDC,( ):=

FDA

FDB

FDC

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

100.0

11.1

155.6

⎛⎜⎜⎝

⎞⎟⎟⎠

N=

Problem 3-49

The crate which of weight F is to be hoisted with constant velocity from the hold of a shipusing the cable arrangement shown. Determine the tension in each of the three cables forequilibrium.

Units Used:

kN 103N:=

Given:

F 2.5kN:=

a 3m:=

b 1m:=

c 0.75m:=

d 1m:=

e 1.5m:=

f 3m:=

Solution:

The initial guesses

FAD 3kN:= FAC 3kN:= FAB 3kN:=

Givenc−

c2 b2+ a2

+

FAD⋅d

d2 e2+ a2

+

FAC⋅+d

d2 f 2+ a2

+

FAB⋅+ 0=

b

c2 b2+ a2

+

FAD⋅e

d2 e2+ a2

+

FAC⋅+f−

d2 f 2+ a2

+

FAB⋅+ 0=

a−

c2 b2+ a2

+

FAD⋅a−

d2 e2+ a2

+

FAC⋅+a−

d2 f 2+ a2

+

FAB⋅+ F+ 0=

FAD

FAC

FAB

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

Find FAD FAC, FAB,( ):=

FAD

FAC

FAB

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

1.548

0.463

0.98

⎛⎜⎜⎝

⎞⎟⎟⎠

kN=

Problem 3-50

The lamp has mass ml and is supported by pole AO and cables AB and AC. If the force in thepole acts along its axis, determine the forces in AO, AB, and AC for equilibrium.

Given:

ml 15kg:= d 1.5m:=

a 6m:= e 4m:=

b 1.5m:= f 1.5m:=

c 2m:= g 9.81m

s2:=

Solution :

The initial guesses:

FAO 100N:=

FAB 200N:=

FAC 300N:=

Given

Equilibrium equations:

c

c2 b2+ a2

+

FAO⋅c e+

c e+( )2 b f+( )2+ a2

+

FAB⋅−c

c2 b d+( )2+ a2

+

FAC⋅− 0=

b

c2 b2+ a2

+

− FAO⋅b f+

c e+( )2 b f+( )2+ a2

+

FAB⋅+b d+

c2 b d+( )2+ a2

+

FAC⋅+ 0=

a

c2 b2+ a2

+

FAO⋅a

c e+( )2 b f+( )2+ a2

+

FAB⋅−a

c2 b d+( )2+ a2

+

FAC⋅− ml g⋅− 0=

FAO

FAB

FAC

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

Find FAO FAB, FAC,( ):=

FAO

FAB

FAC

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

319

110

86

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

N=