ch06f

Problem 6-101 If a force of magnitude P is applied perpendicular to the handle of the mechanism, determine the magnitude of force F for equilibrium. The members are pin-connected at A, B, C, and D. Given: P := 30N a := 625mm b := 100mm c := 125mm d := 100mm e := 125mm f := 125mm g := 750mm Solution: Σ MA = 0;

FBC⋅ b − P ⋅ a = 0 FBC :=

P⋅ a b

FBC = 187.5 N + Σ Fx = 0; →

−Ax + P = 0 Ax := P

+

↑Σ Fy = 0;

Ax = 30 N

−Ay + F BC = 0 Ay := FBC

Σ MD = 0;

Ay = 187.5 N

−e⋅ Ax − Ay⋅ ( b + c) + ( g + b + c) ⋅ F = 0

F :=

e⋅ Ax + Ay⋅ ( b + c) g+b+c

F = 47.1 N

Problem 6-102 The pillar crane is subjected to the load having a mass M. Determine the force developed in the tie rod AB and the horizontal and vertical reactions at the pin support C when the boom is tied in the position shown. Units Used: 3

kN := 10 N Given: M := 500kg a := 1.8m b := 2.4m θ 1 := 10deg θ 2 := 20deg g = 9.81

m 2

s Solution:

initial guesses: F CB := 10kN

F AB := 10kN

Given −M ⋅ g⋅ cos θ 1 − FAB⋅ cos θ 2 + FCB⋅ 2

( )

( )

−M ⋅ g⋅ sin θ 1 − FAB⋅ sin θ 2 + FCB⋅ 2

( )

( )

⎛⎜ FAB ⎞⎟ := Find ( FAB , FCB) ⎜ FCB ⎟ ⎝ ⎠

⎛⎜ Cx ⎞⎟ := ⎜ Cy ⎟ ⎝ ⎠

b 2

= 0

a +b

2

a 2

a +b

− M⋅ g = 0 2

⎛b ⎞ ⎟ 2 2 ⎝a ⎠ a +b F CB

⋅⎜

⎛ FAB ⎞ 9.7 ⎞ ⎜ ⎟ ⎛⎜ ⎟ ⎜ Cx ⎟ = ⎜ 11.53 ⎟ kN ⎜ ⎟ ⎝ 8.65 ⎠ ⎝ Cy ⎠

Problem 6-103 The tower truss has a weight W and a center of gravity at G. The rope system is used to hoist it into the vertical position. If rope CB is attached to the top of the shear leg AC and a second rope CD is attached to the truss, determine the required tension in BC to hold the truss in the position shown. The base of the truss and the shear leg bears against the stake at A, which can be considered as a pin. Also, compute the compressive force acting along the shear leg. Given: W := 2300N θ := 40deg a := 1.5m b := 0.9m c := 3m d := 1.2m e := 2.4m Solution: Entire system:

ΣMA = 0;

TBC⋅ cos ( θ ) ⋅ ( d + e) − TBC⋅ sin ( θ ) ⋅ a − W⋅ ( a + b) = 0 W⋅ ( a + b) TBC := cos ( θ ) ( d + e) − sin ( θ ) ⋅ a TBC = 3.08 kN CA is a two-force member.

At C:

⎛ e ⎞ ⎟ ⎝ b + c⎠

φ := atan ⎜ initial guesses:

F CA := 500N

TCD := 300N

Given

ΣFx = 0;

ΣFy = 0;

F CA⋅

F CA⋅

a 2

a + ( d + e)

2

d+e 2

a + ( d + e)

⎛⎜ FCA ⎞⎟ := Find ( F CA , TCD) ⎜ TCD ⎟ ⎝ ⎠

2

+ TCD⋅ cos ( φ ) − TBC⋅ cos ( θ ) = 0

− TCD⋅ sin ( φ ) − TBC⋅ sin ( θ ) = 0

TCD = 1.43 kN

F CA = 2.96 kN

Problem 6-104 The constant moment M is applied to the crank shaft. Determine the compressive force P that is exerted on the piston for equilibrium as a function of θ. Plot the results of P (ordinate) versus θ (abscissa) for 0deg ≤ θ ≤ 90deg . Given: a := 0.2m b := 0.45m M := 50N⋅ m Solution:

⎛a ⎞ φ = asin ⎜ ⋅ cos ( θ ) ⎟ b ⎝ ⎠

a⋅ cos ( θ ) = b⋅ sin ( φ )

−M + FBC⋅ cos ( θ − φ ) ⋅ a = 0 F BC =

M a⋅ cos ( θ − φ )

P = FBC⋅ cos ( φ ) =

M⋅ cos ( φ ) a⋅ cos ( θ − φ )

This function goes to infinity at θ = 90 deg, so we will only plot it to θ = 80 deg.

θ := 0 , 0.1 .. 80

⎛a ⎞ φ ( θ ) := asin ⎜ ⋅ cos ( θ ⋅ deg) ⎟ b ⎝ ⎠

P ( θ ) :=

M⋅ cos ( φ ( θ ) ) a⋅ cos ( θ ⋅ deg − φ ( θ ) )

Newtons

1500

P (θ )

1000 500 0

0

20

40 θ

Degrees

60

80

Problem 6-105 Five coins are stacked in the smooth plastic container shown. If each coin has weight W, determine the normal reactions of the bottom coin on the container at points A and B. Given: W := 0.094N a := 3 b := 4 Solution: All coins : ΣFy = 0;

NB := 5W NB = 0.470 N

Bottom coin : ΣFy = 0;

b

NB − W − N0⋅

a +b

(

)

N0 := NB − W ⋅

2

NA := N0⋅

b

a 2

a +b

NA = 0.282 N

2

a +b

N0 = 0.47 N

ΣFx = 0;

= 0

2

2

2

Problem 6-106 Determine the horizontal and vertical components of force at pin B and the normal force the pin at C exerts on the smooth slot. Also, determine the moment and horizontal and vertical reactions of force at A. There is a pulley at E. Given: F := 50kN a := 4m b := 3m Solution: Guesses Bx := 1kN By := 1kN NC := 1kN Ax := 1kN

Ay := 1kN

MA := 1kN⋅ m

Given Bx +

By −

a 2

a +b

2

b 2

a +b

2

⋅ NC − F = 0

⋅ NC − F = 0

(F + Bx)⋅ a − (F + By)⋅ b = 0 a

F−

2

a +b b 2

a +b −F⋅ 2a +

2

2

⋅ NC − Ax = 0

⋅ NC − Ay = 0

a 2

a +b

2

⋅ NC⋅ a + MA = 0

⎛ Bx ⎞ ⎜ ⎟ ⎜ By ⎟ ⎜ ⎟ ⎜ NC ⎟ ⎜ ⎟ := Find ( Bx , By , NC , Ax , Ay , MA) A ⎜ x⎟ ⎜A ⎟ ⎜ y⎟ ⎜ MA ⎟ ⎝ ⎠

NC = 20 kN

⎛⎜ Bx ⎞⎟ ⎛ 34 ⎞ = kN ⎜ By ⎟ ⎜⎝ 62 ⎟⎠ ⎝ ⎠ ⎛⎜ Ax ⎞⎟ ⎛ 34 ⎞ = kN ⎜ Ay ⎟ ⎜⎝ 12 ⎟⎠ ⎝ ⎠ MA = 336 kN⋅ m

Problem 6-107 A force F is applied to the handles of the vise grip. Determine the compressive force developed on the smooth bolt shank A at the jaws. Given: F := 20N

b := 25mm

a := 37.5mm

c := 75mm

d := 18.5mm

e := 25mm

θ := 20deg Solution:

From FBD (a) ΣME = 0;

⎤⋅ b = 0 ⎥ 2 2 ⎣ c + ( d + e) ⎦ ⎡

d+e

F ⋅ ( b + c) − F CD⋅ ⎢

2

c + ( d + e)

F CD := F ⋅ ( b + c) ⋅ ΣFx = 0;

b⋅ ( d + e)

2

F CD = 159.5 N

c

Ex := FCD⋅

2

c + ( d + e)

2

Ex = 137.9 N

From FBD (b) ΣMB = 0;

NA⋅ sin ( θ ) ⋅ d + NA⋅ cos ( θ ) ⋅ a − Ex⋅ ( d + e) = 0 NA := Ex⋅

d+e sin ( θ ) ⋅ d + cos ( θ ) ⋅ a

NA = 144.3 N

Problem 6-108 If a force of magnitude P is applied to the grip of the clamp, determine the compressive force F that the wood block exerts on the clamp. Given: P := 40N a := 50mm b := 50mm c := 12mm d := 18mm e := 38mm

Solution: Define

⎛ b⎞ ⎟ ⎝ d⎠

φ := atan ⎜

φ = 70.20 deg

From FBD (a), ΣMB = 0;

F CD⋅ cos ( φ ) ⋅ c − P⋅ ( a + b + c) = 0 F CD :=

+

↑Σ Fy = 0;

P ⋅ ( a + b + c) cos ( φ ) ⋅ ( c)

F CD = 1102.2 N

F CD⋅ sin ( φ ) − By = 0 By := FCD⋅ sin ( φ )

By = 1037.0 N

From FBD (b), ΣMA = 0;

By⋅ d − F⋅ e = 0 F :=

By⋅ d e

F = 491.2 N

Problem 6-109 The hoist supports the engine of mass M. Determine the force in member DB and in the hydraulic cylinder H of member FB. Units Used: 3

kN := 10 N Given: M := 125kg

d := 1m

a := 1m

e := 1m

b := 2m

f := 2m

c := 2m

g = 9.81

m 2

s Solution:

Member GFE : ΣME = 0; −F FB⋅

c+d

b + M⋅ g ⋅ ( a + b) = 0

2

( c + d) + ( b − e)

FFB := M⋅ g⋅

a+b b⋅ ( c + d)

2 2

⋅ ( c + d) + ( b − e)

2

FFB = 1.94 kN

ΣFx = 0;

Ex − FFB⋅

Ex := F FB⋅

b−e

= 0

2

( c + d) + ( b − e)

2

b−e 2

( c + d) + ( b − e)

2

Member EDC: ΣΜ c = 0;

Ex⋅ ( c + d) − F DB⋅

e 2

e +d

⋅d = 0 2

c+d 2 2 FDB := Ex⋅ ⋅ e +d e⋅ d

FDB = 2.6 kN

Problem 6-110 The flat-bed trailer has weight W1 and center of gravity at GT .It is pin-connected to the cab at D. The cab has a weight W2 and center of gravity at GC Determine the range of values x for the position of the load L of weight W3 so that no axle is subjected to a force greater than F Max . The load has a center of gravity at GL. Given: W1 := 35kN

a := 1.2m

W2 := 30kN

b := 1.8m

W3 := 10kN

c := 0.9m

Fmax := 27.5kN d := 3m e := 3.6m Solution: Ay := Fmax

Case 1:

Assume

Guesses

Ay := Fmax x := 1m

Given

By := F max

Cy := Fmax

Dy := Fmax

Ay + By − W2 − Dy = 0 −W2⋅ a − Dy⋅ ( a + b) + By⋅ ( a + b + c) = 0 Dy − W1 − W3 + Cy = 0 W3⋅ x + W1⋅ e − Dy⋅ ( c + d + e) = 0

⎛ By ⎞ ⎜ ⎟ ⎜ Cy ⎟ ⎜ ⎟ := Find ( By , Cy , Dy , x) ⎜ Dy ⎟ ⎜ ⎟ ⎝x ⎠ ⎛ Ay ⎞ 27.5 ⎞ ⎜ ⎟ ⎛⎜ ⎟ ⎜ By ⎟ = ⎜ 31.7 ⎟ kN ⎜ ⎟ ⎝ 15.8 ⎠ ⎝ Cy ⎠

x1 := x

x1 = 9.28 m Since B y > Fmax then this solution is no good.

By := F max

Case 2:

Assume

Guesses

Ay := Fmax

By := F max

Cy := Fmax

x := 1m Given

Dy := Fmax

Ay + By − W2 − Dy = 0 −W2⋅ a − Dy⋅ ( a + b) + By⋅ ( a + b + c) = 0 Dy − W1 − W3 + Cy = 0 W3⋅ x + W1⋅ e − Dy⋅ ( c + d + e) = 0

⎛ Ay ⎞ ⎜ ⎟ ⎜ Cy ⎟ ⎜ ⎟ := Find ( Ay , Cy , Dy , x) ⎜ Dy ⎟ ⎜ ⎟ ⎝x ⎠

⎛ Ay ⎞ 26.2 ⎞ ⎜ ⎟ ⎛⎜ ⎟ ⎜ By ⎟ = ⎜ 27.5 ⎟ kN ⎜ ⎟ ⎝ 21.3 ⎠ ⎝ Cy ⎠

x2 := x

x2 = 5.21 m

Since Ay < Fmax and Cy < Fmax then this solution is good. Cy := Fmax

Case 3:

Assume

Guesses

Ay := Fmax x := 1m

Given

By := F max

Cy := Fmax

Dy := Fmax

Ay + By − W2 − Dy = 0 −W2⋅ a − Dy⋅ ( a + b) + By⋅ ( a + b + c) = 0 Dy − W1 − W3 + Cy = 0 W3⋅ x + W1⋅ e − Dy⋅ ( c + d + e) = 0

⎛ Ay ⎞ ⎜ ⎟ ⎜ By ⎟ ⎜ ⎟ := Find ( Ay , By , Dy , x) ⎜ Dy ⎟ ⎜ ⎟ ⎝x ⎠

⎛ Ay ⎞ 24.8 ⎞ ⎜ ⎟ ⎛⎜ ⎟ ⎜ By ⎟ = ⎜ 22.7 ⎟ kN ⎜ ⎟ ⎝ 27.5 ⎠ ⎝ Cy ⎠

Since Ay < Fmax and By < Fmax then this solution is good. We conclude that x3 = 0.52 m < x < x2 = 5.21 m

x3 := x

x3 = 0.52 m

Problem 6-111 Determine the force created in the hydraulic cylinders EF and AD in order to hold the shovel in equilibrium. The shovel load has a mass W and a center of gravity at G. All joints are pin connected. Units Used: 3

Mg := 10 kg 3

kN := 10 N Given: a := 0.25m

θ 1 := 30deg

b := 0.25m

θ 2 := 10deg

c := 1.5m

θ 3 := 60deg

d := 2m

W := 1.25Mg

e := 0.5m Solution: Assembly FHG :

(

( )) = 0

ΣMH = 0; −[ W⋅ g⋅ ( e) ] + FEF⋅ c⋅ sin θ 1 F EF := W⋅ g⋅

e

( )

c⋅ sin θ 1

F EF = 8.17 kN (T)

Assembly CEFHG :

(

)

( )

ΣMC = 0; F AD⋅ cos θ 1 + θ 2 ⋅ b − W⋅ g⋅ ⎡( a + b + c)cos θ 2 + e⎤ = 0 ⎣ ⎦ F AD := W⋅ g⋅

( )

( )

( )

cos θ 2 ⋅ a + cos θ 2 ⋅ b + cos θ 2 ⋅ c + e

F AD = 158 kN (C)

(

)

cos θ 1 + θ 2 ⋅ b

Problem 6-112 The aircraft-hangar door opens and closes slowly by means of a motor which draws in the cable AB. If the door is made in two sections (bifold) and each section has a uniform weight W and length L, determine the force in the cable as a function of the door's position θ. The sections are pin-connected at C and D and the bottom is attached to a roller that travels along the vertical track.

Solution: L

⎛ θ ⎞ − 2L⋅ sin ⎛ θ ⎞ N = ⎟ ⎜ ⎟ A ⎝2⎠ ⎝2⎠

cos ⎜

ΣMD = 0;

2W

ΣΜ C = 0;

T⋅ L⋅ cos ⎜

2

0

W ⎛θ⎞ NA = cot ⎜ ⎟ 2 ⎝2⎠

⎛ θ ⎞ − N ⋅ L⋅ sin ⎛ θ ⎞ − W⋅ L ⋅ cos ⎛ θ ⎞ = ⎟ ⎜ ⎟ ⎜ ⎟ A 2 ⎝2⎠ ⎝2⎠ ⎝2⎠

0

T= W

Problem 6-113 A man having weight W attempts to lift himself using one of the two methods shown. Determine the total force he must exert on bar AB in each case and the normal reaction he exerts on the platform at C. Neglect the weight of the platform. Given: W := 750N

Solution:

(a) Bar: +

↑Σ Fy = 0;

⎛ F ⎞ − 2⎛ W ⎞ = 0 ⎟ ⎜ ⎟ ⎝ 2⎠ ⎝ 2 ⎠

2⋅ ⎜

F := W

F = 750 N

Man: +

↑Σ Fy = 0;

⎛ F⎞ NC − W − 2⎜ ⎟ = 0 ⎝ 2⎠ NC := W + F

NC = 1500 N

( b) Bar: +

↑Σ Fy = 0;

⎛ W ⎞ − 2⋅ F = 0 ⎟ 2 ⎝4⎠

2⋅ ⎜

F := Man: +

↑Σ Fy = 0;

W

F = 375 N

2

⎛F⎞ NC − W + 2⋅ ⎜ ⎟ = 0 2 ⎝ ⎠

NC := W − F

NC = 375 N

Problem 6-114 A man having weight W1 attempts to lift himself using one of the two methods shown. Determine the total force he must exert on bar AB in each case and the normal reaction he exerts on the platform at C. The platform has weight W2.

Given: W1 := 750N

W2 := 150N

Solution: (a) Bar: +

↑Σ Fy = 0;

2⋅

F

2

(

)

− W1 + W2 = 0

F := W1 + W2 F = 900 N

Man: +

↑

Σ Fy = 0;

F

NC − W1 − 2

2

=0

NC := F + W1 NC = 1650 N

( b)

Bar: +

↑

Σ Fy = 0;

−2 ⋅ ⎛⎜

F⎞

⎟+2 ⎝2⎠

W1 + W2

4

=0

F :=

W1 + W2

2

F = 450 N

Man: +

↑Σ Fy = 0;

NC − W1 + 2

NC := W1 − F NC = 300 N

F

2

=0

Problem 6-115 The piston C moves vertically between the two smooth walls. If the spring has stiffness k and is unstretched when θ = 0, determine the couple M that must be applied to AB to hold the mechanism in equilibrium. Given: k := 15

N cm

θ := 30deg a := 8cm b := 12cm

Solution: Geometry: b sin ( ψ) = a sin ( θ )

⎛ ⎝

ψ := asin ⎜ sin ( θ ) ⋅

a⎞

⎟

b⎠

φ := 180deg − ψ − θ rAC sin ( φ )

=

b sin ( θ )

ψ = 19.47 deg φ = 130.53 deg sin ( φ ) rAC := b⋅ sin ( θ )

rAC = 18.242 cm

Free Body Diagram: The solution for this problem will be simplified if one realizes that member CB is a two force member. Since the spring stretches x := ( a + b) − rAC

x = 1.758 cm Fsp := k⋅ x

the spring force is

Fsp = 26.37 N

Equations of Equilibrium: Using the method of joints +

↑Σ Fy = 0;

FCB⋅ cos ( ψ) − F sp = 0

FCB :=

F sp cos ( ψ)

FCB = 27.97 N

From FBD of bar AB +ΣMA = 0;

FCB⋅ sin ( φ ) ⋅ a − M = 0

M := FCB⋅ sin ( φ ) ⋅ a

M = 1.70 N⋅ m

Problem 6-116 The compound shears are used to cut metal parts. Determine the vertical cutting force exerted on the rod R if a force F is applied at the grip G.The lobe CDE is in smooth contact with the head of the shear blade at E. Given: F := 100N

e := 0.15m

a := 0.42m

f := 0.15m

b := 0.06m

g := 0.15m

c := 0.6m

h := 0.75m

d := 0.225m

θ := 60deg

Solution: Member AG : ΣMA = 0;

F⋅ ( a + b) − F BC⋅ b⋅ sin ( θ ) = 0 F BC := F⋅

a+b b⋅ sin ( θ )

F BC = 923.8 N

Lobe : ΣMD = 0;

F BC⋅ f − NE⋅ e = 0 NE := F BC⋅

f e

NE = 923.8 N

Head : ΣMF = 0;

−NE⋅ ( h + g) + h⋅ NR = 0 NR := NE⋅

h+g h

NR = 1.109 kN

Problem 6-117 The handle of the sector press is fixed to gear G, which in turn is in mesh with the sector gear C. Note that AB is pinned at its ends to gear C and the underside of the table EF, which is allowed to move vertically due to the smooth guides at E and F. If the gears exert tangential forces between them, determine the compressive force developed on the cylinder S when a vertical force F is applied to the handle of the press. Given: F := 40N a := 0.5m b := 0.2m c := 1.2m d := 0.35m e := 0.65m Solution: Member GD : ΣMG = 0;

−F ⋅ a + FCG⋅ b = 0 F CG := F⋅

a

F CG = 100 N

b

Sector gear : ΣMH = 0;

F CG⋅ ( d + e) − FAB⋅

c 2

c +d 2

F AB := FCG⋅ ( d + e) ⋅

⋅d = 0 2

c +d c⋅ d

2

F AB = 297.62 N

Table: ΣFy = 0;

F AB⋅

c 2

c +d

F s := FAB⋅

2

− Fs = 0

c 2

c +d

2

F s = 286 N

Problem 6-118 The mechanism is used to hide kitchen appliances under a cabinet by allowing the shelf to rotate downward. If the mixer has weight W, is centered on the shelf, and has a mass center at G, determine the stretch in the spring necessary to hold the shelf in the equilibrium position shown. There is a similar mechanism on each side of the shelf, so that each mechanism supports half of the load W. The springs each have stiffness k. Given: W := 50N k := 1

a := 50mm

N

b := 100mm

mm

φ := 30deg

c := 375mm

θ := 30deg

d := 150mm

Solution: W

ΣMF = 0;

FED :=

2

⋅ b − a⋅ F ED⋅ cos ( φ ) = 0

W⋅ b 2⋅ a⋅ cos ( φ )

+ Σ Fx = 0; →

−Fx + FED⋅ cos ( φ ) = 0

Fx := F ED⋅ cos ( φ ) +

↑

Σ Fy = 0;

Fy :=

W 2

FED = 57.74 N

−W 2

Fx = 50.00 N + Fy − FED⋅ sin ( φ ) = 0

+ F ED⋅ sin ( φ )

Fy = 53.87 N

Member FBA: ΣMA = 0;

Fs = ks;

Fy⋅ ( c + d) ⋅ cos ( φ ) − Fx⋅ ( c + d) ⋅ sin ( φ ) − Fs⋅ sin ( θ + φ ) ⋅ d = 0 F y⋅ ( c + d) ⋅ cos ( φ ) − F x⋅ ( c + d) ⋅ sin ( φ ) Fs := Fs = 87.50 N d⋅ sin ( θ + φ ) F s = k⋅ x

x :=

Fs k

x = 87.50 mm

Problem 6-119 If each of the three links of the mechanism has a weight W, determine the angle θ for equilibrium.The spring, which always remains horizontal, is unstretched when θ = 0°. Given: W := 25kN k := 60

kN m

a := 4m b := 4m Solution: Guesses

θ := 30deg

Bx := 10kN By := 10kN Cx := 10kN Cy := 10kN Given a −W⋅ ⋅ sin ( θ ) − Cy⋅ a⋅ sin ( θ ) + Cx⋅ a⋅ cos ( θ ) = 0 2 b −W⋅ + Cy⋅ b = 0 2 By + Cy − W = 0 −Bx + Cx = 0 a a a −Bx⋅ a⋅ cos ( θ ) − By⋅ a⋅ sin ( θ ) − W⋅ ⋅ sin ( θ ) + k⋅ ⋅ sin ( θ ) ⋅ ⋅ cos ( θ ) = 0 2 2 2

⎛⎜ Bx ⎞⎟ ⎜ By ⎟ ⎜ ⎟ ⎜ Cx ⎟ := Find ( Bx , By , Cx , Cy , θ ) ⎜ ⎟ ⎜ Cy ⎟ ⎜θ ⎟ ⎝ ⎠ W + 2⋅ Cy 1 Cx = ⋅ sin ( θ ) ⋅ 2 cos ( θ )

⎛a ⎞ T = k⋅ ⎜ ⋅ sin ( θ ) ⎟ ⎝2 ⎠

⎛ Bx ⎞ ⎜ ⎟ ⎛⎜ 16.58 ⎞⎟ ⎜ By ⎟ ⎜ 12.5 ⎟ kN ⎜ ⎟=⎜ ⎟ 16.58 C ⎜ x⎟ ⎜ ⎟ ⎜ C ⎟ ⎝ 12.5 ⎠ ⎝ y⎠

⎛θ ⎞ ⎜ ⎟ ⎜ T ⎟ := Find ( θ , T , Cx) ⎜C ⎟ ⎝ x⎠

θ = 33.6 deg

Problem 6-120 Determine the required force P that must be applied at the blade of the pruning shears so that the blade exerts a normal force F on the twig at E. Given: F := 100N a := 12.5mm b := 100mm c := 18.75mm d := 18.75mm e := 25mm Solution: initial guesses: Ax := 1N

Ay := 1N

Dx := 10N

Dy := 10N

P := 20N

F CB := 20N

Given −P ⋅ ( b + c + d) − Ax⋅ a + F⋅ e = 0 Dy − P − Ay − F = 0 Dx − Ax = 0 −Ay⋅ d − Ax⋅ a + ( b + c) ⋅ P = 0 Ax − FCB⋅

c

= 0

2

2

Ay + P − F CB⋅

a

c +a

2

= 0 2

a +c

⎛ Ax ⎞ ⎜ ⎟ A ⎜ y ⎟ ⎜ ⎟ ⎜ Dx ⎟ := Find A , A , D , D , P , F ( x y x y CB) ⎜D ⎟ y ⎜ ⎟ ⎜ P ⎟ ⎜ ⎟ ⎝ FCB ⎠

⎛ Ax ⎞ ⎜ ⎟ ⎛ 66.67 ⎞ ⎜ Ay ⎟ ⎜ 32.32 ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ Dx ⎟ = ⎜ 66.67 ⎟ N ⎜ ⎟ ⎜ 144.44 ⎟ ⎜ Dy ⎟ ⎜ ⎟ ⎜ F ⎟ ⎝ 80.12 ⎠ ⎝ CB ⎠

P = 12.12 N