Problem 6-101

If a force of magnitude P is applied perpendicular to the handle of the mechanism, determinethe 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=

FBCP a⋅b

:=

FBC 187.5 N=

+→ Ax− P+ 0=Σ Fx = 0;

Ax P:= Ax 30 N=

+↑Ay− FBC+ 0=Σ Fy = 0;

Ay FBC:= Ay 187.5 N=

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

Fe 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 thetie rod AB and the horizontal and vertical reactions at the pin support C when the boom is tiedin the position shown.

Units Used:

kN 103N:=

Given:

M 500kg:=

a 1.8m:=

b 2.4m:=

θ1 10deg:=

θ2 20deg:=

g 9.81m

s2=

Solution:

initial guesses: FCB 10kN:= FAB 10kN:=

Given

M−2

g⋅ cos θ1( )⋅ FAB cos θ2( )⋅− FCBb

a2 b2+

⋅+ 0=

M−2

g⋅ sin θ1( )⋅ FAB sin θ2( )⋅− FCBa

a2 b2+

⋅+ M g⋅− 0=

FAB

FCB

⎛⎜⎜⎝

⎞⎟⎟⎠

Find FAB FCB,( ):=Cx

Cy

⎛⎜⎜⎝

⎞⎟⎟⎠

FCB

a2 b2+

b

a

⎛⎜⎝⎞⎟⎠

⋅:=

FAB

Cx

Cy

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

9.7

11.53

8.65

⎛⎜⎜⎝

⎞⎟⎟⎠

kN=

Problem 6-103

The tower truss has a weight W and a center of gravity at G. The rope system is used tohoist it into the vertical position. If rope CB is attached to the top of the shear legAC and a second rope CD is attached to the truss, determine the required tension in BC tohold the truss in the position shown. The base of the truss and the shear leg bears against thestake at A, which can be considered as a pin. Also, compute the compressive force actingalong 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=

TBCW a b+( )⋅

cos θ( ) d e+( ) sin θ( ) a⋅−:=

TBC 3.08 kN=

CA is a two-force member. At C:

φ atane

b c+⎛⎜⎝

⎞⎟⎠

:=

initial guesses: FCA 500N:= TCD 300N:=

Given

ΣFx = 0; FCAa

a2 d e+( )2+

⋅ TCD cos φ( )⋅+ TBC cos θ( )⋅− 0=

ΣFy = 0; FCAd e+

a2 d e+( )2+

⋅ TCD sin φ( )⋅− TBC sin θ( )⋅− 0=

FCA

TCD

⎛⎜⎜⎝

⎞⎟⎟⎠

Find FCA TCD,( ):= TCD 1.43 kN= FCA 2.96 kN=

Problem 6-104

The constant moment M is applied to the crank shaft. Determine the compressive force P that isexerted 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 cos θ( )⋅ b sin φ( )⋅= φ asinab

cos θ( )⋅⎛⎜⎝

⎞⎟⎠

=

M− FBC cos θ φ−( )⋅ a⋅+ 0=

FBCM

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..:=

φ θ( ) asinab

cos θ deg⋅( )⋅⎛⎜⎝

⎞⎟⎠

:= P θ( )M cos φ θ( )( )⋅

a cos θ deg⋅ φ θ( )−( )⋅:=

0 20 40 60 800

500

1000

1500

Degrees

New

tons

P θ( )

θ

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; NB W− N0b

a2 b2+

⋅− 0=

N0 NB W−( ) a2 b2+

b⋅:=

N0 0.47 N=

ΣFx = 0; NA N0a

a2 b2+

⋅:=

NA 0.282 N=

Problem 6-106

Determine the horizontal and vertical components of force at pin B and the normal force the pin at Cexerts on the smooth slot. Also, determine the moment and horizontal and vertical reactions of forceat 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

Bxa

a2 b2+

NC⋅+ F− 0=

Byb

a2 b2+

NC⋅− F− 0=

F Bx+( ) a⋅ F By+( ) b⋅− 0=

Fa

a2 b2+

NC⋅− Ax− 0=

b

a2 b2+

NC⋅ Ay− 0=

F− 2⋅ aa

a2 b2+

NC⋅ a⋅+ MA+ 0=

Bx

By

NC

Ax

Ay

MA

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎟⎠

Find Bx By, NC, Ax, Ay, MA,( ):= NC 20 kN=

Bx

By

⎛⎜⎜⎝

⎞⎟⎟⎠

34

62⎛⎜⎝

⎞⎟⎠

kN=

Ax

Ay

⎛⎜⎜⎝

⎞⎟⎟⎠

34

12⎛⎜⎝

⎞⎟⎠

kN=

MA 336 kN m⋅=

Problem 6-107

A force F is applied to the handles of the vise grip. Determine the compressive force developedon 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; F b c+( )⋅ FCDd e+

c2 d e+( )2+

⎡⎢⎣

⎤⎥⎦

⋅ b⋅− 0=

FCD F b c+( )⋅c2 d e+( )2

+

b d e+( )⋅⋅:= FCD 159.5 N=

ΣFx = 0; Ex FCDc

c2 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 Exd 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 Fthat the wood block exerts on the clamp.

Given:

P 40N:=

a 50mm:=

b 50mm:=

c 12mm:=

d 18mm:=

e 38mm:=

Solution:

Define φ atanbd⎛⎜⎝⎞⎟⎠

:= φ 70.20 deg=

From FBD (a),

ΣMB = 0; FCD cos φ( )⋅ c⋅ P a b+ c+( )⋅− 0=

FCDP a b+ c+( )⋅

cos φ( ) c( )⋅:= FCD 1102.2 N=

+↑Σ Fy = 0; FCD sin φ( )⋅ By− 0=

By FCD sin φ( )⋅:= By 1037.0 N=

From FBD (b),

ΣMA = 0; By d⋅ F e⋅− 0=

FBy 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 thehydraulic cylinder H of member FB.

Units Used:

kN 103N:=

Given:

M 125kg:= d 1m:=

a 1m:= e 1m:=

b 2m:= f 2m:=

c 2m:= g 9.81m

s2=

Solution:

Member GFE :

ΣME = 0; FFB−c d+

c d+( )2 b e−( )2+

⋅ b M g⋅ a b+( )⋅+ 0=

FFB M g⋅a b+

b c d+( )⋅⋅ c d+( )2 b e−( )2

+⋅:=

FFB 1.94 kN=

ΣFx = 0; Ex FFBb e−

c d+( )2 b e−( )2+

⋅− 0=

Ex FFBb e−

c d+( )2 b e−( )2+

⋅:=

Member EDC:

ΣΜc = 0; Ex c d+( )⋅ FDBe

e2 d2+

⋅ d⋅− 0=

FDB Exc d+

e d⋅⋅ e2 d2

+⋅:= 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 theposition of the load L of weight W3 so that no axle is subjected to a force greater than FMax. Theload 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:

Case 1: Assume Ay Fmax:=

Guesses Ay Fmax:= By Fmax:= Cy Fmax:=

x 1m:= Dy Fmax:=

Given 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

Dy

x

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

Find By Cy, Dy, x,( ):=

Ay

By

Cy

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

27.5

31.7

15.8

⎛⎜⎜⎝

⎞⎟⎟⎠

kN= x1 x:= x1 9.28 m=

Since By > Fmax then this solution is no good.

Case 2: Assume By Fmax:=

Guesses Ay Fmax:= By Fmax:= Cy Fmax:=

x 1m:= Dy Fmax:=

Given 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

Dy

x

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

Find Ay Cy, Dy, x,( ):=

Ay

By

Cy

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

26.2

27.5

21.3

⎛⎜⎜⎝

⎞⎟⎟⎠

kN= x2 x:= x2 5.21 m=

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

Case 3: Assume Cy Fmax:=

Guesses Ay Fmax:= By Fmax:= Cy Fmax:=

x 1m:= Dy Fmax:=

Given 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

Dy

x

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

Find Ay By, Dy, x,( ):=

Ay

By

Cy

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

24.8

22.7

27.5

⎛⎜⎜⎝

⎞⎟⎟⎠

kN= x3 x:= x3 0.52 m=

Since Ay < Fmax and By < Fmax then this solution is good.

We conclude that x3 0.52 m= < x < x2 5.21 m=

Problem 6-111

Determine the force created in the hydraulic cylinders EF and AD in order to hold the shovelin equilibrium. The shovel load has a mass W and a center of gravity at G. All joints are pinconnected.

Units Used:

Mg 103kg:=

kN 103N:=

Given:

a 0.25m:= θ1 30deg:=

b 0.25m:= θ2 10deg:=

c 1.5m:= θ3 60deg:=

d 2m:= W 1.25Mg:=

e 0.5m:=

Solution:

Assembly FHG :

ΣMH = 0; W g⋅ e( )⋅[ ]− FEF c sin θ1( )⋅( )⋅+ 0=

FEF W ge

c sin θ1( )⋅⋅⋅:= FEF 8.17 kN= (T)

Assembly CEFHG :

ΣMC = 0; FAD cos θ1 θ2+( )⋅ b⋅ W g⋅ a b+ c+( )cos θ2( ) e+⎡⎣ ⎤⎦⋅− 0=

FAD W gcos θ2( ) a⋅ cos θ2( ) b⋅+ cos θ2( ) c⋅+ e+

cos θ1 θ2+( ) b⋅⋅⋅:=

FAD 158 kN= (C)

Problem 6-112

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

Solution:

ΣMD = 0; 2WL2

cosθ2⎛⎜⎝⎞⎟⎠

2L sinθ2⎛⎜⎝⎞⎟⎠

⋅ NA− 0= NAW2

cotθ2⎛⎜⎝⎞⎟⎠

=

ΣΜC = 0; T L⋅ cosθ2⎛⎜⎝⎞⎟⎠

⋅ NA L⋅ sinθ2⎛⎜⎝⎞⎟⎠

⋅− WL2

⋅ cosθ2⎛⎜⎝⎞⎟⎠

⋅− 0= T W=

Problem 6-113

A man having weight W attempts to lift himself using one of the two methods shown. Determine thetotal force he must exert on bar AB in each case and the normal reaction he exerts on the platform atC. Neglect the weight of the platform.

Given:

W 750N:=

Solution:

(a)

Bar:+↑Σ Fy = 0; 2

F2

⎛⎜⎝

⎞⎟⎠

⋅ 2W2

⎛⎜⎝

⎞⎟⎠

− 0=

F W:= F 750 N=Man:

+↑Σ Fy = 0; NC W− 2F2

⎛⎜⎝

⎞⎟⎠

− 0=

NC W F+:= NC 1500 N=

b( )Bar:

+↑Σ Fy = 0; 2W4

⎛⎜⎝

⎞⎟⎠

⋅ 2F2

⋅− 0=

FW2

:= F 375 N=

Man:+↑Σ Fy = 0; NC W− 2

F2

⎛⎜⎝

⎞⎟⎠

⋅+ 0=

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 thetotal force he must exert on bar AB in each case and the normal reaction he exerts on the platform atC. The platform has weight W2.

Given:

W1 750N:= W2 150N:=

Solution:

(a)

Bar:

+↑Σ Fy = 0; 2 F2⋅ W1 W2+( )− 0=

F W1 W2+:=

F 900 N=Man:+↑Σ Fy = 0; NC W1− 2

F2

− 0=

NC F W1+:=

NC 1650 N=b( )

Bar:

+↑Σ Fy = 0; 2− F2

⎛⎜⎝

⎞⎟⎠

⋅ 2W1 W2+

4+ 0=

FW1 W2+

2:=

F 450 N=

Man:

+↑Σ Fy = 0; NC W1− 2F2

+ 0=

NC W1 F−:=

NC 300 N=

Problem 6-115

The piston C moves vertically between the two smooth walls. If the spring has stiffness k and isunstretched when θ = 0, determine the couple M that must be applied to AB to hold the mechanismin equilibrium.

Given:

k 15N

cm:=

θ 30deg:=

a 8cm:=

b 12cm:=

Solution:

Geometry:

b sin ψ( ) a sin θ( )=

ψ asin sin θ( )ab

⋅⎛⎜⎝

⎞⎟⎠

:= ψ 19.47 deg=

φ 180deg ψ− θ−:= φ 130.53 deg=

rACsin φ( )

bsin θ( )

= rAC bsin φ( )sin θ( )

⋅:= rAC 18.242 cm=

Free Body Diagram: The solution for this problem will be simplified if one realizes that member CBis a two force member. Since the spring stretches

x a b+( ) rAC−:= x 1.758 cm=

the spring force is Fsp k x⋅:= Fsp 26.37 N=

Equations of Equilibrium: Using the method of joints

+↑FCB cos ψ( )⋅ Fsp− 0= FCB

Fspcos ψ( )

:= FCB 27.97 N=Σ Fy = 0;

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 exertedon the rod R if a force F is applied at the grip G.The lobe CDE is in smooth contact with thehead 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+( )⋅ FBC b⋅ sin θ( )⋅− 0=

FBC Fa b+

b sin θ( )⋅⋅:= FBC 923.8 N=

Lobe :

ΣMD = 0; FBC f⋅ NE e⋅− 0=

NE FBCfe⋅:= NE 923.8 N=

Head :

ΣMF = 0; NE− h g+( )⋅ h NR⋅+ 0=

NR NEh 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 gearC. Note that AB is pinned at its ends to gear C and the underside of the table EF, which isallowed to move vertically due to the smooth guides at E and F. If the gears exert tangentialforces between them, determine the compressive force developed on the cylinder S when avertical 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=

FCG Fab⋅:= FCG 100 N=

Sector gear :

ΣMH = 0; FCG d e+( )⋅ FABc

c2 d2+

⋅ d⋅− 0=

FAB FCG d e+( )c2 d2

+

c d⋅⋅⋅:= FAB 297.62 N=

Table:

ΣFy = 0; FABc

c2 d2+

⋅ Fs− 0=

Fs FABc

c2 d2+

⋅:= Fs 286 N=

Problem 6-118

The mechanism is used to hide kitchen appliances under a cabinet by allowing the shelf to rotatedownward. 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 theload W. The springs each have stiffness k.

Given:

W 50N:= a 50mm:=

k 1N

mm:= b 100mm:=

φ 30deg:= c 375mm:=

θ 30deg:= d 150mm:=

Solution:

ΣMF = 0;W2

b⋅ a FED⋅ cos φ( )⋅− 0=

FEDW b⋅

2 a⋅ cos φ( )⋅:= FED 57.74 N=

+→ Σ Fx = 0; Fx− FED cos φ( )⋅+ 0=

Fx FED cos φ( )⋅:= Fx 50.00 N=

+↑Σ Fy = 0;W−2

Fy+ FED sin φ( )⋅− 0=

FyW2

FED sin φ( )⋅+:= Fy 53.87 N=

Member FBA:

ΣMA = 0; Fy c d+( )⋅ cos φ( )⋅ Fx c d+( )⋅ sin φ( )⋅− Fs sin θ φ+( )⋅ d⋅− 0=

FsFy c d+( )⋅ cos φ( )⋅ Fx c d+( )⋅ sin φ( )⋅−

d sin θ φ+( )⋅:= Fs 87.50 N=

Fs = ks; Fs k x⋅= xFsk

:= x 87.50 mm=

Problem 6-119

If each of the three links of the mechanism has a weight W, determine the angle θ forequilibrium.The spring, which always remains horizontal, is unstretched when θ = 0°.

Given:

W 25kN:=

k 60kNm

:=

a 4m:=

b 4m:=

Solution:

Guesses θ 30deg:=

Bx 10kN:= By 10kN:=

Cx 10kN:= Cy 10kN:=

Given

W−a2⋅ sin θ( )⋅ Cy a⋅ sin θ( )⋅− Cx a⋅ cos θ( )⋅+ 0=

W−b2⋅ Cy b⋅+ 0=

By Cy+ W− 0=

Bx− Cx+ 0=

Bx− a⋅ cos θ( )⋅ By a⋅ sin θ( )⋅− Wa2⋅ sin θ( )⋅− k

a2⋅ sin θ( )⋅

a2⋅ cos θ( )⋅+ 0=

Bx

By

Cx

Cy

θ

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

Find Bx By, Cx, Cy, θ,( ):=

Bx

By

Cx

Cy

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

16.58

12.5

16.58

12.5

⎛⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎠

kN= θ 33.6 deg=

Cx12

sin θ( )W 2 Cy⋅+

cos θ( )⋅⋅= θ

T

Cx

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

Find θ T, Cx,( ):=

θ

T

Cx

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

Find θ T, Cx,( ):=

T ka2

sin θ( )⋅⎛⎜⎝

⎞⎟⎠

⋅=

Problem 6-120

Determine the required force P that must be applied at the blade of the pruning shears so that theblade 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:= FCB 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 FCBc

c2 a2+

⋅− 0=

Ay P+ FCBa

a2 c2+

⋅− 0=

Ax

Ay

Dx

Dy

P

FCB

⎛⎜⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎟⎠

Find Ax Ay, Dx, Dy, P, FCB,( ):=

Ax

Ay

Dx

Dy

FCB

⎛⎜⎜⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎟⎟⎠

66.67

32.32

66.67

144.44

80.12

⎛⎜⎜⎜⎜⎜⎝

⎞⎟⎟⎟⎟⎟⎠

N= P 12.12 N=