978-0133915426 Chapter 6 Part 7

*6–104.

The hydraulic crane is used to lift the 1400-lb load.

Determine the force in the hydraulic cylinder AB and the

force in links AC and AD when the load is held in the

position shown.

8 ft

30

120

70

1 ft

B

AD

C

7 ft

SOLUTION

6–105.

Determine force P on the cable if the spring is compressed

0.025 m when the mechanism is in the position shown. The

spring has a stiffness of k = 6 kN

>

m.

P

150 mm

200 mm

200 mm 800 mm

A

C

D

E

B

F

30

k

6–106.

If and the spring has an unstretched length of 1 ft,

determine the force Frequired for equilibrium.

SOLUTION

Equations of Equilibrium: First, we will analyze the equilibrium of joint B. From the

free-body diagram in Fig. a,

From the free-body diagram in Fig. b, using the result , and

analyzing the equilibrium of joint C, we have

Ans.F =66.14 lb =66.1 lb

2(50 cos 48.59°) –F=0©F

x=0;

:

+

F

CD sin 48.59° –50 sin 48.59° =0

F

CD =50 lb+c©F

y=0;

F

BC =F¿ = 50 lb

F¿ =50 lb

2F¿ sin 48.59° –75 =0+c©F

y=0;

F

AB =F

BC =F¿

F

AB cos 48.59° –F

BC cos 48.59° =0©F

x=0;

:

+

d

=0.75 ft

d

AC

B

1 ft

k 150 lb/ft

FF

6–107.

If a force of is applied to the pads at Aand C,

determine the smallest dimension drequired for equilibrium if

the spring has an unstretched length of 1 ft.

SOLUTION

Spring Force Formula: The elongation of the spring is . Thus, the force

in the spring is given by

Equations of Equilibrium:First, we will analyze the equilibrium of joint B. From the

free-body diagram in Fig. b,

From the free-body diagram in Fig. c, using the result FBC =, and

analyzing the equilibrium of joint C, we have

Solving the above equation using a graphing utility, we obtain Ans.

or d = 0.9334 ft =0.933 ft0.6381 ft =0.638 ftd=

2ca150d–75

dba11–d2bd–50 =0©F

x=0;

:

+

F

CD =150d–75

d

F

CD sin u–a150d–75

db sin u=0+c©F

y=0;

F¿ = 150d–75

d

F¿ = 150d–75

d

2F¿(d)–150(2d–1) =0+c©F

y=0;

F

AB cos u–F

BC cos u=0

F

AB =F

BC =F¿©F

x=0;

:

+

F

sp =kx = 150(2d–1)

x = 2d–1

F

=50 lb

d

AC

B

1 ft

k 150 lb/ft

FF

*6–108.

Theskid steer loader hasamass of 1.18 Mg,and in the

position shown the center of mass is at If there is a300-kg

stone in the bucket, with center of mass at determine the

reactions of each pair of wheels Aand Bon the ground and

the force in the hydraulic cylinder CD and at the pin E.There

isasimilar linkage on each side of the loader.

G2,

G1.

1.25 m

C

D

G

1

G

2

E

30

SOLUTION

6–109.

Determine the force Pon the cable if the spring is

compressed 0.5 in. when the mechanism is in the position

shown. The spring has a stiffness of k=800 lb>ft.

SOLUTION

a

(1)

a

(2)

(3)

a

Thus from Eq. (3)

Using Eqs. (1) and (2):

Ans.P=46.9 lb

2.8867 P+0.6667 P=166.67

Bx=2.8867 P

FCD =3.333 P

+©MB=0; FCD sin 30°(6) –P(10) =0

:

+©Fx=0; –Bx+FCD cos 30° =0

By=0.6667P

+©MD=0; By(6) –P(4) =0

Bx+By=166.67 lb

+©MA=0;

Bx(6) +By(6) –33.33(30) =0

FE=ks =800 a0.5

12 b=33.33 lb

P

6 in.

24 in.

6 in. 6in. 4 in.

A

C

D

E

B

30

k

SOLUTION

6–110.

The spring has an unstretched length of 0.3 m. Determine

the angle

u

for equilibrium if the uniform bars each have a

mass of 20 kg.

2 m

k  150 N/m

A

B

C

u

SOLUTION

6–111.

The spring has an unstretched length of 0.3 m. Determine

the mass m of each uniform bar if each angle

u=30°

for

equilibrium. 2 m

k  150 N/m

A

B

C

u

*6–112.

The piston Cmoves vertically between the two smooth

walls. If the spring has a stiffness of and is

unstretched when determine the couple Mthat

must be applied to AB to hold the mechanism in

equilibrium when u=30°.

u=0°,

k=15 lb

>

in.,

SOLUTION

A

M

u

B

8 in.

12 in.

6–113.

The platform scale consists of a combination of third and first

class levers so that the load on one lever becomes the effort

that moves the next lever.Through this arrangement, a small

weight can balance a massive object. If ,

determine the required mass of the counterweight Srequired

to balance a 90-kg load, L.

x=450 mm

SOLUTION

Using the result of FBG and writing the moment equation of equilibrium about

point Fon the free – body diagram of member EFG in Fig. b,

a

Using the result of FED and writing the moment equation of equilibrium about

point Con the free – body diagram of member CDI in Fig. c,

a

Ans.mS=1.705 kg =1.71 kg

+©MC=0; 158.922(100) –mS(9.81)(950) =0

FED =158.922 N

+©MF=0; FED (250) –264.87(150) =0

FBG =264.87 N

350mm

150 mm

150 mm100 mm

250 mm

B

A

CD

EF

H

G

x

L

S

6–114.

SOLUTION

Using the result of and writing the moment equation of equilibrium about

point Fon the free – body diagram of member EFG in Fig. b,

a

Using the result of and writing the moment equation of equilibrium about

point Con the free – body diagram of member CDI in Fig. c,

a

Ans.mL=105.56 kg =106 kg

+©MC=0; 1.7658mL(100) –2(9.81)(950) =0

FED

FED =1.7658mL

+©MF=0; FED (250) –2.943mL(150) =0

FBG

FBG =2.943 lb

The platform scale consists of a combination of third and

first class levers so that the load on one lever becomes the

effort that moves the next lever.Through this arrangement,

a small weight can balance a massive object. If

and, the mass of the counterweight Sis 2 kg, determine the

mass of the load Lrequired to maintain the balance.

x=450 mm

350mm

150 mm

150 mm100 mm

250 mm

B

A

CD

EF

H

G

x

L

S

6–115.

The four-member “A” frame is supported at Aand Eby

smooth collars and at Gby a pin. All the other joints are

ball-and-sockets. If the pin at Gwill fail when the resultant

force there is 800 N, determine the largest vertical force P

that can be supported by the frame. Also, what are the x, y, z

force components which member BD exerts on members

EDC and ABC?The collars at Aand Eand the pin at G

only exert force components on the frame.

x

y

C

D

BF

G

E

A

z

300 mm

600 mm

SOLUTION

*6–116.

(1)

Ans.

(2)

(3)

From FBD (b)

From Eq.(1)

From Eq.(2) Ans.

From Eq.(3) Ans.

Ans.

Negative sign indicates that acts in the opposite sense to that shown on FBD.MCy

©Mz=0; MCz =0

MCy =-429 N #m

Ay=115 N

Ax=172 N

114.85(3) –Bx(2) =0Bx=172.27 N

©Fz=0; Az=0

The structure is subjected to the loadings shown. Member

AB is supported by a ball-and-socket at Aand smooth

collar at B. Member CD is supported by a pin at C.

Determine the x, y, zcomponents of reaction at Aand C.

1.5 m

A

250 N

D

45

60

z

C

Ans:

A

z

=0

Ax=172 N

A

y

=115 N

Cx=47.3 N

C

y

=61.9 N

C

z

=125 N

M

Cy

=–429 N #m

MC

z

=0

6–117.

SOLUTION

©M

y

=0; –4

5 F

AB

(0.6) +2.5(0.3) =0

The structure is subjected to the loading shown. Member AD

is supported by a cable AB and roller at C and fits through

a smooth circular hole at D. Member ED is supported by a

roller at Dand a pole that fits in a smooth snug circular hole

at E. Determine the x, y, z components of reaction at E and

the tension in cable AB.

z

C

A

D

B

E

0.3 m y

0.5 m

x

0.8 m

6–118.

(2)

From FBD (b),

a(3)

(4)

From FBD (c),

a(5)

(6)

Solving Eqs. (1), (2), (3), (4), (5) and (6) yields,

Ans.F

D

=20.8 lb

F

=14.7 lb

F

A

=24.5 lb

F

E

=36.73 lb

F

C

=22.04 lb

F

B

=61.22 lb

F

A

+F

E

–F

B

=0+c©F

y

=0;

F

E

1102–F

B

162=0+ ©M

A

=0;

F

C

+F

F

–F

E

=0+c©F

y

=0;

F

E

162–F

C

1102=0+ ©M

F

=0;

F

+F

–F

–60 =0+c©F

=0;

The three pin-connected members shown in the top view

support a downward force of 60 lb at G.If only vertical forces

are supported at the connections B, C, E and pad supports A,

D, F, determine the reactions at each pad.

D

G

C

6 ft

2 ft

4 ft