Chapter 3 Homework A concrete foundation mat of 5-m radius supports four

PROBLEM 3.131

A concrete foundation mat of 5-m radius supports four equally spaced

columns, each of which is located 4 m from the center of the mat.

Determine the magnitude and the point of application of the resultant of

the four loads.

SOLUTION

Have:

:

AC DE

Σ +++=F FFFF R

( ) ( ) ( ) ( )

100 kN 125 kN 25kN 75 kN=− − −−R j jjj

PROBLEM 3.132

Determine the magnitude and the point of application of the smallest

additional load that must be applied to the foundation mat of Prob. 3.131

if the resultant of the five loads is to pass through the center of the mat.

SOLUTION

The smallest load P will be placed on the edge of the mat at a radius of 5 m. Therefore

22 2

(5 m) (1)

PP

xz+=

PROBLEM 3.133*

Three forces of the same magnitude P act on a cube of side a as

shown. Replace the three forces by an equivalent wrench and

determine (a) the magnitude and direction of the resultant force

R, (b) the pitch of the wrench, (c) the axis of the wrench.

SOLUTION

Force–couple system at O:

()PPP P= + + = ++R i j k i jk

PROBLEM 3.134*

A piece of sheet metal is bent into the shape shown and is

acted upon by three forces. If the forces have the same

magnitude P, replace them with an equivalent wrench and

determine (a) the magnitude and the direction of the

resultant force R, (b) the pitch of the wrench, (c) the axis

of the wrench.

SOLUTION

First reduce the given forces to an equivalent force–couple system

( )

,

R

O

RM

at the origin.

We have

:PPP RΣ −++ =F jjk

or

P=Rk

PROBLEM 3.134* (Continued)

(c) The components of the wrench are

1

( , ),RM

where

1 1 axis

,M=Mλ

and the axis of the wrench is

assumed to intersect the xy–plane at Point Q, whose coordinates are (x, y, 0). Thus, we require

zQ R

= ×MrR

where

1zO

= ×MMM

Then

55 ()

22

aP aP x y P



−− + − = + +





ij k k i j k

Equating coefficients:

: or

aP yP y a

aP xP x a

−= =−

−=− =

i

j

The axis of the wrench is parallel to the z–axis and intersects the xy–plane at

,.x ay a= = −



PROBLEM 3.135*

The forces and couples shown are applied to two screws

as a piece of sheet metal is fastened to a block of wood.

Reduce the forces and the couples to an equivalent

wrench and determine (a) the resultant force R, (b) the

pitch of the wrench, (c) the point where the axis of the

wrench intersects the xz–plane.

SOLUTION

First, reduce the given force system to a force–couple system.

We have

: (20 N) (15 N) 25 NRΣ− − = =F i jR

We have

:( )

R

O O CO

Σ Σ × +Σ =M rF M M

PROBLEM 3.136*

The forces and couples shown are applied to two screws as a

piece of sheet metal is fastened to a block of wood. Reduce the

forces and the couples to an equivalent wrench and determine

(a) the resultant force R, (b) the pitch of the wrench, (c) the point

where the axis of the wrench intersects the xz–plane.

SOLUTION

PROBLEM 3.136* (Continued)

(c) We have

12

21

(35 lb in.)

R

O

R

O

= +

= −= ⋅

M MM

MMM i

We require

2/

(35 lb in.) ( ) [ (21 lb) ]

35 (21 ) (21 )

QO

xz

= ×

⋅ = + ×−

=−+

Mr R

i ik j

i ki

From i:

35 21

1.66667 in.

z

=

From k:

0 21

0

x

z

= −

=

The axis of the wrench is parallel to the y–axis and intersects the xz–plane at

0, 1.667 in.xz= =



PROBLEM 3.137*

Two bolts at A and B are tightened by applying the forces

and couples shown. Replace the two wrenches with a

single equivalent wrench and determine (a) the resultant

R, (b) the pitch of the single equivalent wrench, (c) the

point where the axis of the wrench intersects the xz–plane.

SOLUTION

PROBLEM 3.137* (Continued)

(c) We have

12

21

[( 15.6 2 82.4 ) (40.102 38.192 )] N m

(15.6 N m) (42.102 N m) (44.208 N m)

R

O

R

O

= +

= − =− +− − − ⋅

=− ⋅+ ⋅− ⋅

M MM

MMM ij k j k

i jk

We require

2/

( 15.6 42.102 44.208 ) ( ) (84 80 )

(84 ) (80 ) (84 )

QO

xz

zxx

= ×

− + − =+× −

=+−

Mr R

i j k ik j k

i jk

From i:

15.6 84

0.185714 m

z

−=

= −

or

0.1857 mz= −

From k:

44.208 84

0.52629 m

x

−=−

=

or

0.526 mx=

The axis of the wrench intersects the xz–plane at

0.526 m 0 0.1857 mx yz= = = −



PROBLEM 3.138*

Two bolts at A and B are tightened by applying the forces

and couples shown. Replace the two wrenches with a single

equivalent wrench and determine (a) the resultant R,

(b) the pitch of the single equivalent wrench, (c) the point

where the axis of the wrench intersects the xz–plane.

SOLUTION

PROBLEM 3.138* (Continued)

and

11

(62.818 lb in.) (117.783 lb in.) (207.30 lb in.)

R

M

λ

= =− ⋅− ⋅− ⋅M i jk

Then pitch

1246.56 lb in. 7.8522 in.

31.4 lb

M

pR

⋅

= = =

or

7.85 in.p=



(c) We have

12

21

(152 210 220 ) ( 62.818 117.783 207.30 )

(214.82 lb in.) (92.217 lb in.) (12.7000 lb in.)

R

B

R

B

= +

= − = − − −− − −

= ⋅− ⋅− ⋅

M MM

MMM i j k i j k

ij k

We require

2/QB

= ×Mr R

214.82 92.217 12.7000 0

8 15 26.4

(15 ) (8 ) (26.4 ) (15 )

xz

zz x x

−− =

−− −

= −+ −

ij k

ij j k

From i:

214.82 15 14.3213 in.zz= =

From k:

12.7000 15 0.84667 in.xx−=− =

The axis of the wrench intersects the xz–plane at

0.847 in. 0 14.32 in.x yz= = =



PROBLEM 3.139*

Two ropes attached at A and B are used to move the trunk

of a fallen tree. Replace the forces exerted by the ropes

with an equivalent wrench and determine (a) the resultant

force R, (b) the pitch of the wrench, (c) the point where the

axis of the wrench intersects the yz–plane.

SOLUTION

(a) First replace the given forces with an equivalent force–couple system

( )

,

R

O

RM

at the origin.

We have

PROBLEM 3.139* (Continued)

(b) We have

22 2

100 (23) (5) (18.5) 2993.7 NR= ++ =

Let

PROBLEM 3.139* (Continued)

Thus, we require

2

()

P

yz=×=+M rR r jk

Substituting:

PROBLEM 3.140*

A flagpole is guyed by three cables. If the tensions in the

cables have the same magnitude P, replace the forces

exerted on the pole with an equivalent wrench and

determine (a) the resultant force R, (b) the pitch of the

wrench, (c) the point where the axis of the wrench

intersects the xz–plane.

SOLUTION

PROBLEM 3.140* (Continued)

(b) We have

1

R

RO

M= ⋅Mλ

where

3 25 1

(2 20 ) (2 20 )

25 27 5 9 5

R

P

RP

== −− = −−

Ri jk i jk

λ

83

( 403 20 406 ) ( ) (2 20 )

675 25

3[20 ( 2 ) 20 ]

25

Pa P

xz

Pz xz x

  

− −− =+× −−

  

  



= ++ −





ij k ik ijk

i jk

From i:

3

8( 403) 20 1.99012

675 25

Pa P

zz a



−= =−





From k:

3

8( 406) 20 2.0049

675 25

Pa P

xx a



−=− =





The axis of the wrench intersects the xz–plane at

2.00 , 1.990x az a= = −



PROBLEM 3.141*

Determine whether the force–and–couple system shown can

be reduced to a single equivalent force R. If it can,

determine R and the point where the line of action of R

intersects the yz–plane. If it cannot be so reduced, replace

the given system with an equivalent wrench and determine

its resultant, its pitch, and the point where its axis intersects

the yz–plane.

SOLUTION

First, reduce the given force system to a force–couple at the origin.

We have

:

AG

Σ +=FF F R

(40 mm) (60 mm) (120 mm)

(50 N) 70 N 140 mm

(20 N) (30 N) (10 N)



+−

= + 



=+−

ij k

Rk

i jk

and

37.417 NR=

PROBLEM 3.141* (Continued)

Then for equivalence,

PROBLEM 3.142*

Determine whether the force–and–couple system shown

can be reduced to a single equivalent force R. If it can,

determine R and the point where the line of action of R

intersects the yz–plane. If it cannot be so reduced, replace

the given system with an equivalent wrench and

determine its resultant, its pitch, and the point where its

axis intersects the yz–plane.

SOLUTION

First determine the resultant of the forces at D. We have

222

22 2

( 12) (9) (8) 17 in.

( 6) (0) ( 8) 10 in.

DA

ED

d

=−++=

= − + +− =

Then