PROBLEM 3.

120
A portion of the flue for a furnace is attached to the ceiling at A. While
supporting the free end of the flue at F, a worker pushes in at E and pulls
out at F to align end E with the furnace. Knowing that the 10-lb force at
F lies in a plane parallel to the yz plane and that α = 60°, (a) replace the
given force system with an equivalent force-couple system at C, (b)
determine whether duct CD will tend to rotate clockwise or
counterclockwise relative to elbow C, as viewed from D to C.

SOLUTION

(a) Have R = ΣF = FF + FE

where FF = 10 lb ( sin 60° ) j + ( cos 60° ) k  = (8.6603 lb ) j + ( 5.0 lb ) k

FE = − ( 5 lb ) k

∴ R = ( 8.6603 lb ) j or R = ( 8.66 lb ) j

Have M CR = Σ ( r × F ) = rF /C × FF + rE/C × FE

where rF /C = ( 9 in.) i − ( 2 in.) j

rE/C = (18 in.) i − (13 in.) j

i j k i j k
∴ M CR = 9 −2 0 lb ⋅ in. + 18 −13 0 lb ⋅ in.
0 8.6603 5.0 0 0 −5

= ( 55 lb ⋅ in.) i + ( 45 lb ⋅ in.) j + ( 77.942 lb ⋅ in.) k

or M CR = ( 55.0 lb ⋅ in.) i + ( 45.0 lb ⋅ in.) j + ( 77.9 lb ⋅ in.) k

(b) To determine which direction duct section CD has a tendency to turn, have
R
M CD = λ DC ⋅ M CR
where
− (18 in.) i + ( 4 in.) j 1
λ DC = = ( −9i + 2 j)
2 85 in. 85
1
Then R
M CD = ( −9i + 2 j) ⋅ ( 55i + 45j + 77.942k ) lb ⋅ in.
85

= ( −53.690 + 9.7619 ) lb ⋅ in.

= −43.928 lb ⋅ in.

Since λ DC ⋅ M CR < 0, duct DC tends to rotate clockwise relative to elbow C as viewed from D to C.
 PROBLEM 3.121
The head-and-motor assembly of a radial drill press was originally
positioned with arm AB parallel to the z axis and the axis of the chuck
and bit parallel to the y axis. The assembly was then rotated 25o about
the y axis and 20o about the centerline of the horizontal arm AB, bringing
it into the position shown. The drilling process was started by switching
on the motor and rotating the handle to bring the bit into contact with the
workpiece. Replace the force and couple exerted by the drill press with an
equivalent force-couple system at the center O of the base of the vertical
column.

SOLUTION
Have R =F

= ( 44 N ) ( sin 20° cos 25° ) i − ( cos 20° ) j − ( sin 20° sin 25° ) k 

= (13.6389 N ) i − ( 41.346 N ) j − ( 6.3599 N ) k

or R = (13.64 N ) i − ( 41.3 N ) j − ( 6.36 N ) k

Have M O = rB/O × F + M C

where

rB/O = ( 0.280 m ) sin 25°  i + ( 0.300 m ) j + ( 0.280 m ) cos 25°  k

= ( 0.118333 m ) i + ( 0.300 m ) j + ( 0.25377 m ) k

M C = ( 7.2 N ⋅ m ) ( sin 20° cos 25° ) i − ( cos 20° ) j − ( sin 20° sin 25° ) k 

= ( 2.2318 N ⋅ m ) i − ( 6.7658 N ⋅ m ) j − (1.04072 N ⋅ m ) k

i j k
∴ MO = 0.118333 0.300 0.25377 N ⋅ m
13.6389 −41.346 −6.3599

+ ( 2.2318i − 6.7658 j − 1.04072k ) N ⋅ m

= (10.8162 N ⋅ m ) i − ( 2.5521 N ⋅ m ) j − (10.0250 N ⋅ m ) k

or M O = (10.82 N ⋅ m ) i − ( 2.55 N ⋅ m ) j − (10.03 N ⋅ m ) k

 PROBLEM 3.122
While a sagging porch is leveled and repaired, a screw jack is used to
support the front of the porch. As the jack is expanded, it exerts on the
porch the force-couple system shown, where R = 300 N and
M = 37.5 N ⋅ m. Replace this force-couple system with an equivalent
force-couple system at C.

SOLUTION

 − ( 0.2 m ) i + (1.4 m ) j − ( 0.5 m ) k 

From R C = R = ( 300 N ) λ AB = 300 N  
 1.50 m 

R C = − ( 40.0 N ) i + ( 280 N ) j − (100 N ) k

From M C = rA/C × R + M

where

rA/C = ( 2.6 m ) i + ( 0.5 m ) k

 ( 0.2 m ) i − (1.4 m ) j + ( 0.5 m ) k 

M = ( 37.5 N ⋅ m ) λ BA = ( 37.5 N ⋅ m )  
 1.50 m 

= ( 5.0 N ⋅ m ) i − ( 35.0 N ⋅ m ) j + (12.5 N ⋅ m ) k

i j k
∴ MC = (10 N ⋅ m ) 2.6 0 0.5 + ( 5.0 N ⋅ m ) i − ( 35.0 N ⋅ m ) j + (12.5 N ⋅ m ) k
−4 28 −10

= ( −140 + 5 ) N ⋅ m  i + ( −20 + 260 − 35 ) N ⋅ m  j + ( 728 + 12.5 ) N ⋅ m  k

or M C = − (135.0 N ⋅ m ) i + ( 205 N ⋅ m ) j + ( 741 N ⋅ m ) k

 PROBLEM 3.123
Three children are standing on a 15 × 15-ft raft. If the weights of the
children at points A, B, and C are 85 lb, 60 lb, and 90 lb, respectively,
determine the magnitude and the point of application of the resultant of
the three weights.

SOLUTION

Have ΣF : FA + FB + FC = R

− ( 85 lb ) j − ( 60 lb ) j − ( 90 lb ) j = R

− ( 235 lb ) j = R or R = 235 lb

Have ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) = R ( z D )

(85 lb)( 9 ft ) + ( 60 lb )(1.5 ft ) + ( 90 lb )(14.25 ft ) = ( 235 lb )( zD )

∴ z D = 9.0957 ft or z D = 9.10 ft

Have ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) = R ( xD )

(85 lb )( 3 ft ) + ( 60 lb )( 4.5 ft ) + ( 90 lb )(14.25 ft ) = ( 235 lb )( xD )

∴ xD = 7.6915 ft or xD = 7.69 ft
 PROBLEM 3.124
Three children are standing on a 15 × 15-ft raft. The weights of the
children at points A, B, and C are 85 lb, 60 lb, and 90 lb, respectively. If a
fourth child of weight 95 lb climbs onto the raft, determine where she
should stand if the other children remain in the positions shown and the
line of action of the resultant of the four weights is to pass through the
center of the raft.

SOLUTION

Have ΣF : FA + FB + FC + FD = R

− ( 85 lb ) j − ( 60 lb ) j − ( 90 lb ) j − ( 95 lb ) j = R

∴ R = − ( 330 lb ) j

Have ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) + FD ( z D ) = R ( z H )

(85 lb )( 9 ft ) + ( 60 lb )(1.5 ft ) + ( 90 lb )(14.25 ft ) + ( 95 lb )( zD ) = ( 330 lb )( 7.5 ft )

∴ z D = 3.5523 ft or z D = 3.55 ft

Have ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) + FD ( xD ) = R ( xH )

(85 lb )( 3 ft ) + ( 60 lb )( 4.5 ft ) + ( 90 lb )(14.25 ft ) + ( 95 lb )( xD ) = ( 330 lb )( 7.5 ft )

∴ xD = 7.0263 ft or xD = 7.03 ft
 PROBLEM 3.125
The forces shown are the resultant downward loads on sections of the flat
roof of a building because of accumulated snow. Determine the
magnitude and the point of application of the resultant of these four
loads.

SOLUTION

Have ΣF : FA + FB + FC + FD = R

− ( 580 kN ) j − ( 2350 kN ) j − ( 330 kN ) j − (140 kN ) j = R

∴ R = − ( 3400 kN ) j R = 3400 kN

Have ΣM x : FA ( z A ) + FB ( z B ) + FC ( zC ) + FD ( z D ) = R ( z E )

( 580 kN )(8 m ) + ( 2350 kN )(16 m ) + ( 330 kN )( 6 m ) + (140 kN )( 33.5 m ) = ( 3400 kN )( zE )

∴ z E = 14.3853 m or z E = 14.39 m

Have ΣM z : FA ( x A ) + FB ( xB ) + FC ( xC ) + FD ( xD ) = R ( xE )

( 580 kN )(10 m ) + ( 2350 kN )( 32 m ) + ( 330 kN )( 54 m ) + (140 kN )( 32 m ) = ( 3400 kN )( xE )

∴ xE = 30.382 m or xE = 30.4 m
 PROBLEM 3.126
The forces shown are the resultant downward loads on sections of the flat
roof of a building because of accumulated snow. If the snow represented
by the 580-kN force is shoveled so that the this load acts at E, determine
a and b knowing that the point of application of the resultant of the four
loads is then at B.

SOLUTION

Have ΣF : FB + FC + FD + FE = R

− ( 2350 kN ) j − ( 330 kN ) j − (140 kN ) j − ( 580 kN ) j = R

∴ R = − ( 3400 kN ) j

Have ΣM x : FB ( z B ) + FC ( zC ) + FD ( z D ) + FE ( z E ) = R ( z B )

( 2350 kN )(16 m ) + ( 330 kN )( 6 m ) + (140 kN )( 33.5 m ) + ( 580 kN )( b ) = ( 3400 kN )(16 m )

∴ b = 17.4655 m or b = 17.47 m

Have ΣM z : FB ( xB ) + FC ( xC ) + FD ( xD ) + FE ( xE ) = R ( xB )

( 2350 kN )( 32 m ) + ( 330 kN )( 54 m ) + (140 kN )( 32 m ) + ( 580 kN )( a ) = ( 3400 kN )( 32 m )

∴ a = 19.4828 m or a = 19.48 m
 PROBLEM 3.127
A group of students loads a 2 × 4-m flatbed trailer with two
0.6 × 0.6 × 0.6-m boxes and one 0.6 × 0.6 × 1.2-m box. Each of the
boxes at the rear of the trailer is positioned so that it is aligned with both
the back and a side of the trailer. Determine the smallest load the students
should place in a second 0.6 × 0.6 × 1.2-m box and where on the trailer
they should secure it, without any part of the box overhanging the sides
of the trailer, if each box is uniformly loaded and the line of action of the
resultant of the weights of the four boxes is to pass through the point of
intersection of the centerlines of the trailer and the axle. (Hint: Keep in
mind that the box may be placed either on its side or on its end.)

SOLUTION

For the smallest weight on the trailer so that the resultant force of the four weights acts over the axle at the
intersection with the center line of the trailer, the added 0.6 × 0.6 × 1.2-m box should be placed adjacent to
one of the edges of the trailer with the 0.6 × 0.6-m side on the bottom. The edges to be considered are based
on the location of the resultant for the three given weights.

Have ΣF : − ( 200 N ) j − ( 400 N ) j − (180 N ) j = R

∴ R = − ( 780 N ) j

Have ΣM z : ( 200 N )( 0.3 m ) + ( 400 N )(1.7 m ) + (180 N )(1.7 m ) = ( 780 N )( x )

∴ x = 1.34103 m

Have ΣM x : ( 200 N )( 0.3 m ) + ( 400 N )( 0.6 m ) + (180 N )( 2.4 m ) = ( 780 N )( z )

∴ z = 0.93846 m

From the statement of the problem, it is known that the resultant of R from the original loading and the
lightest load W passes through G, the point of intersection of the two center lines. Thus, ΣM G = 0.

Further, since the lightest load W is to be as small as possible, the fourth box should be placed as far from G
as possible without the box overhanging the trailer. These two requirements imply

( 0.3 m ≤ x ≤ 1 m ) (1.8 m ≤ z ≤ 3.7 m )

 PROBLEM 3.127 CONTINUED

Let x = 0.3 m, ΣM Gz : ( 200 N )( 0.7 m ) − ( 400 N )( 0.7 m ) − (180 N )( 0.7 m ) + W ( 0.7 m ) = 0

∴ W = 380 N

ΣM Gx : − ( 200 N )(1.5 m ) − ( 400 N )(1.2 m ) + (180 N )( 0.6 m ) + ( 380 N )( z − 1.8 m ) = 0

∴ z = 3.5684 m < 3.7 m ∴ acceptable

Let z = 3.7 m, ΣM Gx : − ( 200 N )(1.5 m ) − ( 400 N )(1.2 m ) + (180 N )( 0.6 m ) + W (1.7 m ) = 0

∴ W = 395.29 N > 380 N

Since the weight W found for x = 0.3 m is less than W found for z = 3.7 m, x = 0.3 m results in the
smallest weight W.

or W = 380 N at ( 0.3 m, 0, 3.57 m )

 PROBLEM 3.128
Solve Problem 3.127 if the students want to place as much weight as
possible in the fourth box and that at least one side of the box must
coincide with a side of the trailer.
Problem 3.127: A group of students loads a 2 × 4-m flatbed trailer with
two 0.6 × 0.6 × 0.6-m boxes and one 0.6 × 0.6 × 1.2-m box. Each of the
boxes at the rear of the trailer is positioned so that it is aligned with both
the back and a side of the trailer. Determine the smallest load the students
should place in a second 0.6 × 0.6 × 1.2-m box and where on the trailer
they should secure it, without any part of the box overhanging the sides
of the trailer, if each box is uniformly loaded and the line of action of the
resultant of the weights of the four boxes is to pass through the point of
intersection of the centerlines of the trailer and the axle. (Hint: Keep in
mind that the box may be placed either on its side or on its end.)

SOLUTION

For the largest additional weight on the trailer with the box having at least one side coinsiding with the side of
the trailer, the box must be as close as possible to point G. For x = 0.6 m, with a small side of the box
touching the z-axis, satisfies this condition.

Let x = 0.6 m, ΣM Gz : ( 200 N )( 0.7 m ) − ( 400 N )( 0.7 m ) − (180 N )( 0.7 m ) + W ( 0.4 m ) = 0

∴ W = 665 N

and ΣM GX : − ( 200 N )(1.5 m ) − ( 400 N )(1.2 m ) + (180 N )( 0.6 m ) + ( 665 N )( z − 1.8 m ) = 0

∴ z = 2.8105 m (2 m < z < 4 m) ∴ acceptable

or W = 665 N at ( 0.6 m, 0, 2.81 m )