PROBLEM 6.

137

The drum lifter shown is used to lift a steel drum. Knowing that the mass of
the drum and its contents is 240 kg, determine the forces exerted at F and H
on member DFH.

SOLUTION

ΣFy = 0;
FBD System:
P −W = 0
P =W

(
= ( 240 kg ) 9.81 m/s 2 )
P = 2354.4 N

FBD machine:
Symmetry: H x = I x

Hy = Iy

ΣFy = 0: P − 2 H y = 0

P
Hy = = 1177.2 N
2

FBD ABC:
Symmetry: C = B

4
ΣFy = 0: P − 2 B = 0
5
5
B= P = 1471.5 N
8
 PROBLEM 6.137 CONTINUED

FBD DFH:
3 
ΣM H = 0: − ( 0.1 m ) F + ( 0.39 m )  1471.5 N 
5 

4 
− ( 0.055 m )  1471.5 N  = 0
5 
F = 3973.05 N F = 3.97 kN W

3
ΣFx = 0: 3973.05 N − 1471.5 N − H x = 0
5
H x = 3090.15 N

H = 3.31 kN 20.9° W
 PROBLEM 6.138

A small barrel having a mass of 72 kg is lifted by a pair of tongs as

shown. Knowing that a = 100 mm, determine the forces exerted at
B and D on tong ABD.

SOLUTION

Notes: From FBD whole, by inspection,

FBB ABD:
(
P = W = mg = ( 72 kg ) 9.81 m/s2 )
P = 706.32 N
BC is a two-force member: Bx = 3By

ΣM D = 0: ( 0.1 m ) Bx + ( 0.06 m ) By − ( 0.18 m ) P = 0

3By + 0.6 By = 1.8P

P
By = = 353.16 N
2
3P
so Bx = = 1059.48 N
2
and B = 1.117 kN 18.43° W
ΣFx = 0: − Bx + Dx = 0

3P
Dx = = 1059.48 N
2
ΣFy = 0: P − By − Dy = 0

P
Dy = P − = 353.16 N
2
so D = 1.117 kN 18.43° W
 PROBLEM 6.139

Determine the magnitude of the ripping forces exerted along

line aa on the nut when two 240-N forces are applied to the handles
as shown. Assume that pins A and D slide freely in slots cut in the
jaws.

SOLUTION

ΣFx = 0: Bx = 0
FBD jaw AB:
ΣM B = 0: ( 0.01 m ) G − ( 0.03 m ) A = 0

G
A=
3

ΣFy = 0: A + G − By = 0

4G
By = A + G =
3
FBD handle ACE:
G
By symmetry and FBD jaw DE: D = A = , Ex = Bx = 0,
3
4G
E y = By =
3

G 4G
ΣM C = 0: ( 0.105 m )( 240 N ) + ( 0.015 m ) − ( 0.015 m ) =0
3 3

G = 1680 N W
 PROBLEM 6.140

In using the bolt cutter shown, a worker applies two 75-lb forces to
the handles. Determine the magnitude of the forces exerted by the
cutter on the bolt.

SOLUTION

FBD Cutter AB: FBD handle BC:

FBD I: ΣFx = 0: Bx = 0 FBD II: ΣM C = 0: ( 0.5 in.) By − (19.5 in.) 75 lb = 0

By = 2925 lb

Then FBD I: ΣM A = 0: ( 4 in.) By − (1 in.) F = 0 F = 4 By

F = 11700 lb = 11.70 kips W

 PROBLEM 6.141

Determine the magnitude of the gripping forces produced when two 50-lb
forces are applied as shown.

SOLUTION

FBD handle CD: 2.8

ΣM D = 0: − ( 4.2 in.)( 50 lb ) − ( 0.2 in.) A
8.84

 1 
+ (1 in.)  A = 0
 8.84 

A = 477.27 8.84 lb

1
( )
FBD handle AD:
ΣM D = 0: ( 4.4 in.)( 50 lb ) − ( 4 in.) 477.27 8.84 lb
8.84

+ (1.2 in.) F = 0

F = 1.408 kips W
 PROBLEM 6.142

The compound-lever pruning shears shown can be adjusted by

placing pin A at various ratchet positions on blade ACE. Knowing
that 1.5-kN vertical forces are required to complete the pruning of a small
branch, determine the magnitude P of the forces that must be applied to the
handles when the shears are adjusted as shown.

SOLUTION

ΣM C = 0: ( 32 mm )1.5 KN − ( 28 mm ) Ay − (10 mm ) Ax = 0
FBD cutter AC:
 11 
10 Ax + 28  Ax  = 48 kN
 13 

Ax = 1.42466 kN

Ay = 1.20548 kN

FBD handle AD:

ΣM D = 0: (15 mm )(1.20548 kN ) − ( 5 mm )(1.42466 kN )

− (70 mm) P = 0

P = 0.1566 kN = 156.6 N W
 PROBLEM 6.143

Determine the force P which must be applied to the toggle BCD

To maintain equilibrium in the position shown.

SOLUTION

Note: θ = 30° + α
FBD joint B:
20
= 30° + tan −1
200
= 30° + 5.711°

= 35.711°

ΣFx′ = 0: FBC cos 35.711° − ( 240 N ) cos 30° = 0

FBC = 255.98 N T

FBD joint C:
By symmetry: FCD = 255.98 N

ΣFx′′ = 0: P − 2 ( 255.98 N ) sin 5.711° = 0

P = 50.9 N 30.0° W
 PROBLEM 6.144

In the locked position shown, the toggle clamp exerts at A a vertical

1.2-kN force on the wooden block, and handle CF rests against the
stop at G. Determine the force P required to release the clamp.
(Hint: To release the clamp, the forces of contact at G must be
zero.)

SOLUTION

FBD BC:

ΣM B = 0: ( 24 mm ) C x − ( 66 mm )(1.2 kN ) = 0

Cx = 3.3 kN

ΣFx = 0: Dx − C x = 0 Dx = 3.3 kN

ΣM C = 0: ( 86 mm ) P

FBD CDF: − 18.5 mm − (16 mm ) tan 40°  ( FDE cos 40° ) = 0

FDE = 22.124 P

ΣFx = 0: Cx − FDE cos 40° + P sin 30° = 0

3.3 kN − ( 22.124 P ) cos 40° + P sin 30° = 0

P = 201 N 60° W
 PROBLEM 6.145

The garden shears shown consist of two blades and two handles.
The two handles are connected by pin C and the two blades are
connected by pin D. The left blade and the right handle are connected
by pin A; the right blade and the left handle are connected by pin B.
Determine the magnitude of the forces exerted on the small branch
E when two 20-lb forces are applied to the handles as shown.

SOLUTION

Note: By symmetry the vertical components of pin forces C and D

are zero.

FBD handle ACF: (not to scale)

ΣFy = 0: Ay = 0

ΣM C = 0: (13.5 in.)( 20 lb ) − (1.5 in.) Ax = 0 Ax = 180 lb

ΣFx = 0: C − Ax − 20 lb = 0 C = (180 + 20 ) lb = 200 lb

FBD blade DE:

ΣM D = 0: ( 9 in.) E − ( 3 in.)(180 lb ) = 0

E = 60.0 lb W