Problem 1:
The beam is adjusted to the horizontal
position by means of a wedge located at its
right support. If the coefficient of static
friction between the wedge and the two
surfaces of contact is μs = 0.25, determine
the horizontal force P required to push the
wedge forward. Neglect the weight and
size of the wedge and the thickness of the
beam.
300 lb
6 ft
2 ft
Ax A
Solution:
If the wedge is on the verge of moving to
the right, then slipping will have to occur
Ay
at both contact surfaces.
FB = μsNB = 0.25NB
FC = μsNC = 0.25NC
B
NB
NB
FB = 0.25 NB
P
FC=0.25NC 70
NC
Beam :
  M
A
0
N B (8)  300(2)  0
 N B  75lb
Wedge :
   Fy  0
N C (sin 70)  0.25 N C (cos 70)  75  0
 N C  87.8lb



 Fx  0
P  0.25(75)  0.25(87.8)(sin 70)  87.8(cos 70)  0
 P  69.4lb
FB=0.25NB
Problem 2:
If the beam AD is loaded as shown,
determine the horizontal force P
which must be applied to the wedge
in order to remove it from under the
beam. The coefficients of static
friction at the wedge’s top and
bottom surfaces are μC-A = 0.25 and
μC-B = 0.35, respectively. If P=0, is
the wedge self-locking? Neglect the
weight and size of the wedge and
the thickness of the beam.
Solution:
The equivalent force for triangular
distributed force is
(1/2)(4)(3)=6
Dx
and also for the uniform one
(4)(4)=16
If the wedge is on the verge of
moving to the right, then slipping
will have to occur at both contact
surfaces.
FA = μC-ANA = 0.25NA
FB = μC-BNB = 0.35NB
(4)(4) =
16 kN
(1/2)(4)(3) =
6 kN
FA= 0.25NA
10
Dy
2m
2m
3m
NA
NA
80
FA= 0.25NA
P
FB = 0.35 NB
NB
Beam :
  M
D
0
N A cos 10(7)  0.25 N A sin 10(7)  (6)( 2)  16(5)  0
 N A  12.78kN
Wedge :
   Fy  0
N B  12.78(sin 80)  0.25(12.78)(sin 10)  0
 N B  13.14kN



 Fx  0
P  (12.78)(cos 80)  0.25(12.78)(cos 10)  0.35(13.14)  0
 P  5.53kN
Since a force P = 5.53 kN > 0 is required to pull out the wedge, the wedge will be self-locking
when P = 0.