Problem 1:
The uniform 20-lb ladder rests on the rough floor for which
the coefficient of static friction is μs = 0.8 and against the
smooth wall at B. Determine the horizontal force P the man
must exert on the ladder in order to cause it to move.
Solution:
Assume the ladder tip s about A
 NB  0
NB
20 lb



 Fx  0
 P  FA  0
 P  FA
   Fy  0
FA
 20  N A  0
 N A  20lb
  M
A
NA
0
20(3)  P (4)  0
 P  15lb
FA  P
 FA  15lb
( FA ) max   s ( N A )  (0.8)( 20)
 ( FA ) max  16lb  15lb
OK
Thus the ladder tip s as assumed.
P  15lb
Problem 2:
A drum has a weight of 100 lb and rests on
the floor for which the coefficient of static
friction is μs = 0.5. If a = 3ft and b = 4 ft,
determine the smallest magnitude of the force
P that will cause impeding motion of the
drum.
Solution:
Assuming that the drum slips, we first draw
FBD and solve for the unknown:



 Fx  0
4
 0 .5 N  P    0
5
8
 N  P 
5
P
3 5
4
100 lb
1.5 ft 1.5 ft
4 ft
   Fy  0
3
 P   100  N  0
5
3
8
  P   100  P   0
5
5
 P  100lb
8
 N  100   160lb
5
  M
O
0
 3
4
160( x)  100 (1.5)  100 (4)  0
5
5
 x  1.44 ft  1.5 ft
OK
 Drum
slips
as assumed
Since x is smaller than the radius of the drum,
the calculated value of P is the smallest
magnitude of force for impending motion of
the drum.
F=sN
x
N
Problem 3:
A force P=20 lb is applied perpendicular to the
handle of the gooseneck wrecking bar as shown. If
the coefficient of static friction between the bar and
the wood is μs = 0.5, determine the normal force of
the tines at A on the upper board. Assume the
surface at C is smooth.
Solution:
We start with the FBD and write the force
equilibrium equations and moment equation about
point C:
 Fx  0
P = 20lb
20 cos 30  FA  0
 FA  17.32lb
 M C  0
NA
FA
N A (3)  (17.32)(1)  20 cos 30( 20)  20 sin 30(6)  0
 N A  129.7lb  130lb
Fmax  (  s )( N A )
 Fmax  0.5(129.7 )  64.8lb  17.32lb
 The bar won' t slip
NC
Problem 4:
The uniform 6-kg slender rod rests on the top center of
the 3-kg block. If the coefficients of static friction at the
points of contact are μA = 0.4, μB = 0.6, and μC = 0.3,
determine the largest couple moment M which can be
applied to the rod without causing motion of the rod.
Solution:
Rod :



 Fx  0
 FB  N C  0
 FB  N C
(1)
   Fy  0
 N B  FC  58.86  0
  M
B
( 2)
0
 FC (0.6)  N C (0.8)  M  58.86(0.3)  0
Box :
FC
(3)
   Fy  0
6(9.81)=58.86N
 N A  N B  29.43  0 (4)



 Fx  0
 FA  FB  0
 FA  FB
  M
O
(5)
0
FB
 FB (0.3)  N B ( x )  29.43( x)  0
(6)
Assuming slipping occurs at point C and the block tips
FC = μs(C) NC = 0.3 NC and x = 0.1 m.
Substituting these values into equations (1) through (6):
M  8.561N .m  8.56 N .m
N B  50.83N
N A  80.26 N
FA  FB  N C  26.75 N
FC  0.3N C  0.3(26.75)  8 N
Since (FA)max = μs(A) NA = 0.4 (80.26) = 32.11 N
(FA)max =32.11 N > FA = 26.75 N,
The block does not slip.
Also, (FB)max = μs(B) NB = 0.6 (50.83) = 30.50 N
(FB)max =30.50 N > FB = 26.75 N,
Hence, slipping does not occur at point B. The
assumption that slipping occurs at C is correct.
NB
NB
FB
3(9.81)=29.43N
0.3m
O
NA
x
FA
NC