Statics Final Exam Practice WITH SOLUTIONS

410 CHAPTER 8 FRICTION



EXAMPLE 8.1
The uniform crate shown in Fig. 8–7a has a mass of 20 kg. If a force
P = 80 N is applied to the crate, determine if it remains in equilibrium.
The coefficient of static friction is ms = 0.3.


0.8 m
P  80 N


30

0.2 m




(a)
Fig. 8–7


196.2 N SOLUTION
P  80 N 0.4 m 0.4 m Free-Body Diagram. As shown in Fig. 8–7b, the resultant normal
force NC must act a distance x from the crate’s center line in order to
30 counteract the tipping effect caused by P. There are three unknowns,
F, N C, and x, which can be determined strictly from the three equations
0.2 m of equilibrium.
O
F
x Equations of Equilibrium.
+ Fx = 0;
S 80 cos 30 N - F = 0
NC
+ c Fy = 0; -80 sin 30 N + N C - 196.2 N = 0
8 (b)
a + M O = 0; 80 sin 30 N(0.4 m) - 80 cos 30 N(0.2 m) + N C (x) = 0

Solving,


F = 69.3 N
N C = 236.2 N
x = -0.00908 m = -9.08 mm


Since x is negative it indicates the resultant normal force acts (slightly)
to the left of the crate’s center line. No tipping will occur since
x 6 0.4 m. Also, the maximum frictional force which can be developed
at the surface of contact is Fmax = ms N C = 0.3(236.2 N) = 70.9 N.
Since F = 69.3 N 6 70.9 N, the crate will not slip, although it is very
close to doing so.

, 8.2 PROBLEMS INVOLVING DRY FRICTION 411


EXAMPLE 8.2
It is observed that when the bed of the dump truck is raised to an
angle of u = 25 the vending machines will begin to slide off the bed,
Fig. 8–8a. Determine the static coefficient of friction between a
vending machine and the surface of the truckbed.




SOLUTION
An idealized model of a vending machine resting on the truckbed is (a)
shown in Fig. 8–8b. The dimensions have been measured and the (© Russell C. Hibbeler)
center of gravity has been located. We will assume that the vending
machine weighs W.
1.5 ft
1.5 ft
Free-Body Diagram. As shown in Fig. 8–8c, the dimension x is used
to locate the position of the resultant normal force N. There are four
unknowns, N, F, ms, and x.

G
Equations of Equilibrium.

+ RFx = 0; W sin 25 - F = 0 (1) 2.5 ft
u  25
+ QFy = 0; N - W cos 25 = 0 (2)
(b)
a + M O = 0; -W sin 25(2.5 ft) + W cos 25(x) = 0 (3)

Since slipping impends at u = 25, using Eqs. 1 and 2, we have 1.5 ft
1.5 ft
8

Fs = ms N; W sin 25 = ms(W cos 25)
W 25
ms = tan 25 = 0.466 Ans.
G
The angle of u = 25 is referred to as the angle of repose, and by
comparison, it is equal to the angle of static friction, u = fs. Notice x
from the calculation that u is independent of the weight of the vending
O 2.5 ft
machine, and so knowing u provides a convenient method for
determining the coefficient of static friction. F
N
(c)
NOTE: From Eq. 3, we find x = 1.17 ft. Since 1.17 ft 6 1.5 ft, indeed
the vending machine will slip before it can tip as observed in Fig. 8–8a. Fig. 8–8

, 412 CHAPTER 8 FRICTION



EXAMPLE 8.3
The uniform 10-kg ladder in Fig. 8–9a rests against the smooth wall
at B, and the end A rests on the rough horizontal plane for which the
coefficient of static friction is ms = 0.3. Determine the angle of
inclination u of the ladder and the normal reaction at B if the ladder is
on the verge of slipping.




B


4m




u
A


(a)


Fig. 8–9


NB SOLUTION
Free-Body Diagram. As shown on the free-body diagram, Fig. 8–9b,
10(9.81) N
the frictional force FA must act to the right since impending motion at A
is to the left.

8 (4 m) sin u Equations of Equilibrium and Friction. Since the ladder is on the
verge of slipping, then FA = msN A = 0.3N A . By inspection, N A can be
obtained directly.
u FA + c Fy = 0; N A - 10(9.81) N = 0 N A = 98.1 N
A Using this result, FA = 0.3(98.1 N) = 29.43 N. Now N B can be found.
NA (2 m) cos u (2 m) cos u + Fx = 0;
S 29.43 N - N B = 0
(b)
N B = 29.43 N = 29.4 N Ans.
Finally, the angle u can be determined by summing moments about
point A.
a + M A = 0; (29.43 N)(4 m) sin u - [10(9.81) N](2 m) cos u = 0
sin u
= tan u = 1.6667
cos u
u = 59.04 = 59.0 Ans.

, 8.2 PROBLEMS INVOLVING DRY FRICTION 413


EXAMPLE 8.4
Beam AB is subjected to a uniform load of 200 N>m and is supported
at B by post BC, Fig. 8–10a. If the coefficients of static friction at B
and C are mB = 0.2 and mC = 0.5, determine the force P needed to 200 N/m
pull the post out from under the beam. Neglect the weight of the A
members and the thickness of the beam. B
4m 0.75 m
P
0.25 m
C
SOLUTION (a)
Free-Body Diagrams. The free-body diagram of the beam is shown
in Fig. 8–10b. Applying M A = 0, we obtain N B = 400 N. This result
is shown on the free-body diagram of the post, Fig. 8–10c. Referring to
this member, the four unknowns FB , P, FC, and N C are determined from
the three equations of equilibrium and one frictional equation applied
either at B or C.

Equations of Equilibrium and Friction.

S+ Fx = 0; P - FB - FC = 0 (1)
+ c Fy = 0; N C - 400 N = 0 (2)
a + M C = 0; -P(0.25 m) + FB (1 m) = 0 (3)

(Post Slips at B and Rotates about C.) This requires FC … mCN C and
FB = mB N B ; FB = 0.2(400 N) = 80 N 800 N

Using this result and solving Eqs. 1 through 3, we obtain Ax
A FB
P = 320 N 2m 2m
Ay NB  400 N
FC = 240 N
(b) 8
N C = 400 N
Since FC = 240 N 7 mCN C = 0.5(400 N) = 200 N, slipping at C
occurs. Thus the other case of movement must be investigated.

(Post Slips at C and Rotates about B.) Here FB … mB N B and
400 N
B
FC = mCN C; FC = 0.5N C (4) FB
0.75 m
Solving Eqs. 1 through 4 yields P
0.25 m
P = 267 N Ans. FC
C
NC = 400 N NC
FC = 200 N
FB = 66.7 N (c)

Obviously, this case occurs first since it requires a smaller value for P. Fig. 8–10