CHAPTER 16
PROBLEM 16.1
A 60–lb uniform thin panel is placed in a truck with end A resting
on a rough horizontal surface and end B supported by a smooth
vertical surface. Knowing that the deceleration of the truck is 12
2
ft/s ,
determine (a) the reactions at ends A and B, (b) the
minimum required coefficient of static friction at end A.
PROBLEM 16.2
A 60–lb uniform thin panel is placed in a truck with end A resting
on a rough horizontal surface and end B supported by a smooth
vertical surface. Knowing that the panel remains in the position
shown, determine (a) the maximum allowable acceleration of the
truck, (b) the corresponding minimum required coefficient of
static friction at end A.
PROBLEM 16.3
Knowing that the coefficient of static friction between the tires
and the road is 0.80 for the automobile shown, determine the
maximum possible acceleration on a level road, assuming (a) four–
wheel drive, (b) rear–wheel drive, (c) front–wheel drive.
PROBLEM 16.3 (Continued)
(c) Front-wheel drive:
eff
( ) : (100 in.) (60 in.) (20 in.)
AA B
M M N W maΣ=Σ − =−
PROBLEM 16.4
The motion of the 2.5–kg rod AB is guided by two small wheels which roll
freely in horizontal slots. If a force P of magnitude 8 N is applied at B,
determine (a) the acceleration of the rod, (b) the reactions at A and B.
PROBLEM 16.5
A uniform rod BC of mass 4 kg is connected to a collar A by a 250–mm
cord AB. Neglecting the mass of the collar and cord, determine (a) the
smallest constant acceleration
A
a
for which the cord and the rod lie in a
straight line, (b) the corresponding tension in the cord.
PROBLEM 16.6
A 2000–kg truck is being used to
lift a 400–kg boulder B that is on a
50–kg pallet A. Knowing the
acceleration of the rear–wheel drive
truck is 1 m/s2, determine (a) the
reaction at each of the front
wheels, (b) the force between the
boulder and the pallet.
PROBLEM 16.6 (Continued)
Truck:
eff
( ):
RR
MM= Σ
(3.4 m) (2.0 m) (0.6 m) (1.0 m)
F T TT
N mg T ma− + −=
2
(2.0 m)(2000 kg)(9.81 m/s ) (0.6 m)(2320 N) (1.0 m)(2000 kg)(1.0 m/s)
3.4 m 3.4 m 3.4 m
10544 N
F
N= −+
=
eff
( ):
yy
FFΣ=Σ
0
F RT
N N mg+− =
2
10544 N (2000 kg)(9.81 m/s ) 0
9076 N
R
R
N
N
+− =
=
eff
2
():
2320 N (2000 kg)(1.0 m/s )
4320 N
x x R TT
R
F F F T ma
F
Σ =Σ −=
= +
=