Page 154 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 11] EQUILIBRIUM 139

Now we calculate the torques produced by F 2y , F 2x , and w, respectively, about the upper end of the ladder:
τ 1 =−F 2y L 1 =−(50 lb)(12 ft)(cos 56 ) =−336 lb·ft
◦
τ 2 = F 2x L 2 = (F 2x )(10 ft) = 10F 2x ft
τ 3 = w L 3 = (50 lb)(6ft)(cos 56 ) = 168 lb·ft
◦
Applying the condition for rotational equilibrium yields
τ = τ 1 + τ 2 + τ 3 = 0
−336 lb·ft + 10F 2x ft + 168 lb·ft = 0
F 2x = 16.8lb

The total force the ground exerts on the ladder is

2
2
F 2 = F + F = 53 lb
2y
2x
The force the ladder exerts on the ground has the same magnitude.
(b) The force the ladder exerts on the wall is equal in magnitude to F 2x , namely, 16.8 lb.
SOLVED PROBLEM 11.13
The front wheels of a truck support 8 kN, and its rear wheels support 14 kN. The axles are 4 m apart.
Where is the center of gravity of the truck located?
With x the distance between the front axle and the center of gravity, as in Fig. 11-14, calculating torques about
the center of gravity yields
τ = w 1 x − w 2 (4m − x) = 0
(8kN)x − 14(4m − x) kN = 0
22x kN = 56 kN·m
x = 2.55 m

Fig. 11-14

FINDING A CENTEROFGRAVITY
The center of gravity (CG) of an object of regular form and uniform composition is located at its geometric
center. In the case of a complex object, the way to find its center of gravity is to consider it as a system of separate
particles and then find the balance point of the system. An example is the massless rod of Fig. 11-15, which has
three particles m 1 , m 2 , and m 3 attached to it. The CG of the system is at a distance X from the end of the rod