Sample Problem 7.5 Determine the largest and smallest values of the force P for which the system in Fig. (a) will be in static equilibrium. The homogeneous bars AB and BC are identical, each having a mass of 100 kg. The coefficient of static friction between the bar at C and the horizontal plane is 0.5. B 1.5 m 3 m 1.5 m 30° 30 , = 0.5 (a)
Sample Problem 7.5 Determine the largest and smallest values of the force P for which the system in Fig. (a) will be in static equilibrium. The homogeneous bars AB and BC are identical, each having a mass of 100 kg. The coefficient of static friction between the bar at C and the horizontal plane is 0.5. B 1.5 m 3 m 1.5 m 30° 30 , = 0.5 (a)
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Read and Analyze the sample problems given. Give a thorough analysis stating what type of situation the author dealt, how the problem was solved, the assumptions that the author considered in solving the problem, and why where these assumptions made.

Transcribed Image Text:Sample Problem 7.5
Determine the largest and smallest values of the force P for which the system
in Fig. (a) will be in static equilibrium. The homogeneous bars AB and BC are
identical, each having a mass of 100 kg. The coefficient of static friction between
the bar at C and the horizontal plane is 0.5.
B
1.5 m
3 m
1.5 m
30°
30°
-4, = 0.5
(a)
Solution
This is a Type II problem because impending sliding at C is implied. However,
finding the largest and smallest values of P are two separate problems.
Note that the weights of the bars have a tendency to slide C to the right.
Therefore, impending sliding of C to the right corresponds to the smallest P. The
largest P occurs when sliding of C impends to the left; in this case, P must over-
come both the friction and the tendency of the weights to slide C to the right.
Consequently, the only difference between the two problems is the direction of
the friction force at C.
The FBD of the system consisting of both bars is shown in Fig. (b); the
two directions of Fc are indicated by dashed lines. The weight of each bar,
W = mg = 100(9.81)= 981 N, is also shown on the diagram.
1.5 m
1.5 m
1.5 m
30
1.5
A, A
981 N
981 N F
FC
(b)
An equation involving only Nc and P is obtained by summing moments about
A in Fig. (b):
EMA = 0 3 Nc(6cos 30°) + P(1.5 sin 30°)
- 981(1.5 cos 30°) – 981(4.5 cos 30°) = 0
(a)
The FBDS of bar BC corresponding to the largest and smallest values of P are
shown in Figs. (c) and (d), respectively. In both cases, Fc is set equal to (Fc)max
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