Can the system in Fig. (a) be in static equilibrium in the position shown? The uniform bar AB weighs 500 lb, and the weight of block C is 300 lb. Friction at A is negligible, and the coefficient of static friction is 0.4 at the other two contact surfaces. 10 ft Solution 30° Regause it ie not knoun uhether motion impende we identifu thie oe o Tvne I 50°
Can the system in Fig. (a) be in static equilibrium in the position shown? The uniform bar AB weighs 500 lb, and the weight of block C is 300 lb. Friction at A is negligible, and the coefficient of static friction is 0.4 at the other two contact surfaces. 10 ft Solution 30° Regause it ie not knoun uhether motion impende we identifu thie oe o Tvne I 50°
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
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.
![Sample Problem
7.4
Can the system in Fig. (a) be in static equilibrium in the position shown? The
uniform bar AB weighs 500 lb, and the weight of block C is 300 lb. Friction at A
is negligible, and the coefficient of static friction is 0.4 at the other two contact
L.
surfaces.
10 ft
Solution
B
30
50°
Because it is not known whether motion impends, we identify this as a Type I
problem. Note that the FBDS of the bar and the block, Figs. (b) and (c), contain
five unknowns: NA, NB, FB, Nc, and Fc.
(a)
Assume Equilibrium
Under this assumption, there are five equilibrium equations: three for the bar
AB and two for the block C. The unknowns may be computed by the following
procedure.
ft
40°
5 ft
FBD of AB [Fig. (b)]
30
500 lb
FB
EMB = 0 3 NẠ sin 40° (10 cos 30°) + Na cos 40°(10 sin 30°)
Ng
-500(5 cos 30°) = 0
(b)
NA = 230.4 lb
ΣF, 0
+ FB - NA cOs 40° = 0
Ng
300 lb
FB = 230.4 cos 40° = 176.50 lb
EF, = 0 1 Ng + NA sin 40° – 500 = 0
NB = -230.4 sin 40° + 500 = 351.9 lb
Ne
FBD of Block C [Fig. (c)]
(c)
EF, = 0 1 Nc – Ng – 300 = 0
Nc = 351.9 + 300 = 651.9 lb
EF, = 0 +> Fc – FB = 0
Fc = FB = 176.50 lb
Check
To check the assumption of equilibrium, we must compare each of the friction
forces against its maximum static value.
(FB)max = 0.4Ng = 0.4(351.9) = 140.76 lb < Fg = 176.50 lb
Answer
(Fc)max = 0.4Nc = 0.4(651.9) = 260.8 lb > Fc = 176.50 lb
We conclude that the system cannot be in equilibrium. Although there is sufficient
friction beneath B, the friction force under C exceeds its limiting value.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F916f8f7d-e1c0-41dd-9d3c-973f8ad5cbd9%2F9fe0b435-b393-4443-82bc-c51ee9c2d555%2F2yexfgm_processed.png&w=3840&q=75)
Transcribed Image Text:Sample Problem
7.4
Can the system in Fig. (a) be in static equilibrium in the position shown? The
uniform bar AB weighs 500 lb, and the weight of block C is 300 lb. Friction at A
is negligible, and the coefficient of static friction is 0.4 at the other two contact
L.
surfaces.
10 ft
Solution
B
30
50°
Because it is not known whether motion impends, we identify this as a Type I
problem. Note that the FBDS of the bar and the block, Figs. (b) and (c), contain
five unknowns: NA, NB, FB, Nc, and Fc.
(a)
Assume Equilibrium
Under this assumption, there are five equilibrium equations: three for the bar
AB and two for the block C. The unknowns may be computed by the following
procedure.
ft
40°
5 ft
FBD of AB [Fig. (b)]
30
500 lb
FB
EMB = 0 3 NẠ sin 40° (10 cos 30°) + Na cos 40°(10 sin 30°)
Ng
-500(5 cos 30°) = 0
(b)
NA = 230.4 lb
ΣF, 0
+ FB - NA cOs 40° = 0
Ng
300 lb
FB = 230.4 cos 40° = 176.50 lb
EF, = 0 1 Ng + NA sin 40° – 500 = 0
NB = -230.4 sin 40° + 500 = 351.9 lb
Ne
FBD of Block C [Fig. (c)]
(c)
EF, = 0 1 Nc – Ng – 300 = 0
Nc = 351.9 + 300 = 651.9 lb
EF, = 0 +> Fc – FB = 0
Fc = FB = 176.50 lb
Check
To check the assumption of equilibrium, we must compare each of the friction
forces against its maximum static value.
(FB)max = 0.4Ng = 0.4(351.9) = 140.76 lb < Fg = 176.50 lb
Answer
(Fc)max = 0.4Nc = 0.4(651.9) = 260.8 lb > Fc = 176.50 lb
We conclude that the system cannot be in equilibrium. Although there is sufficient
friction beneath B, the friction force under C exceeds its limiting value.
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