Sample Problem 7.1 The 100-1b block in Fig. (a) is initially at rest on a horizontal plane. Determine the friction force between the block and the surface after P was gradually increased W- 100 from 0 to 30 lb. Solution A-02 From the problem statement we conclude that this is a Type I problem (impending motion is not specificd). Furthermore, we do not know if the block will even remain at rest in static equilibrium when P = 30 lb. (a) Assume Equilibrium 100 Once we have assumed that the body remains at rest, the equilibrium equations for the free-body diagram in Fig. (b) can be used to calculate the two unknowns (N and F), as follows. P-30 b EF, =0 1 N- 100 = 0 N = 100 Ib EF, =0 P- F = 0 F=P = 30 lb Before we can accept this solution, the assumption of equilibrium must be checked. Check The maximum static friction force is Fa = H.N = 0.5(100) = 50 Ib Because F< Fa, we conclude that the block is in static equilibrium, and the correct value of the friction force is F = 30 lb Answer Comment If the coefficient of static friction had been 0.25, instead of 0.5, the block would not be in equilibrium. The 30-lb friction force required for equilibrium would be greater than Faux = 0.25(100) = 25 lb; therefore, the friction force would be F= F, = HN = 0.2(100) = 20 lb and the block would be sliding to the right.
Sample Problem 7.1 The 100-1b block in Fig. (a) is initially at rest on a horizontal plane. Determine the friction force between the block and the surface after P was gradually increased W- 100 from 0 to 30 lb. Solution A-02 From the problem statement we conclude that this is a Type I problem (impending motion is not specificd). Furthermore, we do not know if the block will even remain at rest in static equilibrium when P = 30 lb. (a) Assume Equilibrium 100 Once we have assumed that the body remains at rest, the equilibrium equations for the free-body diagram in Fig. (b) can be used to calculate the two unknowns (N and F), as follows. P-30 b EF, =0 1 N- 100 = 0 N = 100 Ib EF, =0 P- F = 0 F=P = 30 lb Before we can accept this solution, the assumption of equilibrium must be checked. Check The maximum static friction force is Fa = H.N = 0.5(100) = 50 Ib Because F< Fa, we conclude that the block is in static equilibrium, and the correct value of the friction force is F = 30 lb Answer Comment If the coefficient of static friction had been 0.25, instead of 0.5, the block would not be in equilibrium. The 30-lb friction force required for equilibrium would be greater than Faux = 0.25(100) = 25 lb; therefore, the friction force would be F= F, = HN = 0.2(100) = 20 lb and the block would be sliding to the right.
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 with, how the problem was solved, the assumptions that the author considered in solving the problem, and why where these assumptions were made. It should have 5 sentences for each problem.
Note: You can answer only the 7.1 and 7.2
Transcribed Image Text:Sample Problem 7.1
The 100-lb block in Fig. (a) is initially at rest on a horizontal plane. Determine the
friction force between the block and the surface after P was gradually increased
P
W = 100 Ib
from 0 to 30 lb.
Solution
H, = 0.5
A=0.2
From the problem statement we conclude that this is a Type I problem (impending
motion is not specified). Furthermore, we do not know if the block will even
remain at rest in static equilibrium when P = 30 lb.
(a)
Assume Equilibrium
100 Ib
Once we have assumed that the body remains at rest, the equilibrium equations
for the free-body diagram in Fig. (b) can be used to calculate the two unknowns
(N and F), as follows.
P= 30 lb
EF, =0 1 N – 100 = 0
N = 100 lb
EF, =0
P-F =0
(b)
F = P = 30 lb
Before we can accept this solution, the assumption of equilibrium must be
checked.
Check
The maximum static friction force is
Faax = 4. N = 0.5(100) = 50 lb
Because F < Fmar we conclude that the block is in static equilibrium, and the
correct value of the friction force is
F = 30 lb
Answer
Comment
If the coefficient of static friction had been 0.25, instead of 0.5, the block would
not be in equilibrium. The 30-lb friction force required for equilibrium would be
greater than Fmax = 0.25(100) = 25 Ib; therefore, the friction force would be
F = F, = H,N = 0.2(100) = 20 Ib
and the block would be sliding to the right.
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