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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 W = 100 lb from 0 to 30 lb.

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 W = 100 lb from 0 to 30 lb.

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
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.

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 W = 100 lb from 0 to 30 lb. Solution 4, = 0,5 H = 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 lb 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 IN 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 Fmax = 4,N = 0.5(100) = 50 lb Because F < Fmax, 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 lb; therefore, the friction force would be F = FR = µ&N = 0.2(100) = 20 lb and the block would be sliding to the right.
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 W = 100 lb from 0 to 30 lb. Solution 4, = 0,5 H = 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 lb 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 IN 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 Fmax = 4,N = 0.5(100) = 50 lb Because F < Fmax, 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 lb; therefore, the friction force would be F = FR = µ&N = 0.2(100) = 20 lb and the block would be sliding to the right.
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