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Work Example 8.1 on page 390 of your textbook, but replacing the book values with the following values: The box dimensions are 0.9 m wide x 0.5 m high, the magnitude of the pushing force is 90 N and the angle of the push is 35 deg. The push is applied at 0.26 m above the ground. The coefficient of static friction is μ = 0.32.

Work Example 8.1 on page 390 of your textbook, but replacing the book values with the following values: The box dimensions are 0.9 m wide x 0.5 m high, the magnitude of the pushing force is 90 N and the angle of the push is 35 deg. The push is applied at 0.26 m above the ground. The coefficient of static friction is μ = 0.32.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
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Work Example 8.1 on page 390 of your textbook, but replacing the book values with the following values: The box dimensions are 0.9 m wide x 0.5 m high, the magnitude of the pushing force is 90 N and the angle of the push is 35 deg. The push is applied at 0.26 m above the ground. The coefficient of static friction is μ = 0.32.
Transcribed Image Text:Work Example 8.1 on page 390 of your textbook, but replacing the book values with the following values: The box dimensions are 0.9 m wide x 0.5 m high, the magnitude of the pushing force is 90 N and the angle of the push is 35 deg. The push is applied at 0.26 m above the ground. The coefficient of static friction is μ = 0.32.
8 EXAMPLE P = 80 N 30°/ 0.2 m 8.1 -0.4 m 196.2 N (b) Fig. 8-7 -0.4 m Nc The uniform crate shown in Fig. 8-7a has a mass of 20 kg. If a force P= 80N is applied to the crate, determine if it remains in equilibrium. The coefficient of static friction is μ = 0.3. P = 80 N Touren que 30% Solving, 0.2 m -0.8 m JQESSES (a) Equations of Equilibrium. + ΣΕ = 0; +1ΣΕ = 0; (+ΣΜΟ = 0; L SOLUTION Free-Body Diagram. As shown in Fig. 8-7b, the resultant normal force Nc must act a distance x from the crate's centerline in order to counteract the tipping effect caused by P. There are three unknowns, F, Nc, and x, which can be determined strictly from the three equations of equilibrium. 80 cos 30° N - F = 0 -80 sin 30° N + Nc 196.2 N = 0 nica 80 sin 30° N(0.4 m) - 80 cos 30° N(0.2 m) + Nc(x) = 0 F = 69.3 N Nc = 236.2 N x = -0.00908 m = -9.08 mm wake of Since x is negative it indicates the resultant normal force acts (slightly) to the left of the crate's centerline. Also, the maximum frictional force which can be developed at the surface of contact is Fmax = MsNc = 0.3(236.2 N) = = 70.9 N. Since F = 69.3 N < 70.9 N, the crate will not slip, although it is very close to doing so. Det An idealize SOLUTION is shown in center of Free-Bo Ludoor Equa +2
Transcribed Image Text:8 EXAMPLE P = 80 N 30°/ 0.2 m 8.1 -0.4 m 196.2 N (b) Fig. 8-7 -0.4 m Nc The uniform crate shown in Fig. 8-7a has a mass of 20 kg. If a force P= 80N is applied to the crate, determine if it remains in equilibrium. The coefficient of static friction is μ = 0.3. P = 80 N Touren que 30% Solving, 0.2 m -0.8 m JQESSES (a) Equations of Equilibrium. + ΣΕ = 0; +1ΣΕ = 0; (+ΣΜΟ = 0; L SOLUTION Free-Body Diagram. As shown in Fig. 8-7b, the resultant normal force Nc must act a distance x from the crate's centerline in order to counteract the tipping effect caused by P. There are three unknowns, F, Nc, and x, which can be determined strictly from the three equations of equilibrium. 80 cos 30° N - F = 0 -80 sin 30° N + Nc 196.2 N = 0 nica 80 sin 30° N(0.4 m) - 80 cos 30° N(0.2 m) + Nc(x) = 0 F = 69.3 N Nc = 236.2 N x = -0.00908 m = -9.08 mm wake of Since x is negative it indicates the resultant normal force acts (slightly) to the left of the crate's centerline. Also, the maximum frictional force which can be developed at the surface of contact is Fmax = MsNc = 0.3(236.2 N) = = 70.9 N. Since F = 69.3 N < 70.9 N, the crate will not slip, although it is very close to doing so. Det An idealize SOLUTION is shown in center of Free-Bo Ludoor Equa +2
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