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.

International Edition---engineering Mechanics: Statics, 4th Edition
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Chapter7: Dry Friction
Section: Chapter Questions
Problem 7.50P: The block of weight W is pulled by the force P inclined at the angle to the horizontal. Find the...
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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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