Can you explain why does the moment of x-component of force P is negative? Explain and illustrate how to determine if a component will cause a positive or negative moment.

College Physics
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ISBN:9781938168000
Author:Paul Peter Urone, Roger Hinrichs
Publisher:Paul Peter Urone, Roger Hinrichs
Chapter4: Dynamics: Force And Newton's Laws Of Motion
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Problem 21PE: Show that, as stated in the text, a force F exerted on a flexible medium at its center and...
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Can you explain why does the moment of x-component of force P is negative? Explain and illustrate how to determine if a component will cause a positive or negative moment. 

The uniform crate shown in Fig. 8–7a has a mass of 20 kg. If a force
P = 80 N is applied to the crate, determine if it remains in equilibrium.
The coefficient of static friction is µ, = 0.3.
-0.8 m
P = 80 N
30
0.2 m
(a)
Fig. 8–7
196.2 N
SOLUTION
Free-Body Diagram. As shown in Fig. 8–7b, the resultant normal
force N. must act a distance x from the crate's center line 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.
P = 80 N
-0.4 m
-0.4 m
30°
0.2 m
Equations of Equilibrium.
ΣF, 0
80 cos 30° N – F = 0
+↑£F, = 0;
-80 sin 30° N + Nc – 196.2 N = 0
8
(b)
6+EMo = 0; 80 sin 30° N(0.4 m) – 80 cos 30° N(0.2 m) + Nc(x) = 0
Transcribed Image Text:The uniform crate shown in Fig. 8–7a has a mass of 20 kg. If a force P = 80 N is applied to the crate, determine if it remains in equilibrium. The coefficient of static friction is µ, = 0.3. -0.8 m P = 80 N 30 0.2 m (a) Fig. 8–7 196.2 N SOLUTION Free-Body Diagram. As shown in Fig. 8–7b, the resultant normal force N. must act a distance x from the crate's center line 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. P = 80 N -0.4 m -0.4 m 30° 0.2 m Equations of Equilibrium. ΣF, 0 80 cos 30° N – F = 0 +↑£F, = 0; -80 sin 30° N + Nc – 196.2 N = 0 8 (b) 6+EMo = 0; 80 sin 30° N(0.4 m) – 80 cos 30° N(0.2 m) + Nc(x) = 0
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