GO In Fig. 12-46, a 50.0 kg uniform square sign, of edge length L = 2.00 m, is hung from a horizontal rod of length d h = 3.00 m and negligible mass. A cable is attached to the end of the rod and to a point on the wall at distance d v = 4.00 m above the point where the rod is hinged to the wall. (a) What is the tension in the cable? What are the (b) magnitude and (c) direction (left or right) of the horizontal component of the force on the rod from the wall, and the (d) magnitude and (e) direction (up or down) of the vertical component of this force? Figure 12-46 Problem 30.
GO In Fig. 12-46, a 50.0 kg uniform square sign, of edge length L = 2.00 m, is hung from a horizontal rod of length d h = 3.00 m and negligible mass. A cable is attached to the end of the rod and to a point on the wall at distance d v = 4.00 m above the point where the rod is hinged to the wall. (a) What is the tension in the cable? What are the (b) magnitude and (c) direction (left or right) of the horizontal component of the force on the rod from the wall, and the (d) magnitude and (e) direction (up or down) of the vertical component of this force? Figure 12-46 Problem 30.
GO In Fig. 12-46, a 50.0 kg uniform square sign, of edge length L = 2.00 m, is hung from a horizontal rod of length dh = 3.00 m and negligible mass. A cable is attached to the end of the rod and to a point on the wall at distance dv = 4.00 m above the point where the rod is hinged to the wall. (a) What is the tension in the cable? What are the (b) magnitude and (c) direction (left or right) of the horizontal component of the force on the rod from the wall, and the (d) magnitude and (e) direction (up or down) of the vertical component of this force?
SARET CRKS AUTOWAY
12. A stone is dropped from the top of a cliff. It is seen to hit the ground below
after 3.55 s. How high is the cliff?
13. A ball is dropped from rest at the top of a building that is 320 m tall. Assuming
no air resistance, what is the speed of the ball just before it strikes the ground?
14. Estimate (a) how long it took King Kong to fall straight down from the top
of the Empire State Building (280m high), and (b) his velocity just before
"landing".
Useful equations
For Constant Velocity:
V =>
D
X = V₁t + Xo
For Constant Acceleration:
Vr = V + at
X = Xo+Vot +
v=V+2a(X-Xo)
\prom = V +V
V velocity
t = time
D Distance
X = Final Position
Xo Initial Position
V = Final Velocity
Vo Initial Velocity
a = acceleration
For free fall
Yf
= Final Position
Yo Initial Position
g = 9.80
m
$2
For free fall:
V = V + gt
Y=Yo+Vo t +
+gt
V,² = V₁²+2g (Y-Yo)
V+Vo
Vprom=
2
6
Fundamentals of Anatomy & Physiology (11th Edition)
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