Exercise 8.5 Hints: Getting Started | I'm Stuck Suppose a 427 kg alligator of length 3.4 m is stretched out on a board of the same length weighing 95 N. If the board is supported on the ends as in Figure 8.10, and the scale reads 1967 N, find the x-component of the alligator's center of gravity. X m Xcg, alligator = How many forces are exerted on the board? What is the torque exerted by each one? Since the board is at rest, the net torque must be zero.
Exercise 8.5 Hints: Getting Started | I'm Stuck Suppose a 427 kg alligator of length 3.4 m is stretched out on a board of the same length weighing 95 N. If the board is supported on the ends as in Figure 8.10, and the scale reads 1967 N, find the x-component of the alligator's center of gravity. X m Xcg, alligator = How many forces are exerted on the board? What is the torque exerted by each one? Since the board is at rest, the net torque must be zero.
Physics for Scientists and Engineers, Technology Update (No access codes included)
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Chapter12: Static Equilibrium And Elasticity
Section: Chapter Questions
Problem 12.1OQ: The acceleration due to gravity becomes weaker by about three parts in ten million for each meter of...
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Question
Transcribed Image Text:Exercise 8.5
Hints: Getting Started | I'm Stuck
Suppose a 427 kg alligator of length 3.4 m is stretched out on a board of the same
length weighing 95 N. If the board is supported on the ends as in Figure 8.10, and
the scale reads 1967 N, find the x-component of the alligator's center of gravity.
X m
Xcg, alligator
How many forces are exerted on the board? What is the torque exerted by each
one? Since the board is at rest, the net torque must be zero.
Transcribed Image Text:Example 8.5 Locating Your Lab Partner's Center of Gravity
Goal Use torque to find a center of
gravity.
L/2
Problem In this example, we show
how to find the location of a person's
center of gravity. Suppose your lab
partner has a height L of 176 cm (5
ft, 9 in) and a weight w of 696 N
(156 lb). You can determine the
position of his center of gravity by
having him stretch out on a uniform
board supported at one end by a
scale, as shown in Figure 8.10.
Xeg
9.
Figure 8.10 Determining your lab partner's center of gravity.
If the board's weight wh is 44 N and the scale reading F is 3.06 x 102 N, find the distance of your lab partner's
center of gravity from the left end of the board.
Strategy To find the position xcg of the center of gravity, compute the torques using an axis through O. Set the
sum of the torques equal to zero and solve for xca-
Solution
Apply the second condition of equilibrium. There is
no torque due to the normal force n because its
moment arm is zero.
ΣΤο.
= 0
-wXcg - Wь(L/2) + FL %3D 0
=
Solve for xca and substitute known values.
FL – wb (L/2)
xcg
%3|
w
(306.24 N) (176 cm) – (44 N) (88 cm)
|
696 N
Xcg
72
cm
Remarks The given information is sufficient only to determine the x coordinate of the center of gravity. The
other two coordinates can be estimated, based on the body's symmetry.
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