PRACTICE ITUse the worked example above to help you solve this problem. In this example, we show how to find thelocation of a person's center of gravity. Suppose your lab partner has a height L of 171 cm (5 ft, 7 in) anda weight w of 662 N (149 lb). You can determine the position of his center of gravity by having himstretch out on a uniform board supported at one end by a scale, as shown in the figure. If the board'sweight wis 54 N and the scale reading F is 2.85 x 102 N, find the distance of your lab partner's center ofgravity from the left end of the board.cmEXERCISEHINTS: GETTING STARTED T I'M STUCK!Suppose a 447-kg alligator of length 3.8 m is stretched out on a board of the same length weighing 90 N.If the board is supported on the ends as shown in the figure, and the scale reads 1,840 N, find the x-component of the alligator's center of gravity. (Let x = 0 correspond to the left end of the board.)m EXAMPLE 8.5 Locating Your Lab Partner's Center of GravityGOAL Use torque to find a centerof gravity.PROBLEM In this example weshow how to find the location of aperson's center of gravity. Supposeyour lab partner has a height L of173 cm (5 ft, 8 in) and a weight w of715 N (160 lb). You can determinethe position of his center of gravityby having him stretch out on auniform board supported at one endby a scale, as shown in the figure. If the board's weight w, is 49 N and the scale reading F is 3.50 × 10²N, find the distance of your lab partner's center of gravity from the left end of the board.taSOLUTIONApply the second condition ofequilibrium.Substitute expressions for the torques.Solve for xcg and substitute knownvalues.L/2XcgDetermining your lab partner's center of gravity.cgSTRATEGY To find the position x of the center of gravity, compute the torques using an axis throughO. There is no torque due to the normal force n because its moment arm is zero about an axis through O.Set the sum of the torques equal to zero and solve for x,*cg.Στ; = tn + Tw + Twb + TF: = 0Xcg0 - wxcg-w(L/2) + FL = 0FL - W (L/2)W(350 N) (173 cm) - (49 N) (86.5 cm) = 79 cm715 N

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Description: PRACTICE ITUse the worked example above to help you solve this problem. In this example, we show how to find thelocation of a person's center of gravity. Suppose your lab partner has a height L of 171 cm (5 ft, 7 in) anda weight w of 662 N (149 lb).

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PRACTICE IT Use the worked example above to help you solve this 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 171 cm (5 ft, 7 in) and a weight w of 662 N (149 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 the figure. If the board's weight w, is 54 N and the scale reading F is 2.85 x 102 N, find the distance of your lab partner's center of gravity from the left end of the board. cm EXERCISE HINTS: GETTING STARTED I'M STUCK! Suppose a 447-kg alligator of length 3.8 m is stretched out on a board of the same length weighing 90 N. If the board is supported on the ends as shown in the figure, and the scale reads 1,840 N, find the x- component of the alligator's center of gravity. (Let x = 0 correspond to the left end of the board.) m

PRACTICE IT Use the worked example above to help you solve this 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 171 cm (5 ft, 7 in) and a weight w of 662 N (149 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 the figure. If the board's weight w, is 54 N and the scale reading F is 2.85 x 102 N, find the distance of your lab partner's center of gravity from the left end of the board. cm EXERCISE HINTS: GETTING STARTED I'M STUCK! Suppose a 447-kg alligator of length 3.8 m is stretched out on a board of the same length weighing 90 N. If the board is supported on the ends as shown in the figure, and the scale reads 1,840 N, find the x- component of the alligator's center of gravity. (Let x = 0 correspond to the left end of the board.) m

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

PRACTICE IT
Use the worked example above to help you solve this 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 171 cm (5 ft, 7 in) and
a weight w of 662 N (149 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 the figure. If the board's
weight w
is 54 N and the scale reading F is 2.85 x 102 N, find the distance of your lab partner's center of
gravity from the left end of the board.
cm
EXERCISE
HINTS: GETTING STARTED T I'M STUCK!
Suppose a 447-kg alligator of length 3.8 m is stretched out on a board of the same length weighing 90 N.
If the board is supported on the ends as shown in the figure, and the scale reads 1,840 N, find the x-
component of the alligator's center of gravity. (Let x = 0 correspond to the left end of the board.)
m

EXAMPLE 8.5 Locating Your Lab Partner's Center of Gravity
GOAL Use torque to find a center
of gravity.
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
173 cm (5 ft, 8 in) and a weight w of
715 N (160 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 the figure. If the board's weight w, is 49 N and the scale reading F is 3.50 × 10²
N, find the distance of your lab partner's center of gravity from the left end of the board.
ta
SOLUTION
Apply the second condition of
equilibrium.
Substitute expressions for the torques.
Solve for xcg and substitute known
values.
L/2
Xcg
Determining your lab partner's center of gravity.
cg
STRATEGY To find the position x of the center of gravity, compute the torques using an axis through
O. There is no torque due to the normal force n because its moment arm is zero about an axis through O.
Set the sum of the torques equal to zero and solve for x,
*cg.
Στ; = tn + Tw + Twb + TF: = 0
Xcg
0 - wxcg-w(L/2) + FL = 0
FL - W (L/2)
W
(350 N) (173 cm) - (49 N) (86.5 cm) = 79 cm
715 N

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