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Unformatted text preview: f the forces in the vertical direction must be zero ( ΣFy = 0) . These two conditions will allow us to determine the magnitudes of F1 and F2.
F1 +y F2
Axis Wh +τ
+x 1
5 L 1
2 L Wb SOLUTION
a. We will begin by taking the axis of rotation about the right end of the bridge. The torque
produced by F2 is zero, since its lever arm is zero. When we set the sum of the torques equal
to zero, the resulting equation will have only one unknown, F1, in it. Setting the sum of the
torques produced by the three forces equal to zero gives
Στ = − F1L + Wh ( 4 L ) + Wb ( 1 L ) = 0
5
2 Algebraically eliminating the length L of the bridge from this equation and solving for F1
gives
4
4
F1 = 5 Wh + 1 Wb = 5 ( 985 N ) + 1 ( 3610 N ) = 2590 N
2
2
b. Since the bridge is in equilibrium, the sum of the forces in the vertical direction must be
zero:
Σ Fy = F1 − Wh − Wb + F2 = 0 Solving for F2 gives
F2 = − F1 + Wh + Wb = −2590 N + 985 N + 3610 N = 2010 N 14. REASONING The net torque is the sum of the torques produced by the three forces:
Στ = τA + τ B + τ D . The magnitude of a torque is the magnitude of the force times the lever
arm of the force, according to Equation 9.1. The lever arm is the perpendicular distance
between the line of action of the force and the axis. A torque that tends to produce a 444 ROTATIONAL DYNAMICS counterclockwise rotation about the axis is a positive torque. Since the piece of wood is at
equilibrium, the net torque is equal to zero.
SOLUTION Let L be the length of the short side of the rectangle, so that the length of the
long side is 2L. The counterclockwise torque produced by the force at corner B is + F l B ,
and the clockwise torque produced by the force at corner D is − F l D . Assuming that the
force at A (directed along the short side of the rectangle) points toward corner B, the
counterclockwise torque produced by this force is + FAl A . Setting the net torque equal to
zero gives:
Στ = FA l A + F l B − F l D = 0 FA L + (12 N ) ( 1 L ) − (12 N ) L = 0
2 The length L can be eliminated algebraically from this result, which can then be solved for
FA:
FA = − (12 N ) ( 1 ) + 12 N =
2 6.0 N (pointing toward corner B) Since the value calculated for FA is positive, our assumption that FA points toward corner B
must have been correct. Otherwise, the result for FA would have been negative. 15. REASONING Since the forearm is in equilibrium, the sum of the torques about any axis of
rotation must be zero ( Στ = 0 ) . For convenience, we will take the elbow joint to be the axis
of rotation.
SOLUTION Let M and F be the magnitudes of the forces that the flexor muscle and test
apparatus, respectively, exert on the forearm, and let l M and l F be the respective lever arms
about the elbow joint. Setting the sum of the torques about the elbow joint equal to zero
(with counterclockwise torques being taken as positive), we have
Στ = M l M − F l F = 0 Solving for M yields
M= F l F (190 N ) ( 0.34 m )
=
= 1200 N
lM
0.054 m The direction of the force is to the left . Chapter 9 Problems 16. REASONING The truck is subject to three
vertical forces only (see the freebody
diagram), and is in equilibrium. Therefore,
the conditions ΣFy = 0 (Equation 4.9b) and
Στ = 0 (Equation 9.2) apply to the forces
and torques acting on it. The ground exerts
upward forces FR on the rear wheels and FF
on the front wheels, and the earth exerts a
downward weight force W on the truck’s
center of gravity. The net vertical force on
the truck must be zero, so we find
FR + FF − W = 0 445 Freebody diagram of the truck W FF FR Rotation axis lR (1) lF where we have assumed upward to be the positive direction. To apply the zero net torque
condition (Equation 9.2), we choose a rotation axis located on the ground, directly below the
truck’s center of gravity (see the diagram). The weight force W has no lever arm about this
axis, and so it generates no torque. The force FF exerts a counterclockwise torque about this
axis, and the force FR exerts a clockwise torque about this axis. Since counterclockwise is
the positive direction, we find from Equation 9.2 that
Στ = FFl F − FR l R = 0 FFl F = FR l R or (2) We will use Equations (1) and (2) to find the two unknown forces. SOLUTION
a. In order to find the force magnitude FF, we must eliminate the unknown force magnitude
FR from Equations (1) and (2). Solving Equation (1) for FR yields FR = W − FF = mg − FF .
Substituting this expression into Equation (2), we obtain FFl F = ( mg − FF ) l R = mgl R − FFl R or FF ( l F + l R ) = mgl R (3) Solving Equation (3) for the force magnitude FF and noting from the drawing in the text that
l F = 2.30 m and lR = 0.63 m, we find that FF = ( ) ( 7460 kg ) 9.80 m/s 2 ( 0.63 m )
mg l R
=
= 1.57 × 104 N
lF + lR
2.30 m + 0.63 m 446 ROTATIONAL DYNAMICS b. Returning to Equation (1), we obtain the magnitude of the force on the rear wheels: ( ) FR = W − FF = mg − FF = ( 7460 kg ) 9.80 m/s 2 − 1.57 × 10 4 N = 5.74 × 10 4 N 17. REASONING MultipleConcept Example 8 discusses the static stability factor (SSF) and
rollover. In that example, it is determined tha...
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 Spring '13
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 Physics, The Lottery

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