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Unformatted text preview: e about the axis of rotation. The force of static friction
that the ground applies to the wheel is labeled as fs. This force
produces a counterclockwise torque τ about the axis of rotation,
which is given by Equation 9.1 as τ = fs l , where l is the lever
arm. Using this relation we can find the magnitude fs of the
static frictional force. Axis of
rotation r
fs SOLUTION The countertorque is given as τ = fs l , where fs is the magnitude of the static
frictional force and l is the lever arm. The lever arm is the distance between the line of
action of the force and the axis of rotation; in this case the lever arm is just the radius r of
the tire. Solving for fs gives
τ 295 N ⋅ m
fs = =
= 843 N
l 0.350 m 3. SSM REASONING According to Equation 9.1, we have
Magnitude of torque = F l
where F is the magnitude of the applied force and l is the lever arm. From the figure in the
text, the lever arm is given by l = (0.28 m) sin 50.0° . Since both the magnitude of the
torque and l are known, Equation 9.1 can be solved for F. Chapter 9 Problems 437 SOLUTION Solving Equation 9.1 for F, we have 4. Magnitude of torque
45 N ⋅ m
=
= 2.1×102 N
l
(0.28 m) sin 50.0° REASONING The net torque on the branch is the
sum of the torques exerted by the children. Each
individual torque τ is given by τ = F l (Equation
9.1), where F is the magnitude of the force exerted
on the branch by a child, and l is the lever arm (see
the diagram). The branch supports each child’s
weight, so, by Newton’s third law, the magnitude F
of the force exerted on the branch by each child has
the same magnitude as the child’s weight: F = mg.
Both forces are directed downwards. The lever arm
for each force is the perpendicular distance between
the axis and the force’s line of action, so we have
l = d cos θ (see the diagram). d Tree F= F θ
l Axis of
rotation SOLUTION The mass of the first child is m1 = 44.0 kg. This child is a distance d1 = 1.30 m from the tree trunk. The mass of the second child, hanging d2 = 2.10 m from the tree trunk,
is m2 = 35.0 kg. Both children produce positive (counterclockwise) torques. The net torque
exerted on the branch by the two children is then
Στ = τ1 + τ 2 = F1l1 + F2l 2 = m1 g d1 cos θ + m2 g d 2 cos θ = g cos θ ( m1d1 + m2 d1 )
{ 124 { 1 24
43
43
F1 l1 F2 l2 Substituting the given values, we obtain ( )( ) Στ = 9.80 m/s 2 cos 27.0o ( 44.0 kg )(1.30 m ) + ( 35.0 kg )( 2.10 m ) = 1140 N ⋅ m 438 ROTATIONAL DYNAMICS 5. SSM REASONING In both parts of the problem, the
magnitude of the torque is given by Equation 9.1 as the
magnitude F of the force times the lever arm l . In part (a),
the lever arm is just the distance of 0.55 m given in the
drawing. However, in part (b), the lever arm is less than
the given distance and must be expressed using
trigonometry as l = ( 0.55 m ) sin θ . See the drawing at the
right. axis θ 0.55 m lever arm
49 N SOLUTION
a. Using Equation 9.1, we find that Magnitude of torque = F l = ( 49 N )( 0.55 m ) = 27 N ⋅ m
b. Again using Equation 9.1, this time with a lever arm of l = ( 0.55 m ) sin θ , we obtain
Magnitude of torque = 15 N ⋅ m = F l = ( 49 N )( 0.55 m ) sin θ 6. 15 N ⋅ m
( 49 N )( 0.55 m ) 15 N ⋅ m = 34° ( 49 N )( 0.55 m ) θ = sin −1 or REASONING The maximum torque will occur when the force is
applied perpendicular to the diagonal of the square as shown. The
lever arm l is half the length of the diagonal. From the
Pythagorean theorem, the lever arm is, therefore,
l= 1
2 2 2 (0.40 m) + (0.40 m) = 0.28 m Since the lever arm is now known, we can use Equation 9.1 to
obtain the desired result directly.
SOLUTION Equation 9.1 gives τ = F l = (15 N)(0.28 m) = 4.2 N ⋅ m 0.40 m 0.40 m sin θ = axis F Chapter 9 Problems 7. 439 SSM REASONING Each of the two forces produces a torque about the axis of rotation,
one clockwise and the other counterclockwise. By setting the sum of the torques equal to
zero, we will be able to determine the angle θ in the drawing. F2 = 55.0 N x θ
θ Rod F1 = 38.0 N l2 Hinge (axis
of rotation) 90º Table (Top view)
SOLUTION The two forces act on the rod at a distance x from the hinge. The torque τ1 produced by the force F1 is given by τ1 = +F1 l 1 (see Equation 9.1), where F1 is the magnitude of the force and l 1 is the lever arm. It is a positive torque, since it tends to
produce a counterclockwise rotation. Since F1 is applied perpendicular to the rod, l1 = x.
The torque τ2 produced by F2 is τ2 = −F2 l 2 , where l 2 = x sin θ (see the drawing). It is a
negative torque, since it tends to produce a clockwise rotation. Setting the sum of the
torques equal to zero, we have + F1l1 + ( − F2l 2 ) = 0 +F1 x − F2 ( x sin θ ) = 0
1 24
43
l2 or The distance x in this relation can be eliminated algebraically. Solving for the angle θ gives sin θ = 8. F1
F2 or F1 −1 38.0 N = sin = 43.7° 55.0 N F2 θ = sin −1 REASONING We know that the torques generated by the two applied forces sum to zero.
In order for this to be true, one of the forces must tend to produ...
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 Spring '13
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 Physics, The Lottery

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