Unformatted text preview: ce a clockwise rotation about
the pinned end of the meter stick, and the other must tend to produce a counterclockwise
rotation about the pinned end. We will assume that the first force, applied perpendicular to 440 ROTATIONAL DYNAMICS the length of the meter stick at its free end, tends to produce a counterclockwise rotation
(see the drawing). Counterclockwise is the positive direction.
F2 pinned
end d θ
θ
F1 l2
l1
TOP VIEW With this assumption, we obtain
F1l1 − F2l 2 = 0 or F1l1 = F2l 2 (1) The drawing shows that the lever arm l1 of the force F1 is equal to the length of the meter
stick: l1 = 1.00 m. The force F2 is applied a distance d from the pinned end of the meter
stick. The lever arm l 2 of the second force is
l 2 = d sin θ (2) SOLUTION Substituting Equation (2) into Equation (1) and solving for the distance d, we
obtain
F1l1 = F2 ( d sin θ ) 9. or d= F1l1
( 2.00 N ) (1.00 m ) = 0.667 m
=
F2 sin θ ( 6.00 N ) sin 30.0o REASONING AND SOLUTION The torque produced by each force of magnitude F is
given by Equation 9.1, τ = F l , where l is the lever arm and the torque is positive since
each force causes a counterclockwise rotation. In each case, the torque produced by the
couple is equal to the sum of the individual torques produced by each member of the couple.
a. When the axis passes through point A, the torque due to the force at A is zero. The lever
arm for the force at C is L. Therefore, taking counterclockwise as the positive direction, we
have
τ = τ A + τ C = 0 + FL = FL Chapter 9 Problems 441 b. Each force produces a counterclockwise rotation. The magnitude of each force is F and
each force has a lever arm of L / 2 . Taking counterclockwise as the positive direction, we
have
L
L
τ = τ A + τ C = F + F = FL
2
2 c. When the axis passes through point C, the torque due to the force at C is zero. The lever
arm for the force at A is L. Therefore, taking counterclockwise as the positive direction, we
have
τ = τ A + τ C = FL + 0 = FL
Note that the value of the torque produced by the couple is the same in all three cases; in
other words, when the couple acts on the tire wrench, the couple produces a torque that does
not depend on the location of the axis. 10. REASONING AND SOLUTION The net torque about the axis in text drawing (a) is
Στ = τ1 + τ2 = F1b − F2a = 0 Considering that F2 = 3F1, we have b – 3a = 0. The net torque in drawing (b) is then
Στ = F1(1.00 m − a) − F2b = 0 or 1.00 m − a − 3b = 0 Solving the first equation for b, substituting into the second equation and rearranging, gives
a = 0.100 m and b = 0.300 m 11. REASONING Although this arrangement of body parts is vertical, we can apply Equation
9.3 to locate the overall center of gravity by simply replacing the horizontal position x by
the vertical position y, as measured relative to the floor. SOLUTION Using Equation 9.3, we have
y cg =
= W1 y 1 + W2 y 2 + W3 y 3
W1 + W2 + W3
438
+ 87 b
b N g1.28 mgb N g0.760 m gb N g0.250 mg
b + 144 b
=
438 N + 144 N + 87 N 1.03 m 442 ROTATIONAL DYNAMICS 12. REASONING At every instant before the plank begins to tip, it is in equilibrium, and the
net torque on it is zero: Στ = 0 (Equation 9.2). When the person reaches the maximum
distance x along the overhanging part, the plank is just about to rotate about the right
support. At that instant, the plank loses contact with the left support, which consequently
exerts no force on it. This leaves only three vertical forces acting on the plank: the weight
W of the plank, the force FR due to the right support, and the force P due to the person (see
the freebody diagram of the plank). The force FR acts at the right support, which we take as
the axis, so its lever arm is zero. The lever arm for the force P is the distance x. Since
counterclockwise is the positive direction, Equation 9.2 gives
Στ = W l W − Px = 0 x= or WlW (1) P FR
d
lW L/2 x W
Axis P Freebody diagram of the plank SOLUTION The weight W = 225 N of the plank is known, and the force P due to the
person is equal to the person’s weight: P = 450 N. This is because the plank supports the
person against the pull of gravity, and Newton’s third law tells us that the person and the
plank exert forces of equal magnitude on each other. The plank’s weight W acts at the center
of the uniform plank, so we have (see the drawing)
lW + d = 1 L
2 or lW = 1 L − d
2 (2) where d = 1.1 m is the length of the overhanging part of the plank, and L = 5.0 m is the
length of the entire plank.
Substituting Equation (2) into Equation (1), we obtain x= W 2 ( 1 L − d ) = ( 225 N ) 1 ( 5.0 m ) −1.1 m = 0.70 m
2
P 450 N Chapter 9 Problems 443 13. SSM REASONING The drawing shows the bridge and the four forces that act on it: the
upward force F1 exerted on the left end by the support, the force due to the weight Wh of the
hiker, the weight Wb of the bridge, and the upward force F2 exerted on the right side by the
support. Since the bridge is in equilibrium, the sum of the torques about any axis of rotation
must be zero ( Στ = 0 ) , and the sum o...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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