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Goal Solution G: If the tension in each chain is 350 N at the lowest point, then the force of the seat on the child
should just be twice this force or 700 N. The child’s speed is not as easy to determine, butsomewhere between 0 and 10 m/s would be reasonable for the situation described.
O: We should first draw a free body diagram that shows the forces acting on the seat and applyNewton’s laws to solve the problem.
A: We can see from the diagram that the only forces acting on the system of child+seat are thetension in the two chains and the weight of the boy:
∑F = 2T – mg = ma where a = v2
r is the centripetal acceleration
F = Fnet = 2(350 N) – (40.0 kg)(9.80 m/s2) = 308 N upwards
v = Fmaxr
m =
(308 N)(3.00 m)40.0 kg = 4.81 m/s ◊
The child feels a normal force exerted by the seat equal to the total tension in the chains.n = 2(350 N) = 700 N (upwards) ◊
L: Our answers agree with our predictions. It may seem strange that there is a net upward forceon the boy yet he does not move upwards. We must remember that a net force causes anacceleration, but not necessarily a motion in the direction of the force. In this case, theacceleration is due to a change in the direction of the motion. It is also interesting to note thatthe boy feels about twice as heavy as normal, so he is experiencing an acceleration of about2g’s.
6.18 (a ) Consider the forces acting on the system consisting of the child and the seat:
Thus, the normal force would have to point away from the center of the curve. Unlessthey have belts, the riders will fall from the cars. To be safe we must require n1 to bepositive. Then ar > g. We need
(b) If v = const, a = 0, so T = 0 (This is also an equilibrium situation.)
(c) Someone in the car (noninertial observer) claims that the forces on the mass along x are Tand a fictitious force (–Ma). Someone at rest outside the car (inertial observer) claimsthat T is the only force on M in the x-direction.
Goal Solution G: If the horizontal acceleration were zero, then the angle would be 0, and if a = g, then the
angle would be 45°, but since the acceleration is 3.00 m/s2, a reasonable estimate of the angleis about 20°. Similarly, the tension in the string should be slightly more than the weight ofthe object, which is about 5 N.
O: We will apply Newton’s second law to solve the problem.
A: The only forces acting on the suspended object are the force of gravity mg and the force oftension T, as shown in the free-body diagram. Applying Newton's second law in the x and ydirections,
mg
T cos
T sin
θ
θ
∑Fx = T sin θ = ma (1)
∑Fy = T cos θ – mg = 0
or T cos θ = mg (2)
(a ) Dividing equation (1) by (2) gives
tan θ = ag
= 3.00 m/s2
9.80 m/s2 = 0.306
Solving for θ, θ = 17.0°
(b) From Equation (1),
T = m a
sin θ = (0.500 kg)(3.00 m/s2)
sin (17.0°) = 5.12 N
L: Our answers agree with our original estimates. This problem is very similar to Prob. 5.30, sothe same concept seems to apply to various situations.
*6.28 In an inertial reference frame, the girl is accelerating horizontally inward at
v2
r =
(5.70 m/s)2
2.40 m = 13.5 m/s2
In her own non-inertial frame, her head feels a horizontally outward fictitious force equal to itsmass times this acceleration. Together this force and the weight of her head add to have amagnitude equal to the mass of her head times an acceleration of
g2 + (v2/r)2 = (9.80)2 + (13.5)2 m/s2 = 16.7 m/s2
This is larger than g by a factor of 16.79.80 = 1.71.
Thus, the force required to lift her head is larger by this factor, or the required force is
a = g – bv = 9.80 m/s2 – (32.7 s–1)(0.150 m/s) = 4.90 m/s2 down
*6.32 (a ) ρ = mV
; A = 0.0201 m2; R = 12 ρADv
2t = mg
m = ρV = (0.830 g/cm3)
4
3 π (8.00 cm)3 = 1.78 kg
Assuming a drag coefficient of D = 0.500 for this spherical object,
vt = 2(1.78 kg)(9.80 m/s2)
0.500(1.20 kg/m3)(0.0201 m2) = 53.8 m/s
(b) v2f = v2
i + 2gh = 0 + 2gh
h = v
2f
2g =
(53.8 m/s)2
2(9.80 m/s2) = 148 m
*6.33 Since the upward velocity is constant, the resultant force on the ball is zero. Thus, the upwardapplied force equals the sum of the gravitational and drag forces (both downward):F = mg + bv.
The mass of the copper ball is
m = 4πρr3
3 =
4
3 π
8.92 × 103
kgm3 (2.00 × 10–2 m) 3 = 0.299 kg
The applied force is then
F = mg + bv = (0.299)(9.80) + (0.950)(9.00 × 10–2) = 3.01 N
6.34 ∑Fy = may
+T cos 40.0° – mg = 0
T = (620 kg)(9.80 m/s2)
cos 40.0° = 7.93 × 103 N
∑Fx = max
–R + T sin 40.0° = 0
R = (7.93 × 103 N) sin 40.0° = 5.10 × 103 N = 12 DρAv2
. . . we list the result after each tenth iteration
0.5 1.990 –0.393 –17.6 –5.87
0.1 1.965 –0.629 –10.5 –3.51
0.15 1.930 –0.770 –6.31 –2.10
0.2 1.889 –0.854 –3.78 –1.26
0.25 1.845 –0.904 –2.26 –0.754
0.3 1.799 –0.935 –1.35 –0.451
0.35 1.752 –0.953 –0.811 –0.270
0.4 1.704 –0.964 –0.486 –0.162
0.45 1.65 –0.970 –0.291 –0.0969
0.5 1.61 –0.974 –0.174 –0.0580
0.55 1.56 –0.977 –0.110 –0.0347
0.6 1.51 –0.978 –0.0624 –0.0208
0.65 1.46 –0.979 –0.0374 –0.0125
Terminal velocity is never reached. The leaf is at 99.9% of vt after 0.67 s. The fall to theground takes about 2.14 s. Repeating with ∆t = 0.001 s, we find the fall takes 2.14 s.
*6.41 (a ) When v = vt, a = 0, ΣF = –mg + Cv2t = 0
(b) We set up a spreadsheet to calculate the motion, try different initial speeds, and home in
on 53 m/s as that required for horizontal range of 155 m, thus:
(c) Similarly, the initial speed is 42 m/s . The motion proceeds thus:
The trajectory in (c) reaches maximum height 39 m, as opposed to 33 m in (b). In both, the ballreaches maximum height when it has covered about 57% of its range. Its speed is a minimumsomewhat later. The impact speeds are both about 30 m/s.
6.51 (a ) Since the 1.00 kg mass is in equilibrium, we have for the tension in the string,
T = mg = (1.00)(9.80) = 9.80 N
(b) The centripetal force is provided by the tension in the string. Hence,
Fr = T = 9.80 N
(c) Using Fr = mpuckv2
r , we have v =
rFr
mpuck =
(1.00)(9.80)0.250 = 6.26 m/s
6.52 (a ) Since the mass m2 is in equilibrium,
∑Fy = T – m2g = 0
or T = m2g
(b) The tension in the string provides the required centripetal force for the puck.
Thus, Fr = T = m2g
(c) From Fr = m1v2
R , we have v =
RFr
m1 =
m2
m1 gR
6.53 (a ) Since the centripetal acceleration of a person is downward (toward the axis of the earth),it is equivalent to the effect of a falling elevator.
Therefore, F'g = Fg – mv2
r or Fg > F 'g
(b) At the poles v = 0, and F'g = Fg = mg = (75.0)(9.80) = 735 N down
At the equator, F'g = Fg – mar = 735 N – (75.0)(0.0337) N = 732 N down
Goal Solution G: Since the centripetal acceleration is a small fraction (~0.3%) of g, we should expect that a
person would have an apparent weight that is just slightly less at the equator than at thepoles due to the rotation of the Earth.
O: We will apply Newton’s second law and the equation for centripetal acceleration.A: (a ) Let n represent the force exerted on the person by a scale, which is the "apparent
weight." The true weight is mg. Summing up forces on the object in the direction towardsthe Earth's center gives
mg – n = mac (1)
where ac = v2
Rx = 0.0337 m/s2
is the centripetal acceleration directed toward the center of the Earth.
Thus, we see that n = m(g – ac ) < mg
or mg = n + mac > n ◊ (2)
(b) If m = 75.0 kg, ac = 0.0337 m/s2, and g = 9.800 m/s2 ,
at the Equator: n = m(g – ac) = (75.0 kg)(9.800 m/s2 – 0.0337 m/s2) = 732.5 N ◊
at the Poles: n = mg = (75.0 kg)(9.800 m/s2) = 735.0 N ◊ (ac = 0)
L: As we expected, the person does appear to weigh about 0.3% less at the equator than thepoles. We might extend this problem to consider the effect of the earth’s bulge on a person’sweight. Since the earth is fatter at the equator than the poles, g is less than 9.80 m/s2 at theequator and slightly more at the poles, but the difference is not as significant as from thecentripetal acceleration. (Can you prove this?)
*6.60 Standing on the inner surface of the rim, and moving with it, each person will feel a normalforce exerted by the rim. This inward force supplies the centripetal force to cause the 3.00 m/s2
centripetal acceleration:
ar = v2
r
v = arr = (3.00 m/s2)(60.0 m) = 13.4 m/s
The period of rotation comes from v = 2π rT
T = 2π rv
= 2π(60.0 m)13.4 m/s = 28.1 s
so the frequency of rotation is
f = 1T
= 1
28.1 s = 1
28.1 s
60 s
1 min = 2.14 rev/min
6.61 (a ) The mass at the end of the chain is in vertical equilibrium.
*6.67 (a ) The bead moves in a circle with radius r = R sin θ at a speed of
v = 2πrT
= 2πR sin θ
T
The normal force has an inward radial component of n sin θand upward component ofn cos θ.
∑Fy = may ⇒ n cos θ – mg = 0 or n = mg
cos θ
Then ∑Fx = n sin θ = m v2
r becomes
mg
cos θ sin θ = m
R sin θ
2πR sin θ
T 2, which reduces to
g sin θcos θ =
4π2R sin θT2 . This has two solutions: (1) sin θ = 0 ⇒ θ = 0°, and (2) cos θ =
gT2
4π2R .
If R = 15.0 cm and T = 0.450 s, the second solution yields
cos θ = (9.80 m/s2)(0.450 s)2
4π2(0.150 m) = 0.335 and θ = 70.4°
Thus, in this case, the bead can ride at two positions θ = 70.4° and θ = 0° .
(b) At this slower rotation, solution (2) above becomes
cos θ = (9.80 m/s2)(0.850 s)2
4π2(0.150 m) = 1.20, which is impossible.
In this case, the bead can ride only at the bottom of the loop, θ = 0°. The loop’s rotation must be faster than a certain threshold valuein order for the bead to move away from the lowest position.
6.68 At terminal velocity, the accelerating force of gravity is balanced byfrictional drag:
mg = arv + br2v2
(a ) mg = 3.10 × 10–9 v + 0.870 × 10–10 v2
For water, m = ρV = (1000 kg/m8)
4
3 π (10–5) 3
4.11 × 10–11 = (3.10 × 10–9)v + (0.870 × 10–10)v2
Assuming v is small, ignore the second term: v = 0.0132 m/s