INTRODUCTION to BIOMECHANICS for HUMAN MOTION ANALYSIS,
THIRD EDITION
SOLUTIONS to ODD-NUMBERED PROBLEMS
by
D. Gordon E. Robertson, PhD, FCSB
School of Human Kinetics, University of Ottawa
© Copyright 2013 (revised 11 December 2013)
INTRODUCTION (p. 12)
Conversion factors are taken from Table 1.3 on page 8.
1.
(a)
350 × 0.4536 × 9.81 = 1557 N
(b)
6.50 / 0.4536 = 14.33 lbs.
(c)
168.5 × 2.54 = 428 cm
(d)
(10 × 100) /2.54 = 394 inches
(e)
70.0
miles 1.609 km 1 hour 1000 m
×
×
×
= 31.3 m/s
hour
1 mile
3600 s 1 km
(f)
80.0
miles 1.609 km
×
= 128.7 m/s
hour
1 mile
(g)
8.35 × 12 × 2.54 = 255 cm
(h)
440 × 0.9144 = 402 m
(i)
(800 / 0.9144) × 3 = 2620 feet
(j)
50.0 × 1.609 = 80.5 km
(k)
25.0 / 1.609 = 15.54 miles
(l)
3.00 × 9.81 = 29.4 newtons
3.
lbs 0.4536 kg
×
= 113.4 kg
1
1 lbs
W = mg = 113.4 × 9.81 = 1112.5 N
Thus, the 1200 N person weighs more than the 250 lbs. person.
250
5.
45 ft. = 45 ×12 = 540 in. = 540 × 2.54 = 1371.6 cm = 13.716 m
Thus, 13.75 m is longer than 45 feet.
2
FUNDAMENTAL CONCEPTS (p. 26)
x = r cos θ
r = x2 + y2
y = r sin θ
⎛ y⎞
θ = tan −1 ⎜ ⎟
⎝x⎠
1.
(a)
(c)
(e)
r = 250 2 + 350 2 = 430 cm
(b)
θ = tan −1 (350 / 250) = 54.5o
r = − 10.00 2 + −20.0 2 = 22.4 m/s
θ = tan −1 (−20.0 / − 10.00) = −116.6o
r = 25.0 2 + (−35.0) 2 = 43.0 m/s 2
θ = tan −1 (350 / 250) = −54.5o
(d)
(f)
r = 2.00 2 + 1.000 2 = 2.24 kN
θ = tan −1 (1.000 / 2.00) = 26.6 o
r = 1000 2 + 2000 2 = 2240 N
θ = tan −1 (2000 / 1000) = 63.4o
r = − 5.00 2 + 15.00 2 = 15.81 kN
θ = tan −1 (15.00 / − 5.00) = 108.4o
3.
(a)
(c)
(e)
x = 25.0 cos 25.0o = 22.7 m/s
(b)
y = 25.0 sin 25.0o = 10.57 m/s
x = 100.0 cos− 45.0o = 70.7 m/s 2
y = 100.0 sin − 45.0 = −70.7 m/s
o
2
x = 25.0 cos160.0o = −23.5 N
y = 25.0 sin 160.0o = 8.55 N
(d)
(f)
x = 10.00 cos 30.0o = 8.66 N
y = 10.00 sin 30.0o = 5.00 N
x = 5.00 cos(π / 2) = 0.0 m
y = 5.00 sin( π / 2) = 5.00 m
x = 25.0 cos( π / 4) = 17.68 m/s
y = 25.0 sin( π / 4) = 17.68 m/s
5.
Sum of sides is always greater than length of hypotenuse.
7.
(a)
(30.0 − 20.0, 50.0 − 40.0) = (10.00, 10.00) cm
(b)
(400 − 500, 700 − 400) = (−100.0, 300) N
9.
R = (750, 1000) N = 1250 N at 53.1 deg
11.
r = 2.75 2 + 5.25 2 = 5.93 km
3
SPATIAL (3D) COORDINATE SYSTEMS (p. 30)
P = Px2 + Py2 + Pz2
cos α = Px / P
cos β = Py / P
cos γ = Pz / P
cos 2 α + cos 2 β + cos 2 γ = 1
r = x2 + y2
θ = tan −1 ( y / x)
z=z
1.
(a)
(c)
(e)
3.
(a)
P = 250 2 + 350 2 + 450 2 = 622.5
cos α = 250 / 622 = 0.402
(b)
cos β = 350 / 622 = 0.562
cos β = 1 / 3.74 = 0.267
cos γ = 450 / 622 = 0.723
cos γ = 3 / 3.74 = 0.802
P = − 10 2 + (−20) 2 + 34 2 = 40.7
P = 10002 + 20002 + 50002 = 5477
cos α = 1000 / 5477 = 0.1826
cos α = −10 / 40.7 = −0.246
(d)
cos β = −20 / 40.7 = −0.492
cos β = 2000 / 5477 = 0.365
cos γ = 34 / 40.7 = 0.836
cos γ = 5000 / 5477 = 0.913
P = 252 + (−35) 2 + 552 = 69.8
P = − 52 + 152 + 202 = 25.5
cos α = −5 / 25.5 = −0.1961
cos α = 25 / 69.8 = 0.358
(f)
cos β = −35 / 69.8 = −0.501
cos β = 15 / 25.5 = 0.588
cos γ = 55 / 69.8 = 0.788
cos γ = 20 / 25.5 = 0.785
r = 250 2 + 350 2 = 430 cm
r = 2 2 + 12 = 2.24 m
θ = tan −1 (350 / 250) = 54.5o
z = 450 cm
(b)
r = − 10 2 + (−20) 2 = 22.4 m
(c)
P = 2 2 + 12 + 32 = 3.74
cos α = 2 / 3.74 = 0.535
−1
θ = tan (−20 / − 10) = 63.4 + 180
= 243.4
o
z = 34.0 m
o
θ = tan −1 (1 / 2) = 26.6o
z = 3.00 m
r = 1000 2 + 2000 2 = 2240 mm
o
(d)
θ = tan −1 (2000 / 1000) = 63.4 o
z = 5000 mm
4
r = − 52 + 152 = 15.81 cm
r = 252 + (−35) 2 = 43.0 cm
(e)
θ = tan −1 (−35 / 25) = −54.5o
z = 55.0 cm
(f)
x = 25.0 cos 35.0o = 20.5 cm
5.
(a)
(c)
y = 25.0 sin 35.0o = 14.34 cm
θ = tan −1 (15 / − 5) = −71.6o + 180.0o
= 108.4o
z = 20.0 cm
x = 1.100 cos 35.0o = −0.376 m
(b)
y = 1.100 sin 35.0o = 1.034 m
z = 45.0 cm
z = 0.500 m
x = 67.5 cos 45.0o = 47.7 cm
x = 85.0 cos− 120.0o = −42.5 cm
y = 67.5 sin 45.0o = 47.7 cm
z = −20.0 cm
(d)
y = 85.0 sin 120.0o = −73.6 cm
z = 55.0 cm
5
RESOLUTION of FORCES into COMPONENTS (p. 37)
W = weight = mg
Fx = F cos θ
1.
Gm1m2
r2
Fy = F sin θ
FG =
W = mg = 50.5 × 9.81 = 495 N
3.
1 kg
= 68.2 kg
2.2045 lbs
W = mg = 68.2 × 9.81 = 669 N
150 lbs. ×
5.
m=
7.
(a)
(b)
(c )
(d )
W 625
=
= 63.7 kg
g 9.81
R = (25 + 10 + 4.53, 30 + 12 − 1.56) = (39.5, 40.4)
R = (3 × 4.53 − 1.25, 3 × −1.56 − 5) = (12.34, − 9.68)
R = (0 − 10 − 1.25, 10 − 12 − 5) = (−11.25, − 7.00)
R = (4.53 + 2(−1.25) − 0, − 1.56 + 2(−5) − 10) = (2.03, − 21.6)
1
⎡1
⎤
R = ⎢ (4.53) + 5.6, (−1.56 − 2)⎥ = (7.87, − 2.78)
2
⎣2
⎦
1
⎡1
⎤
( f ) R = ⎢ (2.53 + 10), (30 + 12)⎥ = (8.75, 10.50)
4
⎣4
⎦
( g ) R = ( −1.25 − 10 + 0, − 5 − 12 + 10) = (−11.25, − 7.00)
(h) R = −[(10 − 25), (12 − 30)] = (15.00, 18.00)
( e)
9.
1
Fmoon = (9.81)60.0 = 98.1 N
6
mspace = 60.0 kg
Fspace = 0.00 N
11.
g=
G × mmars
2
rmars
= 6.673 ×10 −11 ×
= 3.71 m/s 2
6.419 ×10 23
(3.396 × 106 ) 2
6
MOMENT of FORCE (p. 46–7)
M = Fd
M = rF sin θ
1.
F=
3. (a)
(b)
M
500
=
= 303 N
d 1.650
R = ΣF = −50.0 − 25.0 + 125.0 − 50.0 = 0.00 N
M R = ΣFi d i = 50.0(0.0) − 25.0(0.05) + 125.0(0.15) − 50.0(0.25)
= 0 − 1.25 + 18.75 − 12.50 = 5.00 N.m
(c)
Σ MA = 0
∴125.0(d C ) − 50.0(0.0) − 25.0(0.05) − 50.0(0.25) = 0
125.0(d C ) = 0.0 + 25.0(0.05) + 50.0(0.25)
d C = (1.25 + 12.50)/125.0 = 0.1100
The force at C should be moved 11.00 cm to the right of A.
5. (a)
ΣM = 200(2.50) − 250(2.20) = −50.0
Therefore, Cathy will go up.
(b) Jill must move 20.0 cm towards Cathy so the two moments are equal.
(c)
7. (a)
ΣM = 0
200(2.50) − 250(2.20) − 150.0(d ) = 0
− 50.0
= 0.333
d=
− 150.0
Therefore, Ian must sit 33.3 cm from fulcrum on Cathy’s side.
M A = Fd = −300(0.25) = −75.0 N.m
(b) M A = FC cos θ(0.20) = −200(cos 20O )(0.20) = −35.6 N.m
Note, vertical component at C cancels vertical component at D and horizontal component
at D has no moment about axis at A.
9.
M = Fd = 56.5(0.650) = 36.7 N.m
7
LAWS of STATICS (p. 62–4)
ΣFx = 0
M = r × F = (rx Fy − ry Fx )k
ΣFy = 0
ΣF = 0
ΣM A = 0
Σ M = Σ( r × F ) = 0
1. (a)
M = (0.35 × 80.0 − 0.20 × 50.0) = 18.00k N.m
(b)
M = (0.35 × 20.0 − 0.20 × −30.0) = 13.00k N.m
(c)
M = [−0.10 × (80.0 + 20.0) − 0.30 × (50.0 − 30.0)] = −16.00k N.m
(d)
(e)
M = (−0.10 × 20.0 − 0.30 × 20.0) = 7.00 N.m
M = (1.250 × −50.0 − 2.50 × 85.0) = −275 N.m
(f)
M = (−0.10 × −50.0 − 0.30 × 85.0) = −20.5k N.m
3.
F = ( F cos θ, F sin θ) = (56.3 cos 34.3o , 56.3 sin 34.3o ) = (46.51, 31.73)
M = (0.1366 × 31.73 − 0.205 × 46.51) = −5.20k N.m
5. (a)
(b)
(c)
(d)
M = (4.00 × 35.0 − 8.00 × 25.0) = −60.0 N.cm = −0.600 N.m
M = (−4.00 × 35.0 − 8.00 × 25.0) = −340 N.cm = −3.40 N.m
M = (4.00 × 35.0 − (−8.00) × 25.0) = 340 N.cm = −3.40 N.m
M = (−4.00 × 35.0 − (−8.00) × 25.0) = 60.0 N.cm = 0.600 N.m
7.
C=
9.
L × FA 2.00 × 300
=
= 1.019 m = 101.9 cm
mg
60.0(9.81)
ΣFx = Fg x + Fkneex = 0
ΣFy = Fg y + Fknee y − mg = 0
ΣM cg = M knee + (r knee × F knee ) + (r g × F g ) = 0
8
15.
ΣFy = 0 : 2 Fcable − mg = 0
Fcable =
80.0(9.81)
= 392 N per cable
2
17. (a)
jacking up a car, prying with a bottle opener, shoveling
(b)
throwing a dart, kicking, jumping
21.
ΣFx = 0 : − F1x + F2 x = 0
F1x = F2 x
F2 x = 200 cos 30o = 173.2 N
ΣFy = 0 : F1 y + F2 y − W = 0
F1x = F2 x + mg = −200 sin 30o + 400 = 300 N
23.
ΣFx = 0 : Fknee − Fload x = 0
x
Fkneex = Fload x
Fkneex = 250 cos 30o = 217 N
ΣFy = 0 : Fknee y − Fload y − W = 0
Fkneex = Fload x + mg = 250 sin 30o + 40.0 = 165 N
9
SPATIAL (3D) LAWS of STATICS (p. 67)
Σ F = 0 Σ M = Σ( r × F ) = 0
ΣFx = 0
ΣFy = 0
ΣFZ = 0
ΣM A x = 0
ΣM A y = 0
ΣM A z = 0
i
r × F = rx
Fx
1. (a)
j
ry
Fy
k
rz = (ry Fz − rz Fy )i − (rx Fz − rz Fx ) j + (rx Fy − ry Fx )k
Fz
M x = 20.0 × 90.0 − 40.0 × 80.0 = −1400 N.cm = −14.00 N.m
M y = −(35.0 × 90.0 − 40.0 × 50.0) = −1150 N.cm = −11.50 N.m
M z = 35.0 × 80.0 − 20.0 × 50.0 = 1800 N.cm = 18.00 N.m
(b)
M x = 20.0 × 45.0 − 40.0 × 20.0 = 100.0 N.cm = 1.000 N.m
M y = −(35.0 × 45.0 − 40.0 × −30.0) = −2780 N.cm = −27.8 N.m
M z = 35.0 × 20.0 − 20.0 × −30.0 = 1300 N.cm = 13.00 N.m
(c)
C + D = (50.0 − 30.0, 80.0 + 20.0, 90.0 + 45.0) = (20.0, 100.0, 135.0)
M x = 30.0 × 135.0 − 0.0 × 100.0 = 4050 N.cm = 40.5 N.m
M y = −(−10.00 × 135.0) − 0.0 × 20.0) = 1350 N.cm = 13.50 N.m
M z = −10.00 × 100.0 − 30.0 × 20.0 = −1600 N.cm = − 16.00 N.m
(d)
5b = 5(−10.00, 30.0, 0.0) = (−50.0, 150.0, 0.0)
M x = 150.0 × 90.0 − 0.0 × 80.0 = 13500 N.cm = 135.0 N.m
M y = −(−50.0 × 90.0 − 0.0 × 50.0) = 4500 N.cm = 45.0 N.m
M z = −50.0 × 80.0 − 150.0 × 50.0 = − 11500 N.cm = − 115.0 N.m
(e)
a + b = (35.0 − 10.00, 20.0 + 30.0, 40.0 + 0.0) = (25.0, 50.0, 40.0)
M x = 50.0 × 90.0 − 40.0 × 80.0 = 1300 N.cm = 13.00 N.m
M y = −(25.0 × 90.0 − 40.0 × 50.0) = −250 N.cm = −2.50 N.m
M z = 25.0 × 80.0 − 50.0 × 50.0 = − 500 N.cm = − 5.00 N.m
(f)
10 D = 10(−30.0, 20.0, 45.0) = (−300, 200, 450)
M x = 30.0 × 450 − 0.0 × 200 = 13500 N.cm = 135.0 N.m
M y = −(−10.00 × 450) − 0.0 × −300) = 4500 N.cm = 45.0 N.m
M z = −10.00 × 200 − 30.0 × −300 = 7000 N.cm = 70.0 N.m
10
3.
Σ F = 0 = F 1 + F 2 − 600k N
ΣFx = 0 = F1x + 150.0 N
∴ F1x = −150.0 N
ΣFy = 0 = F1 y + 65.0 N
∴ F1 y = −65.0 N
ΣFz = 0 = F1z + 350 N − 600 N
∴ F1z = 250 N
DRY FRICTION (pp. 78–9)
Fstatic = µ static Fnormal
Fkinetic = µ kinetic Fnormal
1.
ΣFn = 0 : Fnormal − mg = 0
Fnormal = mg = 50 × 9.81 = 490.5
Fstatic = µ static Fnormal = 0.95 × 490.5 = 466 N
Fkinetic = µ kinetic Fnormal = 0.90 × 490.5 = 441 N
3.
ΣFn = 0 : Fnormal − mg = 0
Fnormal = mg = 35 × 9.81 = 343.4
Since object is in motion :
Fkinetic = µ kinetic Fnormal = 0.90 × 343.4 = 188.8 N in negative direction
5.
Since there are only three forces and they must add to zero for statics you can apply
the triangle rule :
ΣF = 0 : F friction + F normal + W = 0
be constructing the triangle we see that :
F friction = W sin 15 = 50 × 9.81× 0.25882 = 127.0 N up the incline
7.
ΣFn = 0 :
Fnormal − mg cos 12 o + Fapplied sin45 o = 0
Fnormal = 30 × 9.81 × cos 12 o − 200 sin45 o = 146.4
Fstatic = µ static Fnormal = 0.80 × 146 .4 = 117.2
Assume ΣFt = 0 :
Fequilbrium − mg sin 12 o + Fapplied cos45 o = 0
Fequilbrium = 30 × 9.81 × sin 12 o − 200 cos45 o = −80.2
since this absulute value is smaller th an Fstatic friction is 80.2 N down the incline.
11
9.
ΣFn = 0 : Fnormal − mg − Fapplied sin13o = 0
Fnormal = 50 × 9.81 + 500 sin13o = 603.0
Fstatic = µ static Fnormal = 0.80 × 603 = 482.4 N
Assume ΣFt = 0 : Fequilbrium + Fapplied cos13o = 0
Fequilbrium = −500 cos13o = −487.2
since absolute value is greater than Fstatic the body is moving (friction = Fkinetic ).
11.
ΣFn = 0 : Fnormal − Wcos 10o = 0
Fnormal = 250 cos10o = 246.2
Fstatic = µ static Fnormal = 0.500 × 246.2 = 123.10 N
To get the box moving the friction must equal Fstatic .
ΣFt = 0 : − Fstatic − Wcos 10o + Fapplied = 0
Fapplied = Fstatic + 250 sin 10o = 123.10 + 43.41 = 166.5 N
13.
Fstatic = µ static Fnormal = 0.950(150.0 × 9.81) = 1398 N
Fkinetic = µ kinetic Fnormal = 0.900(150.0 × 9.81) = 1324 N
15.
Fnormal = 10.00 × 9.81 = 98.1 N
µ static =
µ kinetic =
Fstatic
25.0
=
= 0.255
Fnormal 98.1
Fkinetic 20.0
=
= 0.204
Fnormal 98.1
17.
Fnormal = 30.0 × 9.81 = 294.3 N
µ static =
µ kinetic =
19.
Fstatic
225
=
= 0.765
Fnormal 294.3
Fkinetic
215
=
= 0.731
Fnormal 294.3
No. By definition the coefficient of static friction only occurs at the instant the slipping
occurs when the maximum friction also occurs.
12
LINEAR KINEMATICS (pp. 94–5)
(b)
v f = vi + at
s f = si + vi t + ½ at
s f = si + vi t + ½ at 2
2
= 0 + 3.5 ×10 2 + ½ (0.005)10 2
= 35 + 0.25 = 35.3 m
v 2f = vi2 + 2a( s f − si )
s f = si + ½(vi + v f )t
5. (a)
1. (a)
0 to 40 m: a =
v 2f − vi2
t=
2( s f − si )
40 to 70 m:
70 to 100 m:
½(−0.3)(7.33)2 = 8.07 m
2
(30 − 36)
= −0.750 m/s 2
16
(b)
t0− 40 =
v f − vi
=
0 − 2.20
= 7.33 s
− 0.300
= 0 + 2.20(7.333) +
a = 0 m/s2
s f = si + vt + ½at 2
100 = 70 + 9(4) + ½ a (4)
a
=
(b)
s f = si + vi (7.333) + ½ at 2
92 − 0
92
=
2(40 − 0) 2(40)
81
=
= 1.0125 m/s 2
80
a=
a=2
v f − vi
9−0
= 8.89 s
1.0125
a
70 − 40
t 40 − 70 =
= 3.33 s
9
t 70 − 100 = 4.00 s
ttotal = 8.89 + 3.33 + 4 = 16.22 s
(c)
⎛ 81 ⎞
s f = 0 + 0 + ½⎜ ⎟52 = 12.66 m
⎝ 80 ⎠
(d)
⎛3⎞
v100 = 9 − ⎜ ⎟4 = 6.00 m/s
⎝4⎠
3. (a)
v f 2 = ± 3.52 + 2(0.005)(25 − 0)
= ±3.536 m/s
v f − vi + 3.536 − 3.50
t=
=
= 7.11 s
a
0.005
7.
v = 150 km/h ÷ 3.6 = 41.67 m/s
1st catcher :
v 2f − vi2
0 − 41.67 2
=
a=
2( s f − si ) 2(0.300 − 0)
= −2894
v f − vi 0 − 41.67
=
t=
− 2894
a
= 0.01440 s = 14.40 ms
2 nd catcher :
a=
v 2f − vi2
2( s f − si )
=
0 − 41.67 2
2(0.500 − 0)
= −1736
v f − vi 0 − 41.67
=
t=
− 1736
a
= 0.0240 s = 24.0 ms
The difference in the accelerations is
1158 m/s2. The difference in the times is
9.60 ms.
13
9.
2
⎛ 7.32 ⎞
s = 12.00 + ⎜
⎟ = 12.546 m
⎝ 2 ⎠
v = s / t = 12.546 / 1.500 = 8.36 m/s
2
Ball speed must be greater than 8.36 m/s.
11. (a)
v=
s 100 m 1 min
=
×
= 0.417 m/s
t 4 min 60 s
(b)
sdown river = vcurrent t = 4.00
km
1h
1000 m
×
×
× 4.00 min = 266.7 m
h 60 min 1 km
stotal = 266.7 2 + 100.0 2 = 285 m
(c)
v = stotal / t = 284.8 /( 4.00 × 60.0) = 1.187 m/s
13.
0 = vi2 + 2a ( s f − si )
vi2 = −2a ( s f − si ) = −2(−2.00)(6.00 − 0)
vi = ± 24.0 = ±4.90
The initial velocity must be 4.90 m/s.
14
PROJECTILE MOTION (pp. 101–2)
5. (a)
v fy = viy − gt
s fy = siy − gt
s fy = siy +
v = v − 2 g ( s fy − siy )
2
fy
2
iy
(b)
km 1000 m
1h
×
×
= 25.0 m/s
h
1 km 3600 s
v x = v cos θ = 25 cos 5° = 24.90 m/s
90
t=
v f = ± 232.2 = ±15.24 m/s
select the negative velocity
10 − 0
= 0.402 s
24.9
t=
+ siy
s x = v xt = 0.500(2.16) = 1.082 m
7.
θ vx
30̊ 8.66
45̊ 7.07
60̊ 5.00
0 − (2.179)
+ 0 = 0.242 m = 24.2 cm
− 2(9.81)
vx =
10 − 0
= 28.6 m/s
0.350
time
1.019
1.442
1.765
v fy = ± 9.81 = −3.13
t=
v f − vi
=
− 3.13 − 0
= 0.319 s
− 9.81
−g
10
vx =
= 31.3 m/s
0.319
v f = −19.81 m/s 2
(c)
stx = six − v xt = 0 + 2(2.02) = 4.04 m
xmax
8.83
10.19
8.83
2
= ± 392.4 = ±19.81 m/s
− 19.81 − 0
= 2.02 s
=
t=
− 9.81
−g
ymax
1.270
2.55
3.82
v fy = 0 − 2 g (1.0 − 1.5) = 9.81
v fy = ± 0 2(9.81)(−20 − 0)
v f − vi
vy
5.00
7.07
8.66
9.
3. (a)
2
2
v fy = viy − 2 g ( sfy − siy )
(b)
− 15.24 − 6
= 2.16 s
− 9.81
(c)
2
(d)
2
= 6 2 − 2(9.81)(0 − 10)
(c)
2
2
v fy = viy − 2 g (s fy − siy )
=
−g
2
(b)
s fx = six − v xt
− 2g
v f − vi
v f = vi − 2 g ( s f − si )
v y = v sin θ = 25 sin 5° = 2.179 m/s
s fy =
2
6.00 2
= 0+
=
= 1.835 m
+ 2g
2(9.81)
1. (a)
v 2fy −viy2
− 2g
0 + viy
s fy = siy + ½ (viy + v fy )t
t=
v fy 2 − viy 2
11.
2
v f = 0 − 2(9.81)(−0.795)
v f = −3.94 m/s
t=
vf
g
v fps =
= 0.4026
12
= 29.8 fps
0.4206
15
ANGULAR KINEMATICS (p. 106)
ω f = ωi + αt
θ f = θ i + ωi t + ½ αt 2
ω2f = ωi2 + 2α(θ f − θ i )
θ f = θ i + ½(ωi + ω f )t
1. (a)
ω f − ωi 5 − 2
=
= 1.500 r/s 2
t
2
2
= 3π rad/s = 9.42 rad/s 2
α=
(b)
θ f = θ i + ωit + ½αt 2
= 0 + 2(2) + ½(1.5)22 = 7.00 revolutions
3.
1
θ = 0 + 0 (0.35)36 2
2
= 227 revolutions
5.
ω f = ω i + αt
= 6 + 0.750(5) = 9.75 rad/s
7.
ω=
3r
6π
=
= 21.2 rad/s
0.89 0.89
9.
ω final = ωinitial + αt = 5.60 + (−0.200)3.00 = 5.00 rad/s
11.
t=
ω final − ωinitial
α
=
0 − 20.0
= 13.33 s
− 1.500
16
RELATIONSHIP between LINEAR and ANGULAR MEASURES (pp. 111–2)
7. (a)
vtfeet = rω = 2.30(10) = 23.0 m/s
vt = rω
at = rα
vtcg = rω = 1.30(10) = 13.00 m/s
(b)
v 10 − 0
ω= =
= 5.00 rad/s
r
2
v2
a r = rω = t
r
2
a = ar2 + at2
(c)
s feet = 2πr = 2 π(2.30) = 14.45 m
1. (a)
vt = rω = 0.75(15) = 11.25 m/s
scg = 2πr = 2π(1.3) = 8.16 m
(b)
at = rα = 0.75(150) = 117.5
= 14.45 − 8.16 = 6.29 m
ar = rω 2 = 0.75(15) 2 = 168.75
2
(d)
ar feet = rω 2 = 2.30(10.00) 2
2
a = at + ar = 117.52 + 168.752
= 230 m/s 2
= 203 m/s 2
9. (a)
rtotal = 1.00 + .70 = 1.700 m
3. (a)
deg ⎞ π rad
⎛
vt = rω = 0.900⎜ 573
⎟×
s ⎠ 180 deg
⎝
= 9.00 m/s
(b)
ωf − ωi 10 − 0
α=
=
= 6.67 rad/s 2
t
1.5
(c)
at = rα = 0.9(10) = 9.00
vt = rω = 1.70(8.75) = 14.875 m/s
v
,
ω
v + 1 .5
1.70 + r = t
ω
16.375
r=
− 1.7 = 0.1714 m
8.75
v = rω; r =
ar = rω 2 = 0.9(10) 2 = 90.0
2
(b)
ar = rω 2 = 0.1714(8.75) 2
2
a = at + ar = 9.00 2 + 90.0 2
= 90.4 m/s 2
5.
vt = rω
= 13.12 m/s 2
11.
r
2π rad 1 min
×
×
min
1r
60 s
= 6.81 rad/s
vrim = rω = 0.330(6.81) = 2.25 m/s
= 0.75(10.00)
= 7.50 m/s
ar = rω 2 = 0.75(10) 2 = 75.0 m/s 2
ω = 65
at = rα = 0.75(2.00) = 1.500 m/s 2
v pushwheel = rω = 0.100(6.81)
a = at 2 + ar 2 = 752 + 1.52
= 0.681 m/s
= 75.0 m/s 2
17
LAW of ACCELERATION (p. 117)
ΣF = m a
Σ Fx = max
Σ Fy = ma y
1.
⎛ v 2f − vi2 ⎞ ⎛ 0 − 9 2 ⎞ − 81
⎟=⎜
⎟=
a=⎜
= −1.350 m/s 2
⎜ 2( x − x ) ⎟ ⎜ 2(30 − 0) ⎟ 60
⎝
⎠
f
i ⎠
⎝
F = ma = 60.0(−1.350) = −81.0 N
3.
Σ F y = ma y = Flifter − mg
⎛ v yf − v yi ⎞ 2 − 0
⎟⎟ =
a y = ⎜⎜
= 2 .00
t
1
⎝
⎠
Flifter = ma y + mg = 40 ( 2 ) + 40 (9 .81) = 472 N
5.
Fnormal = mg = 70(9.81) = 686.7
Fkinetic = Fnormal µ kinetic = 686.7 × 0.6 = 412 N
ΣFx = ma x = − Fkinetic
− Fkinetic
= −5.886 m/s 2
70
2
⎛ v 2fx −v ix2 ⎞
⎛
⎞
⎟ = 0 + ⎜ 0 − 10 ⎟ = 8.49 m
Thus, x f = xi + ⎜
⎜
⎟
⎜ 2a ⎟
⎝ 2(−5.886) ⎠
x
⎝
⎠
ax =
7.
ΣF = ma = 1000 = 900a
a = 1000 / 900 = 1.111 m/s 2
t=
9.
v f − vi
a
v f − vi
=
3−0
= 2.70 s
1.111
0 − 20
= −66.67 m/s 2
t
0 .3
F = ma = 0.180(−66.67) = −12.00 N
a=
=
11.
ΣFx = ma x = 35 + 58 − 52 = 41.0 N
ΣF y = ma y = 25 − 20 + 23 − mg = 28 − 196.2 = −168.2 N
a x = 41 / 20 = 2.05 m/s 2
a y = −168.2 / 20 = −8.41 m/s 2
18
MOMENT of FORCE (p. 122)
M a = [r × F ]z = (rx Fy − ry Fx )
M R = ΣM a = I a α
M = Fd
1.
α=
3.
M
= 35 / 2 = 17.50 rad/s 2
I
M = Fd = 150(0.35) = 52.5 N.m
α = M / I = 2.10 rad/s 2
5.
ω f − ωi
2.35 − 4.45
= −0.420
t
5.00
M = Iα = 32.0(−0.42) = −13.44 N.m
α=
7.
9.
=
M = Fd = 400 × 0.35 = 140.0 N.m
M = Fd = 500 × 0.40 = 200 N.m
I = M / α = 200 / 20 = 10.00 kg.m 2
19
MOMENT of INERTIA (p. 129)
Kcg = k cg /L
I cg = mkcg2
I axis = I cg + mr 2
1.
kcg =
I cg
m
=
0.5
= 0.250 m
8
I axis = I cg + mr 2 = 0.5 + 8(0.25) 2
= 1.000 kg.m 2
3.
k = KL = 0.60(0.95) = 0.57 m
I cg = mk 2 = 12.00(0.57) 2 = 3.90 kg.m 2
I hip = I cg + mr 2
= 3.90 + 12.00(0.50) 2 = 6.90 kg.m 2
5. (a)
I bar = I cg + mr 2 = 15.80 + 80.0(1.450) 2
= 15.8 + 168.2 = 184.0 kg.m 2
(b)
k=
7.
15.80
= 0.1975 = 0.444 m
80.0
M = Fd = 300(0.15) = 45.0 N.m
α = (ω f − ωi ) / t = 20 / 1 = 20.0 rad/s 2
I = M/a = 45 / 20 = 2.25 kg/m 2
k = I / m = 2.25 / 90.0 = 0.1581 m
9.
I proximal = I cg + mr 2 = 0.1489 + 7.05(0.1925) 2
= 0.410 kg.m 2
k proximal = I proximal / m = 0.453 / 7.05 = 0.241 m = 2.41 cm
20
REACTION FORCES (p. 136)
F = −R
Fcentripetal = mrw2 = mv 2 / r
1.
ΣFx = Fgx = ma x
a x = Fgx / m = 227 / 53.8 = 4.22 m/s 2
ΣFy = Fgy − mg = ma y
a y = (1345 − 53.8 × 9.81) / 53.8 = 15.19 m/s 2
3.
s = rθ
r = s / θ = 100 / π = 31.83
⎛ v2 ⎞
112
θ = tan −1 ⎜⎜ ⎟⎟ =
⎝ rg ⎠ 31.83 × 9.81
= tan −1 (0.3875) = 21.2 deg
5.
v2
r
2
v
2502
r=
=
= 6371 m = 6.37 km
g
9.81
mg = m
7.
ΣFy = ma y = Fgy − mg
Fgy = ma y + mg = 65.0(0.75 + 9.81)
= 686 N
9.
mar = −mvt2 / r = −mrω 2
= −3.50(2.00)52 = −175.0 N
11.
F = ma = 75(20 × 9.81) = 14 720 N
21
LINEAR IMPULSE and MOMENTUM (p. 150)
momentum = p = mv
impulse = ∫ Fdt = Ft
∫ Fdt = mv
1.
3.
5.
f
− mvi
⎛ 0 − 12.00 ⎞
F = ma = 60.0⎜
⎟ = 60(−10.909) = −655 N
⎝ 1.100 ⎠
Impulse = Ft = mv f − mvi = 55.0(2.50) − 0 = 137.5 N.s
mv fx = mvix + ∫ Fx dt = 0 + 200
v fx = 200 / 70.0 = 2.86 m/s
mv fy = mviy + ∫ Fy dt − Wt
= 0 + 1200 − 70(9.81)1.2 = 375 .86
v fy = 375 .86 / 70.0 = 5.37 m/s
7.
v 2fy = viy2 − 2 g ( y f − yi )
= 0 − 2(9.81)(−1.350 − 0) = 26.487
v fy = 26.476 = −5.1247 m/s
alanding =
Flanding
v f − vi
0 − (−5.127)
= 12.87 m/s 2
0.400
= 60.0(12.87) = 772 N
=
t
= malanding
9.
Ft = mv f − mvi
F=
− mvi − 0.005(400)
=
= −40.0 N
t
0.050
11.
v f = vi +
Ft
= 0 + 24.0(1) / 4 = 6.00 m/s
m
22
ANGULAR IMPULSE and MOMENTUM (p. 155)
angular momentum = L = Iω
angular impulse = ∫ Mdt = M t
∫ Mdt = Iω
1.
f
− Iωi
L = Iω = 5.00(5.52)
= 27.6 kg.m 2 /s
3.
r 2π rad ⎞
⎛
L = Iω = 5.65⎜ 2.25 ×
⎟
s
1r ⎠
⎝
= 5.65(14.137) = 79.9 kg.m 2 /s
5.
M t = ( F d )t = 250(1.350)0.500
= 168.8 N.m.s
7.
M t = ( F d )t = 68.0(0.320)5
= 108.8 N.m.s
9.
k = Kl = 0.326(1.235) = 0.4026 m
I = mk 2 = 11.50(0.4026) 2 = 1.8641 kg.m 2
L = Iω = 1.8641(2.55) = 4.75 kg.m 2 /s
23
CONSERVATION of MOMENTUM (p. 163)
p f = pi = mv = constant
L f = Li = Iω = constant
1. (a)
L = Iω = 2(20) = 40.0 kg.m 2 /s
(b)
I top ωtop = Lstart = Ltop = Lland = 40.0 kg.m 2 / s
I top = L / 30.0 = 1.333 kg.m 2
I land = L / 26.6 = 1.504 kg.m 2
3. (a)
L = Iω
8.25 = 0.430ω
ω = 8.25 / 0.43 = 19.19 rad/s
(b)
⎛ 19.19 − 0 ⎞
M = Iα = 0.43⎜
⎟ = 16.50 N.m
⎝ 0.5 ⎠
5.
ω = L / I = 8.95 / 16.26 = 0.5504
θ − θi
ω= f
t
θ −0
2p
t= f
=
= 11.42 s
ω
0.5504
7.
I = L / ω = 35.6 / 3.25 = 10.95 kg.m 2
24
WORK-ENERGY THEOREM (p. 174)
work = W = ∆E = E f − Ei
energy = E = mgy + ½ mv 2 + ½ Iω2
1.
E = mgy + ½ mv 2 = 70(9.81)1.150 + ½ (70)12.00 2
= 789.7 + 5040 = 5830 joules = 5.83 kJ
3.
E = ½ mv 2 = ½ (60.0)2.00 2 = 120.0 J
5.
r
2π 1min ⎞
⎛
E = ½Iw 2 = ½ (2.00)⎜ 300
×
×
⎟
min 1r 60 s ⎠
⎝
= ½ (2.00)31.416 2 = 987 J
2
7.
y = v sin 20 o t = 6.00 sin 20 o (60.0 s) = 123.1 m
W = Ef − Ei = mgy f − mgyi
= 50(9.81)123.1 − 0 = 490.5(123.1) = 60.4 kJ
9.
W = E f − Ei = 0 − Ei = 0 − ½ mv 2 = −½(25.0)52 = −313 J
11. (a)
vf − vi 0 − 3
=
= −0.750
t
4
v 2f − vi2
0 − 32
− 9.0
s f = si +
=
=
= 6.00 m
2a
2(−0.75) − 1.50
a=
(b)
F = ma = 20.0(−0.75) = −15.00 N
(c)
W = Fs = −15.00(6.00) = −90.0 J
25
WORK of a FORCE or MOMENT of FORCE (p. 180–1)
W force = Fs cos φ
W force = F ⋅ s = Fx s x + Fy s y
Wmoment = M θ = ( Fr sin φ)θ
1. (a)
9.
W = Fs = ( Lg ) s
W = Fs = 358 × 0.517 = 185.1 J
= 3.50(9.81)4000
(b)
= 137 340 J = 137.3 kJ
W = Fx s x + Fy s y
3.
= 25.3 × 1.325 + 63.2 × 2.92
= 218 J
11.
W force = Fs cosφ
13. (a)
W = mgy = 60(9.81)0.350
= 206 J
= 35.0 (23.0) cos 30o = 697 J
5.
= 441 J
(b)
Fnormal = mg = 25.0(9.81)
= 245.25 N
Fkinetic = Fnormal µ kinetic
Wmoment = Mθ = ( Fd )θ
= (90.0 × 0.600)(1 rad) = 54.0 J
(c)
= 245.25 × 0.800 = 196.2 N
Wµ =0.8 = Fkinetic s = 196.2 × 10.00
= 1962 J
Wµ =0.2 = 245.25 × 0.2 ×10.00
W = Fs = ( Lg ) s
= 20(9.81)10(6.00)
= 196.2(60.0) = 11772 J
= 11.77 kJ
n = Wtotal / Wmoment
=
441
= 8 and 1/6 = 8.17 cycles
54
15. (a)
W = Ef − Ei = 0 − ½ I ω2
= 491 J
7.
Wtotal = mgy = 300(9.81)0.1500
= −½ (0.450) 20.0 2 = −90.0 J
(b)
F friction = Fnormal µ kinetic
= −12.50(0.800) = −10.00 N
M Fd 10.00 × 0.300
α=
=
=
I
I
0.450
2
= −6.667 rad/s
ω f − ωi 0 − 20.0
t=
=
= 3.00 s
α
− 6.667
26
POWER (p. 185)
P =W /t = ∆ E /t
Pforce = Fv cosφ
Pforce = F ⋅ v = Fx v x + Fy v y
Pmoment = Mω
1.
s = 2 × 70.0 × 6.00 = 840 metres
WL = 10.00 × 9.81 = 98.1 joules
P = WL s / t = 98.1(840) /( 2 × 60)
= 687 W
3.
P = (WL g ) s / t = (3.5 × 9.81) × 4000 /(6 × 60) = 382 W
5.
P=0
An isometric contraction does no mechanical work.
7.
deg 2 π rad
×
= 3.49
s 360 deg
P = Mω = 50.0 × 3.39 = 174.5 W
ω = 200
9.
r 2π rad
ω=3 ×
= 18.85 rad/s
1r
s
P = Mω = 55.0 × 18.85 = 1037 W
11.
P = Fv = 225 × 0.555 = 124.9 W
W = Pt = 124.9 × 3.50 = 437 J
27
CONSERVATION of MECHANICAL ENERGY (p. 190)
E f = Ei = constant
1.
WG = mgy = 75.0 × 9.81 × 3.00 = 2210 J
½ mv 2 = mgy
v = 2 gy = 2(9.81)3 = 7.67 m/s
3.
E f = Ei
½ mv 2f + mgy f = ½ mvi2 + mgyi
½ mv 2f + 0 = ½(65.0)4.15 2 + 65.0(9.81)10.00
= 559.7 + 6376.5 = 6936 J
vf =
2 × 6936
= 213.4 = 14.61 m/s
65.0
5.
mgy = ½ mv 2
y=
v2
5.50 2
30.25
=
=
= 1.542 m
2 g 2(9.81) 19.62
7.
2
W = mgytop = ½ mvtakeoff
W = 60.0(9.81)0.452 = 266 J
2
½ mvtakeoff
= 266 J
vtakeoff =
2 × 266
= 8.868 = 2.98 m/s
60.0
9.
mgy = ½ mv 2
mgy = (15.00 + 0.100)(9.81)(0.35) = 51.85 J
½ mv 2 = 51.85 J
v=
2 × 51.85
= 1036.9 = 32.2 m/s
0.100