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Rotational Equilibrium and Dynamics
Chapter
8
I
Practice 8A, p. 282
Givens
1. F = 3.0 N
Solutions
t = Fd(sin q) = (3.0 N)(0.25 m)(sin 90.0°) = 0.75 N • m
d = 0.25 m
q = 90.0°
2. m = 3.0 kg
a. t = Fd(sin q) = mgd(sin q)
d = 2.0 m
t = (3.0 kg)(9.81 m/s2)(2.0 m)(sin 5.0°) = 5.1 N • m
q = 5.0°
g = 9.81 m/s2
q = 15.0°
3. t = 40.0 N • m
d = 30.0 cm
b. t = mgd(sin q) = (3.0 kg)(9.81 m/s2)(2.0 m)(sin 15.0°) = 15 N • m
For a given torque, the minimum force must be applied perpendicular to the lever
arm, or sin q = 1. Therefore,
t 40.0 N • m
F =  =  = 133 N
d
0.300 m
Section Review, p. 282
3. F30 = 30.0 N
q30 = 45°
Copyright © by Holt, Rinehart and Winston. All rights reserved.
d30 = 0 m
F25 = 25.0 N
q25 = 59°
t30 = F30d30(sin q30) = (30.0 N)(0 m)(sin 45°) = 0 N • m
t25 = F25d25(sin q25) = (25.0 N)(2.0 m)(sin 59°) = 43 N • m
t10 = F10d10(sin q10) = (10.0 N)(4.0 m)(sin 23°) = 16 N • m
The bar will rotate counterclockwise because t25 > t10 .
d25 = 2.0 m
F10 = 10.0 N
q10 = 23°
d10 = 4.0 m
Section One—Pupil’s Edition Solutions
I Ch. 8–1
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Practice 8B, p. 288
Givens
Solutions
1. L = 5.00 m
I
Apply the first condition of equilibrium.
Fg,b = 315 N
x-component: Fx = Rx − FT (cos q) = 0
Fg,p = 545 N
y-component: Fy = Ry + FT (sin q) − Fg,p − Fg,b = 0
Ry = −FT (sin q) + Fg,p + Fg,b
q = 53°
d = 1.50 m
Apply the second condition of equilibrium.
L
L
L
FT (sin q )  + Fg,p  − d − Ry  = 0
2
2
2
Substitute the Ry value from the 1st condition y-component equation.
L
L
L
FT (sin q )  + Fg,b  − d + [FT (sin q) − Fg,p − Fg,b]  = 0
2
2
2
Fg,p + Fg,b = 545 N + 315 N = 8.60 × 102 N
FT (sin 53°)(2.50 m) + (545 N)(1.00 m) + [FT (sin 53°) − 8.60 × 102 N](2.50 m) = 0
FT (2.0 m) + 545 N • m + FT (2.0 m) − 2150 N • m = 0
(4.0 m)FT − 1610 N • m = 0
1610 N • m
FT =  = 4.0 × 102 N
4.0 m
Solve for R:
Ry = −FT (sin q ) + Fg,p + Fg,b
Ry = −(4.0 × 102 N)(sin 53°) + 8.60 × 102 N = −320 N + 8.60 × 102 N = 540 N
Rx = FT (cos q ) = (4.0 × 102 N)(cos 53°) = 240 N
2 + R 2 = 240 N)2 + (540)2 = 58 000 N2 + 290 000 N2
R= R
x y
2. Fg,b = 4.00 × 105 N
Apply the first condition of equilibrium.
Fp,1 + Fp,2 − Fg,b − Fg,c = 0
db = 0 m
Fg,c = 1.96 × 10 N
Fp,1 = Fg,b + Fg,c − Fp,2
dc = 10.0 m − 8.00 m = 2.0 m
Fp,1 = 4.00 × 105 N + 1.96 × 104 N − Fp,2 = 4.20 × 105 N − Fp,2
d1 = 10.0 m − 3.00 m = 7.0 m
Apply the second condition of equilibrium using the center of mass of the bridge as
the pivet point.
4
d2 = 7.0 m
Fp,1d1 + Fg,c dc − Fp,2d2 = 0
Substitute the Fp,1 value from the first-condition equation.
(4.20 × 105 N − Fp,2)d1 + Fg,c dc − Fp,2 d2 = 0
(4.20 × 105 N)(7.0 m) − Fp,2 (7.0 m) + (1.96 × 104 N)(2.0 m) − Fp,2 (7.0 m) = 0
2.9 × 106 N • m
2.9 × 106 N • m + 3.9 × 104 N • m
Fp,2 =  =  = 2.1 × 105 N
14.0 m
14.0 m
Fp,1 = 4.20 × 105 N − 2.1 × 105 N = 2.1 × 105 N
I Ch. 8–2
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
R = 35
000
0N
2 = 590 N
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Givens
3. Fg,w = 700.0 N
Solutions
Apply the first condition of equilibrium.
Fg,s = 200.0 N
FT,1 + FT,2 − Fg,w − Fg,s = 0
dw = 1.00 m
FT,1 = Fg,w + Fg,s − FT,2
ds = 1.50 m
d2 = 3.00 m
Choose the end of side 1 as the pivot point and apply the second condition of
equilibrium.
I
FT,2 d2 − Fg,w dw − Fg,s ds = 0
Fg,w dw + Fg,s ds
FT,2 = 
d2
(700.0 N)(1.00 m) + (200.0 N)(1.50 m)
FT,2 = 
3.00 m
7.00 102 N • m + 3.00 102 N • m
10.00 102 N • m
FT,2 =  =  = 333 N
3.00 m
3.00 m
FT,1 = Fg,w + Fg,s − FT,2 = 700.0 N + 200.0 N − 333 N = 567 N
4. Fg,1 = 400.0 N
a. Using the pivot point as the axis:
Fg,2 = 300.0 N
Fg,1d1 − Fg,2d2 = Fg,1d1 − Fg,2 (2.0 m – d1) = 0
d2 = 2.0 m − d1
(400.0 N)d1 − (300.0 N)(2.0 m) + (300.0 N)d1 = 0
(700.0 N)d1 − 6.0 102 N • m = 0
6.0 102 N • m
d1 =  = 0.86 m from the 400.0 N child
700.0 N
d1 = 0.86 m
b. Using the pivot point as the axis point:
Fg,3 = 225 N
Fg,1d1 + Fg,3d3 − Fg,2d2 − Fg,4d4 = 0
d3 = 0.86 m − 0.200 m
= 0.66 m
Fg,1d1 + Fg,3d3 − Fg,2d2
d4 = 
Fg,4
(400.0 N)(0.86 m) + (225 N)(0.66 m) − (300.0 N)(1.1 m)
d4 = 
325 N
340 N • m + 150 N • m − 330 N • m
160 N • m
d4 =  = 
325 N
325 N
d2 = 2.0 m − 0.86 m = 1.1 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fg,4 = 325 N
d4 = 0.49 m from the pivot point on the same side as the 300.0 N child
Section Review, p. 289
5. Fg ,b = 40.0 N
Fg ,1 = 510 N
Fg ,2 = 350 N
d1 = 1.50 m
a. Choose the center of mass as the pivot point and apply the second condition of
equilibrium.
Fg ,1d1 − Fg ,2 d2 = 0
Fg ,1d1
(510 N)(1.50 m)
d2 =  =  = 2.2 m from center
Fg , 2
350 N
b. Apply the first condition of equilibrium.
Fs − Fg ,b − Fg ,1 − Fg ,2 = 0
Fs = Fg ,b + Fg ,1 + Fg ,2
Fs = 40.0 N + 510 N + 350 N = 9.0 × 102 N
Section One—Pupil’s Edition Solutions
I Ch. 8–3
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Practice 8C, p. 291
Givens
Solutions
1. R = 0.50 m
I
M = 100.0 kg
(0 rev/min − 50.0 rev/min)(2p rad/rev)(1 min/60 s)
wf − w
a. α = i =  = −0.87 rad/s2
6.0 s
∆t
wi = 50.0 rev/min
b. t = Ia = (2MR2)a
1
wf = 0 rev/min
1
t = 2(100.0 kg)(0.50 m)2(−0.87 rad/s2) = −11 N • m
∆t = 6.0 s
2. R = 0.33 m
wf − w
t = Ia = (MR2) i
∆t
M = 1.5 kg
wi = 98.7 rad/s
wf = 0 rad/s
(0 rad/s − 98.7 rad/s)
t = (1.5 kg)(0.33 m)2 = −8.1 N • m
2.0 s
∆t = 2.0 s
3. R = 0.075 m
M = 0.500 kg
1
a. t = Ia = (2MR2)a
Because F is perpenddicular to d,
∆s = 4.00 m
t = Fd = (mg)(R)
m = 5.00 kg
1
MR2a
2
2
g = 9.81 m/s
= mgR
2mg (2)(5.00 kg)(9.81 m/s2)
a =  =  = 2.6 × 103 rad/s2
(0.500 kg)(0.0075 m)
MR
wi = 0 rad/s
(2)(2.6 × 10 rad/s )(4.00 m)
w = w
+ 
+2aR∆s = (0rad/s)
(0.075 m)
∆s
b. wf 2 = wi2 + 2a∆q = wi2 + 2a 
R
f
i
2
2
3
2
Practice 8D, p. 294
1. m = 80.0 kg
1
Li = Lc,i + Lm,i = Tcwi + Imwi = 2MR2wi + mri2wi
1
M = 6.50 × 102 kg
Lf = Lc,f + Lm,f = Icwf + Imwf = 2MR2wf + mrf2wf
R = 2.00 m
Li = Lf
ri = 2.0 m
2MR2 + mri2wi = 2MR2 + mrf2wf
1
1
2MR2 + mri2wi 2(6.50 × 102 kg)(2.00 m)2 + (80.0 kg)(2.0 m)2(0.30 rad/s)
rf = 1.0 m
wi = 0.30 rad/s
1
1
wf = 
= 
1
1
MR2 + mr 2
(6.50 × 102 kg)(2.00 m)2 + (80.0 kg)(1.0 m)2
f
2
2
(1.30 × 103 kg • m2 + 320 kg • m2)(0.30 rad/s)
(1620 kg • m2)(0.30 rad/s)

wf = 
=
1.30 × 103 kg • m2 + 8.0 × 101 kg • m2
1380 kg • m2
wf = 0.35 rad/s
I Ch. 8–4
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
wf = 5.3 × 102 rad/s
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Givens
Solutions
2. M = 2.0 kg
Li = Lw,i + Lr,i = Iwwi + Irwi = MR2wi + mri2wi
R = 0.30 m
wi = 25 rad/s
m = 0.30 kg
Lf = Lw,f + Lr,f = Iwwf + Irwf = MR2wf + mrf2wf
Li = Lf
(MR2 + mri2)wi = (MR2 + mrf2)wf
ri = 0.19 m
rf = 0.25 m
wf =
2
(MR + mri2)wi


MR2 + mrf2
I
2
2
[(2.0 kg)(0.30 m) + (0.30 kg)(0.19 m) ](25 rad/s)
= 
(2.0 kg)(0.30 m)2 + (0.30 kg)(0.25 m)2
(0.18 kg • m2 + 1.1 × 10−2 kg • m2)(25 rad/s)
wf = 
0.18 kg • m2 + 1.9 × 10−2 kg • m2
(0.19 kg • m2)(25 rad/s)
wf = 
0.20 kg • m2
wf = 24 rad/s
3. M = 10.0 kg
1
Li = Icwi = 2MR2wi
R = 1.00 m
Lf = (Ic + Ip)wf = 2MR2 + mr2wf
1
wi = 7.00 rad/s
m = 0.250 kg
r = 0.900 m
Li = Lf
1
MR2w
i
2
= 2MR2 + mr2wf
1
1
1
(10.0 kg)(1.00 m)2 (7.00 rad/s)
MR2w
i
2
2
wf = 
= 
1
1
(10.0 kg)(1.00 m)2 + (0.250 kg)(0.900 m)2
MR2 + mr2
2
2
35.0 kg • m2/s
35.0 kg • m2/s

wf = 
=
5.20 kg • m2
5.00 kg • m2 + 0.202 kg • m2
wf = 6.73 rad/s
4. ri = 8.8 × 1010 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
rf = 5.2 × 1012 m
4
vf = 5.4 × 10 m/s
Li = Lf
Iiwi = If w f
mri2wi = mrf 2wf
vf
v
wi = i and w f = , so
rf
ri
rv
(8.8 × 1010 m)(5.4 × 104 m/s)
= 9.1 × 102 m/s
vf = i i = 
rf
5.2 × 1012 m
5. ri = 0.54 m
Li = Lf
rf = 0.040 m
Iiwi = Ifwf
m = 25 g
mri2wi = mrf2wf
wi = 0.35 rev/s
ri2wi (0.54 m)2 (0.35 rev/s)(2p rad/rev)
 = 
wf = 
rf2
(0.040 m)2
wf = 4.0 × 102 rad/s
Section One—Pupil’s Edition Solutions
I Ch. 8–5
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Practice 8E, p. 297
Givens
Solutions
MEi = MEf
1. h = 2.00 m
2
g = 9.81 m/s
I
Ivf 2 1 2
mr 2 1 2 3
1
mgh = 2 mvf 2 + 
= 2 vf m + 
= 2 vf 2 m
2
r
2r 2
4 mgh 4
vf 2 =  = 3 gh
3m
vf =
2. m = 1.5 kg
r = 0.33 m
h = 14.8 m
g = 9.81 m/s2
gh = (9.81m/s)(2.00m) = 5.11 m/s
4

3
4

3
2
The sphere has the greater speed, and so would win the race.
MEi = MEf
2
= mv (1 + 1) = mv
v
1
1
1
mgh = 2 mvf2 + 2 Iwf2 = 2 mvf 2 + (mr2) f
r
1

2
2
f
f
2
vf = gh
= (9
.8
1m
/s
2)(1
4.
8m
) = 12.0 m/s
3. r = 25 cm
MEi = MEf
d = 4.0 m
mgh = 2 mvf2 + Iwf2 = 2 mvf2 + 3mr2
q = 30.0°
mgh = mgd(sin q) = 6mvf2
g = 9.81 m/s2
vi = 0 m/s
1
1
2
5
vf =
2
=
vf

r
1 + 3 = 2 mvf23
1
 mv 2
f
2
2
1
5
in q)
6g
d(s5
1
d = 2 (vi + vf)∆t
2d
2d
∆t =  = 
6g
d(sin q)
vi + vf
vi + 
5
∆t =
(2)(4.0 m)
0 m/s +
(6)(9.81 m/s2)(4.0 m)(sin 30.0°)

5
=
(2)(4.0 m)
(6)(9.81 m/s2)(4.0 m)(sin 30.0°)

5
∆t = 1.6 s
Section Review, p. 297
1. m = 3.0 kg
Li = Iiw i = (2mri 2 + I)w i
w i = 0.75 rad/s
Lf = If w f = (2mrf 2 + I)w f
ri = 1.0 m
Li = L f
rf = 0.30 m
(2 mri2 + I)wi = (2mrf2 + I)wf
I = 3.0 kg • m2
(2 mr 2 + I)wi [(2)(3.0 kg)(1.0 m)2 + 3.0 kg • m2](0.75 rad/s)
wf = i 
= 
2 mrf2 + I
(2)(3.0 kg)(0.30 m)2 + 3.0 kg • m2
(9.0 kg • m2)(0.75 rad/s)
(6.0 kg • m2 + 3.0 kg • m2)(0.75 rad/s)
wf = 
= 
2
2
3.5 kg • m2
0.54 kg • m + 3.0 kg • m
w f = 1.9 rad/s
I Ch. 8–6
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
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Givens
2. mm = 4.0 kg
Solutions
MEi = MEf
1
1
1
mw = 8.0 kg
mm gh = 2mm v f2 + 2m w v f2 + 2Iw f2
rw = 0.50 m
1
1
1 v
mm gh = 2mmv f2 + 2mw v f2 + 2I f
rw
I = 2.0 kg • m2
h = 2.0 m
g = 9.81 m/s2
vf =
v=
3. F = 40.0 N
d = 0.15 m
r = 50.0 cm
∆v = 2.25 m/s
∆t = 3.0 s
2
2mm gh

=
I
mm + mw + 2
rw
I
(2)(4.0 kg)(9.81 m/s2)(2.0 m)

.0 kg • m2
4.0 kg + 8.0 kg + 2

(0.50 m)2
(2)(4.0 kg)(9.81 m/s2)(2.0 m)
 =
12.0 kg + 8.0 kg
(2)(4.0 kg)(9.81 m/s2)(2.0 m)
 = 2.8 m/s
20.0 kg
t = Ia
Because F is perpendicular to d, t = Fd.
t Fd∆t Fdr∆t
I =  =  = 
∆w
∆v
a
(40.0 N)(0.15 m)(0.500 m)(3.0 s)
I =  = 4.0 kg • m2
2.25 m/s
Section Review, p. 301
2. eff = 0.73
din = 18.0 m
dout = 3.0 m
m = 58 kg
Copyright © by Holt, Rinehart and Winston. All rights reserved.
g = 9.81 m/s2
3. Fg = 950 N
Fapplied = 350 N
W ut
eff = o
Win
Fout dout
 where Fout = mg
eff = 
Findin
mg dout
(58 kg)(9.81 m/s2)(3.0 m)
Fin =  
=  = 1.3 × 102 N
ef f din
(0.73)(18.0 m)
F ut
Fg
950 N
= 
=  = 2.7
MA = o
Fin Fapplied 350 N
Chapter Review and Assess, pp. 305–312
9. m = 54 kg
r = 0.050 m
t = Fd(sin q) = mgr(sin q)
t = (54 kg)(9.81 m/s2)(0.050 m)(sin 90°) = 26 N • m
g = 9.81 m/s2
q = 90°
10. q = 90.0° − 8.0° = 82.0°
m = 1130 kg
d = 3.05 m − 1.12 m −
0.40 m = 1.53 m
g = 9.81 m/s2
tnet = tg + tjack = 0
mgd(sin q) + tjack = 0
tjack = −mgd(sin q) = −(1130 kg)(9.81 m/s2)(1.53 m)(sin 82.0°)
tjack = −1.68 × 104 N • m = 1.68 × 104 N • m clockwise
Section One—Pupil’s Edition Solutions
I Ch. 8–7
Givens
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11. d = 15.0 m
q = 90.0° − 20.0° = 70.0°
Solutions
a. tmax = Fg,maxd(sin q) = (450 N)(15.0 m)(sin 70.0°) = 6.3 × 103 N • m
Fg,max = 450 N
I
q = 90.0° − 40.0° = 50.0°
20. Fg,s = 205 N
l
= 3.00 m
Fg,w = 675 N
t ax
6.3 × 103 N • m
b. Fg = m
=  = 5.5 × 102 N
d(sin q) (15.0 m)(sin 50.0°)
Apply the first condition of equilibrium.
FT,1 + FT,2 − Fg,w − Fg,s = 0
FT,1 = Fg,w + Fg,s − FT,2 = 675 N + 205 N − FT,2 = 8.80 × 102 N − FT,2
dw = 1.00 m
Choose the end of the scaffold closest to the person as the pivot point.
Apply the second condition of equilibrium.
FT,2l − Fg,wdw − Fg,s l = 0
2
FT,2 =
Fg,wdw + Fg,s l
2
l
3.00 m
(675 N)(1.00 m) + (205 N) 
2
FT,2 = 
3.00 m
675 N • m + 308 N • m 983 N • m
FT,2 =  =  = 328 N
3.00 m
3.00 m
FT,1 = 8.80 × 102 N − FT,2 = 8.80 × 102 N − 328 N = 552 N
a. Apply the first condition of equilibrium in the x and y direction.
g = 9.81 m/s2
Fx = Rx − FT (cos q) = 0
q = 30.0°
Fy = Ry + FT (sin q) − mg = 0
To solve for FT, apply the second condition of equilibrium, using the end of the
beam at the pole as the pivot point. Use l to represent the length of the beam.
FT(sin q)l − mg l = 0
mg l
(20.0 kg)(9.81 m/s2)
FT =  =  = 392 N
sin q
sin 30.0°
b. Substitute the value for FT into the two first-condition equations to solve for R.
Rx = FT (cos q) = (392 N)(cos 30.0°) = 339 N
Ry = mg − FT (sin q) = (20.0 kg)(9.81 m/s2) − (392)(sin 30.0°) = 196 N − 196 N
= 0N
I Ch. 8–8
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
21. m = 20.0 kg
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Givens
Solutions
22. Fg,b = 1200.0 N
Apply the first condition of equilibrium in the x and y directions.
Fg,m = 2000.0 N
Fx = Rx,base − FT (cos qc) = 0
qb = 65°
Fy = Ry,base − Fg,b − Fg,m + FT (sin qc) = 0
qc = 25°
To solve for FT, apply the second condition of equilibrium, using the base of the
beam as the pivot point.
I
3
L
FT L − Fg,b  (cos qb) − Fg,m L(cos qb) = 0
4
2
3
F
4 T
= 2Fg,b(cos qb) + Fg,m(cos qb) = 2Fg,b + Fg,m(cos qb)
1
1
FT = 32Fg,b + Fg,m(cos qb) = 3(1200.0 N) + 3(2000.0 N)(cos 65°)
4 1
2
4
FT = (800.00 N + 2666.7 N)(cos 65°) = (3466.7 N)(cos 65°)
FT = 1.5 × 103 N
Substitute the value for FT into the two first-condition equations to solve for Rbase.
Rx,base = FT (cos qc ) = (1.5 × 103 N)(cos 25°) = 1.4 × 103 N
Ry,base = Fg,b + Fg,m − FT (sin qc) = 1200.0 N + 2000.0 N − (1.5 × 103 N)(sin 25°)
Ry,base = 3200.0 N − 630 N = 2.6 × 103 N
23. Fg = 10.0 N
Apply the first condition of equilibrium in the x and y directions.
dg = 15 cm
Fx = FT,1(cos q) − F = 0
d1,x = 15 cm
Fy = FT,1(sin q) + FT,2 − Fg = 0
d1,y = 30.0 cm
Choose the lower left-hand corner as the pivot point and apply the second condition
of equilibrium.
q = 50.0°
−Fg dg − FT,1d1,x(cos q) + FT,1d1,y (sin q) = 0
−(10.0 N)(0.15 m) − FT,1(0.15 m)(cos 50.0°) + FT,1(0.300 m)(sin 50.0°) = 0
Copyright © by Holt, Rinehart and Winston. All rights reserved.
1.5 N • m
1.5 N • m
FT,1 =  =  = 11 N
0.230 m − 0.096 m
0.134 m
Substitute the value for FT,1 into the two first-condition equations and solve for the
unknown.
FT,2 = Fg − FT,1(sin q) = 10.0 N − (11 N)(sin 50.0°)
FT,2 = 10.0 N − 8.4 N = 1.6 N
F = FT,1(cos q) = (11 N)(cos 50.0°) = 7.1 N
27. M = 30.0
R = 0.180
t = Ia = 2MR2a = 2(30.0 kg)(0.180 m)2(2.30 × 10−2 rad/s2) = 1.12 × 10−2 N • m
1
1
a = 2.30 × 10−2 rad/s2
28. M = 350 kg
R = 1.5 m
wf = 3.14 rad/s
wf − w
3.14 rad/s − 0 rad/s
1
1
t = Ia = 2MR2 i = 2(350 kg)(1.5 m)2  = 620 N • m
∆t
2.00 s
wi = 0 rad/s
∆t = 2.00 s
Section One—Pupil’s Edition Solutions
I Ch. 8–9
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Givens
Solutions
35. Mt = 15.0 kg
1
Mdi = 9.0 kg
1
Lf = Itwf = 2MtR2wf
R = 25 cm
I
1
Li = Lt + Ldi = Itwi + Idiwi = 2MtR2wi + 2MdiR2wi
Li = Lf
wi = 0.75 rad/s
1
(M
t
2
1
+ Mdi)R2wi = 2MtR2wf
(Mt + Mdi)wi

wf = 
Mt
(15.0 kg + 9.0 kg)(0.75 rad/s) (24.0 kg)(0.75 rad/s)
wf =  =  = 1.2 rad/s
15.0 kg
15.0 kg
Li = Lf = 0, so
It = 1.5 × 10 kg • m
Lf = Iwww + Itwt = 0
r = 2.0 m
2
−I ww −(mr2)ww −(65 kg)(2.0 m) (−0.75 rad/s)
wt = w
=  = 
1.5 × 103 kg • m2
It
It
3
2
ww = 0.75 rad/s, clockwise
= −0.75 rad/s
37. m = 35 kg
r = 13 cm
h = 3.5 m
g = 9.81 m/s2
38. Fg = 240 N
wt = 0.13 rad/s, counterclockwise
MEi = MEf
2
vf
1
1
1
12
1
2
1
7
mgh = 2mvf2 + 2Iwf2 = 2mvf2 + 25 mr2  = 2mvf21 + 5 = 2mvf25
r
10 gh
(10)(9.81 m/s2)(3.5 m)
vf =  =  = 7.0 m/s
7
7
MEi = MEf
1
1
r = 0.20 m
mgh = 2mv f2 + 2Iw f2
d = 6.0 m
mgd(sin q ) = 2m(w f r)2 + 25mr 2w f2
q = 37°
1
12
1
g = 9.81 m/s2
wf =
45. m = 75 kg
r = 0.075 m
d = 0.25 m
1
7
gd(sin q ) = 2w f2r 2 + 5r 2w f2 = 10r 2w f2
in q )
(10)(9.81 m/s )(6.0 m)(sin 37°)
 = 36 rad/s
10
gd7(rs
= (7)(0.20 m)
2
2
2
For a force perpendicular to d, t = Fd.
t mgr (75 kg)(9.81 m/s2)(0.075 m)
F =  =  =  = 2.2 × 102 N
d
d
0.25 m
g = 9.81 m/s2
46. t = 58 N • m
d = 0.35 m
q = 56°
47. d = 1.4 m
F = 1600 N
t = Fd(sin q)
58 N • m
t
F =  =  = 2.0 × 102 N
(0.35 m)(sin 56°)
d(sin q)
t = Fd(sin q) = (1600 N)(1.4 m)(sin 53.5°) = 1800 N • m
q = 53.5°
I Ch. 8–10
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
36. m = 65 kg
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Givens
48.
Solutions
l s = 23.0 cm
Apply the second condition of equilibrium.
d = 2.00 cm
t net = Fs(l s − d) − Fd = 0
Fs = 84.3 N
Fs(l s − d) (84.3 N)(0.230 m − 0.0200 m)
F= 
= 
0.0200 m
d
I
(84.3 N)(0.210 m)
F =  = 885 N
0.0200 m
49. wf = 220 rad/s
wi = 0 rad/s
Wnet = 3000.0 J
50. mms = 0.100 kg
1
1
1
Wnet = ∆KE = 2Iwf2 − 2Iwi2 = 2I(wf2 − wi2)
(2)(3000.0 J)
2Wnet
 = 
= 0.12 kg• m2
I= 
(220 rad/s)2 − (0 rad/s)2
wf2 − wi2
a. Apply the first condition for equilibrium.
dms = 50.0 cm
Fs − m1g − mms g − m2 g = 0
m1 = 0.700 kg
Fs − g(m1 + mms ) 19.6 N − (9.81 m/s2)(0.700 kg + 0.100 kg)
m2 =  = 
g
9.81 m/s2
d1 = 5.00 cm
Fs = 19.6 N
ds = 40.0 cm
g = 9.81 m/s2
19.6 N − (9.81 m/s2)(0.800 kg)
m2 = 
9.81 m/s2
19.6 N − 7.85 N
11.8 N
m2 = 
= 2 = 1.20 kg
9.81 m/s2
9.81 m/s
b. Choose the zero mark as the pivot point and apply the second condition of equilibrium.
m1gd1 + m2gd2 + mmsgdms − Fsds = 0
m2gd2 = Fsds − m1gd1 − mmsgdms
Copyright © by Holt, Rinehart and Winston. All rights reserved.
Fsds − m1gd1 − mmsgdms
d2 = 
m2g
(19.6 N)(0.400 m) − (0.700 kg)(9.81 m/s2)(0.0500 m) − (0.100 kg)(9.81 m/s2)(0.500 m)
d2 = 
(1.20 kg)(9.81 m/s2)
7.84 N• m − 0.343 N • m − 0.490 N• m
7.01 N • m
d2 = 
= 
(1.20 kg)(9.81 m/s2)
(1.20 kg)(9.81 m/s2)
d2 = 0.595 m
Section One—Pupil’s Edition Solutions
I Ch. 8–11
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Givens
Solutions
51. Fg,l = 200.0 N
Apply the first condition of equilibrium in the x and y directions.
L = 8.00 m
Fx = Fs − Fwall = msRy,base − Fwall = 0
ms = 0.600
Fy = Ry,base − Fg,L − Fg,p = 0
q = 50.0°
Ry,base = Fg,L + Fg,p = 200.0 N + 800.0 N = 1000.0 N
Fg,p = 800.0 N
Fwall = msRy,base = (0.600)(1000.0 N) = 6.00 × 102 N
Choose the base of the ladder as the pivot point and apply the second condition of
equilibrium.
L
Fwall L(sin q) − Fg,L  (cos q) − Fg,pd1(cos q) = 0
2
L
FwallL(sin q) − Fg,l  (cos q)
2FwallL(tan q) − Fg,l)L
2
dp =  = 
2Fg,p
Fg,p(cos q)
(2)(6.00 × 102 N)(8.00 m)(tan 50.0°) − (200.0 N)(8.00 m)
dp = 
(2)(800.0 N)
1.14 × 104 N • m − 1.60 × 103 N • m
dp = 
1.600 × 103 N
9.8 × 103 N • m
dp = 
= 6.1 m
1.600 × 103 N • m
0.0200 m
52. r =  = 0.0100 m
2
wi = 45.0 rad/s
g = 9.81 m/s2
MEi = MEf
1
1
mv 2 + Iw 2 = mgh
i
2
2 i
1
1 1
2
m(rw ) +  mr2 w 2 = mgh
i
i
2
2 2
1
1
1
3
2
2
mr w 1 +  = mr2w 2  =
i
i 2
2
2
2
3
gh = 4r2wi2
mgh
3r2wi2 (3)(0.0100 m)2(45.0 rad/s)2
h = 
= 
4g
(4)(9.81 m/s2)
53. m = 4.0 kg
r = 2.0 m
wi = 6.0 rad/s
q = 15°
g = 9.81 m/s2
I Ch. 8–12
MEi = MEf
1
1
mv 2 + Iw 2 = mgh
i
2
2 i
1
1
m(rw )2 + (mr2)w 2 = mgd(sin q)
i
i
2
2
1
mr2w 2(1 + 1) = mr2w 2 = mgd(sin
i
1
2
q)
r2wi2
(2.0 m)2(6.0 rad/s)2
d = 
= 
= 57 m
g(sin q) (9.81 m/s2)(sin 15°)
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
h = 1.55 × 10−2 m = 1.55 cm
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Givens
Solutions
54. m = 12 kg
a. Apply Newton’s second law along the incline.
r = 10.0 cm
2
mg(sin q) − FT = ma
a = 2.0 m/s
FT = mg(sin q ) − ma = (12 kg)(9.81 m/s2)(sin 37°) − (12 kg)(2.0 m/s2)
q = 37°
FT = 71 N − 24 N = 47 N
I
g = 9.81 m/s2
b. t = Ia
Because FT is perpendicular to r, t = FT r.
t
F r
F r2
(47 N)(0.100 m)2
I =  = T = T = 
= 0.24 kg • m2
a
(a/r)
a
2.0 m/s2
∆t = 2.0 s
w i = 0 rad/s
55. Fg = Fn = 700.0 N
a
c. wf = w i + a∆t = w i +  ∆t
r
(2.0 m/s2)(2.0 s)
w f = 0 rad/s +  = 4.0 × 101 rad/s
0.100 m
First, apply the second condition of equilibrium, choosing the toe as the pivot point,
q = 21.2°
TdT − RdR = 0
f = 15.0°
dT = 25.0 cm
Td T
R= 
dR
dR = 18.0 cm
Apply the first condition of equilibrium in the y direction.
Fn − R(cos f) + T(cos q) = 0
TdT(cos f)
 + I(cos q) = 0
Fg − 
dR
Fg
700.0 N
T =  = 
d
0.250 m
T (cos f) − (cos q)
 (cos 15.0°) − (cos 21.2°)
dR
0.180 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
700.0 N
700.0 N
T =  =  = 1.7 × 103 N
1.34 − 0.932
0.41
TdT
(1.7 × 103 N)(0.250 m)
R =  =  = 2.4 × 103 N
0.180 m
dR
56. M = 0.85 kg
R = 4.0 cm
tr = 1.3 N • m
a = 66 rad/s2
a. Because F is perpendicular to r, t = Fr.
Ia = FR − tr
1
mr 2a + t
Ia + t
r
F = r = 2
R
R
t 1
1.3 N • m
1
F = 2 mra + r = 2(0.85 kg)(0.040 m)(66 rad/s2) + 
R
0.040 m
F = 1.1 N + 32 N = 33 N
∆t = 0.50 s
w i = 0 rad/s
1
b. ∆q = w i ∆t + 2a∆t2
1
∆q = (0 rad/s)(0.50 s) + 2(66 rad/s2)(0.50 s)2 = 8.2 rad
∆s = r∆q = (0.040 m)(8.2 rad) = 0.33 m
Section One—Pupil’s Edition Solutions
I Ch. 8–13
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Givens
Solutions
57. t net = 36 N • m
a. t net = Ia
∆t1 = 6.0 s
t
wf - w
 where a = i
I = net
∆t1
a
w i = 0 rad/s
I
w f = 12 rad/s
∆t2 = 65 s
w i = 12 rad/s
w f = 0 rad/s
(36 N • m)(6.0 s)
tnet∆t1
 =  = 18 kg • m2
I= 
12 rad/s − 0 rad/s
wf − wi
I(wf − wi)
(18 kg • m2)(0 rad/s −12 rad/s)
b. t f = Ia =  =  = −3.3 N • m
∆t2
65 s
(w i + w f )
(0 rad/s + 12 rad/s)
c. q1 =  ∆t1 =  (6.0 s) = 36 rad
2
2
(12 rad/s + 0 rad/s)
q2 =  (65 s) = 390 rad
2
q1 + q2
430 rad
36 rad + 390 rad
 =  =  = 68 rev
N= 
2p rad/rev
2p rad/rev
2p rad/rev
58. F1 = 120.0 N
Because F is perpendicular to the pulley’s radius, t = FR.
F2 = 100.0 N
tnet = t1 − t2 = F1R − F2R = (F1 − F2)R
M = 2.1 kg
t et (F1 − F2)R
2(F1 − F2)

a = n
= 
= 
1
 MR2
I
MR
2
R = 0.81 m
(2)(120.0 N − 100.0 N)
(2)(20.0 N)
a =  =  = 24 rad/s2
(2.1 kg)(0.81 m)
(2.1 kg)(0.81 m)
1
mv 2
i
2
MEi = MEf
r = 3.0 m
w i = 3.0 rad/s
q = 20.0°
1
m(rw )2
i
2
g = 9.81 m/s2
1
+ 2(mr 2)w i 2 = mgd(sin q)
1
mr 2w 2(1
i
2
2
1
+ 2Iw i 2 = mgh
+ 1) = mr 2w i 2 = mgd(sin q)
2
r wi
d = 
g(sin q)
(3.0 m)2(3.0 rad/s)2
d = 
= 24 m
(9.81 m/s2)(sin 20.0°)
60. M = 5.00 kg
a. Apply Newton’s second law for the bucket.
R = 0.600 m
FT − mg = −ma
m = 3.00 kg
Because FT is perpendicular to R,
∆t = 4.00 s
2
g = 9.81 m/s
vi = 0 m/s
a
1
I 
2MR2a 1
Ia
t
R
= 2mp a
FT =  =  =  = 
R
R
R
R2
1
Ma
2
− mg = −ma
mg
(3.00 kg)(9.81 m/s2)
a= 
= 
1
1
M + m
(5.00 kg) + 3.00 kg
2
2
(3.00 kg)(9.81 m/s2)
a =  = 5.35 m/s2
5.50 kg
1
1
b. ∆y = vi∆t + 2 a∆t2 = (0 m/s)(4.00 s) + 2(5.35 m/s2)(4.00 s)2 = 42.8 m
a 5.35 m/s2
c. a =  =  = 8.92 rad/s2
R
0.600 m
I Ch. 8–14
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
59. m = 5.0 kg
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Solutions
61. Lh = 2.7 m
Consider the total mass of each hand to be at the midpoint of that hand.
Lm = 4.5 m
mh = 60.0 kg
mm = 100.0 kg
Lm
Lh
tnet = −mh g  (sin qh ) − mm g  (sin q m)
2
2
qh = 30.0° from 6:00
2.7 m
4.5 m
tnet = −(60.0 kg)(9.81 m/s2)  (sin 30.0°) − (100.0 kg)(9.81 m/s2)  (sin 60.0°)
2
2
qm = 60.0° from 6:00
tnet = −4.0 × 102 N • m − 1.9 × 103 N • m = −2.3 × 103 N • m
I
g = 9.81 m/s2
3.00 cm
62. r =  = 1.50 cm
2
q = 30.0°
wi = 60.0 rad/s
g = 9.81 m/s2
63.
MEi = MEf
1
mv 2
i
2
1
+ 2Iwi2 = mgh
1
m(rw )2
i
2
+ 22mr2wi2 = mgd(sin q)
11
1 + 2 = 2mr2wi22 = mgd(sin q)
1
mr2w 2
i
2
1
1
3
(3)(1.50 × 10−2 m)2 (60.0 rad/s)2
3r2wi2
d = 
= 
= 0.124 m = 12.4 cm
(4)(9.81 m/s2)(sin 30.0°)
4g(sin q)
2
v
1
12
1
KErot = 2Iw2 = 25 mr2  = 5mv 2
r
1
1
7
KEtot = KErot + KEtrans = 5mv2 + 2mv 2 = 10mv 2
1
2
KErot 5 mv
1 10
 =  = 57 =
7
2
KEtot
mv
10
Copyright © by Holt, Rinehart and Winston. All rights reserved.
64. Fg = 800.0 N
F is perpendicular to R, so
R = 1.5 m
t = FR = Ia
F = 50.0 N
Because wi = 0 rad/s,
wf − w wf w
a = i =  = 
∆t
∆t ∆t
∆t = 3.0 s
wi = 0 rad/s
2

7
2
2
g = 9.81 m/s
2
1
1
1 t∆t
1 FR∆t
KE = 2Iw2 = 2I(a∆t)2 = 2I  = 2I 
I
I
(FR∆t)2 (FR∆t)2
(F∆t)2


KE =  = 
=
1
2I
M
2(2MR2)
(F∆t)2g
(F∆t)2
 = 
Fg
Fg

g
[(50.0 N)(3.0 s)]2 (9.81 m/s2)
KE =  = 2.8 × 102 J
800.0 N
Section One—Pupil’s Edition Solutions
I Ch. 8–15
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Solutions
65. I = 4.00 × 10−4 kg • m2
wi = 0 rad/s
1
Because F is in the direction of d, Wnet = Fd.
F = 5.57 N
I
1
Wnet = ∆KE = KEf − KEi = 2Iwf2 − 2Iwi2
1
1
Wnet = Fd = 2Iwf2 − 2Iwi2
d = 80.0 cm
wi = 0 rad/s, so
1
Fd = 2Iwf2
wf =
66. RE = 6.37 × 106 m
24
(2)(5.57 N)(0.800 m)
 = 149 rad/s
2FId = 4.00 × 10 kg m
−4
•
2
a. L = IEw
ME = 5.98 × 10 kg
From Conceptual Question 4 on p. 305,
w = 1 rev/day
IE = 0.331 MERE2
L = 0.331 MERE2w
2p rad 1 day 1 h
L = (0.331)(5.98 × 1024 kg)(6.37 × 106 m)2(1 rev/day)   
1 rev 24 h 3600 s
L = 5.84 × 1033 kg • m2/s
r = 1.496 × 1011 m
w = 1 rev/365.25 days
b. L = Iw = MEr2w
2p rad 1 day 1 h
L = (5.98 × 1024 kg)(1.496 × 1011 m)2 (1 rev/365.25 days)   
1 rev 24 h 3600 s
L = 2.66 × 1040 kg • m2/s
67. w i = 12.0 rad/s
a. Li = Lf
2
Ii w i = If w f
2
Iw
(41 kg • m2)(12.0 rad/s)
w f = i i = 
= 14 rad/s
If
36 kg • m2
Ii = 41 kg • m
If = 36 kg • m
1
1
1
1
KEf = 2If w f 2 = 2(36 kg • m2)(14 rad/s)2 = 3.5 × 103 J
I Ch. 8–16
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
KEi = 2Ii w i 2 = 2(41 kg • m2)(12.0 rad/s)2 = 3.0 × 103 J
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Givens
Solutions
68. I = 5.0 kg • m2
a. Apply Newton’s second law for the 2.0 kg mass.
R = 0.50 m
FT,1 − m1g = m1a
m1 = 2.0 kg
FT,1 − (2.0 kg)(9.81 m/s2) = (2.0 kg)a
m2 = 5.0 kg
FT,1 − 2.0 × 101 N = (2.0 kg)a
g = 9.81 m/s2
FT,1 = (2.0 kg)a + 2.0 × 101 N
I
Apply Newton’s second law for the 5.0 kg mass.
m2 g − FT,2 = m2 a
FT,2 = m2 g − m2 a
FT,2 = (5.0 kg)(9.81 m/s2) − (5.0 kg)a
FT,2 = 49 N − (5.0 kg)a
Find an expression for the torque on the pulley, noting that Fnet is perpendicular
to the pulley’s radius.
t net = Fnetd(sin θ) = Ia
a
a
= I  = (5.0 kg m ) 
R (0.50 m) a
(FT,2 − FT,1)r = I 
R
FT,2 − FT,1
•
2
2
2
1
FT,2 − FT,1 = (2.0 × 10 kg)a
Substitute the values for FT,1 and FT,2 from the equations above.
[49 N − (5.0 kg)a] − [(2.0 kg)a + 2.0 × 101 N] = (2.0 × 101 kg)a
29 N = (2.0 × 101 kg)a + (7.0 kg)a
29 N
a =  = 1.1 m/s2
27 kg
Copyright © by Holt, Rinehart and Winston. All rights reserved.
m2 accelerates downward at 1.1 m/s2 .
m1 accelerates at the same rate in the opposite direction: −1.1 m/s2
b. FT,1 = (2.0 kg)a + 2.0 × 101 N = (2.0 kg)(1.1 m/s2) + 2.0 × 101 N = 2.2 N
+ 2.0 × 101 N = 22 N
FT,2 = 49 N − (5.0 kg)a = 49 N − (5.0 kg)(1.1 m/s2) = 49 N − 5.5 N = 43 N
Section One—Pupil’s Edition Solutions
I Ch. 8–17
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Givens
Solutions
69. m1 = 4.0 kg
a. Apply Newton’s second law for the 3.0 kg mass.
m2 = 3.0 kg
F2 = m2 a = (3.0 kg)a
2
I = 0.50 kg • m
I
R = 0.30 m
g = 9.81 m/s2
Apply Newton’s second law for the 4.0 kg mass.
m1g − F1 = m1a
F1 = m1g − m1a = (4.0 kg)(9.81 m/s2) − (4.0 kg)a = 39 N − (4.0 kg)a
Find an expression for the torque on the pulley, noting that Fnet is perpendicular
to the pulley’s radius.
t net = Fnetd(sin q) = Ia
a
a
F − F = I  = (0.50 kg m ) 
R (0.30 m) a
(F1 − F2 )r = I 
R
1
2
2
•
2
2
F1 − F2 = (5.6 kg)a
Substitute the values for F1 and F2 from above.
[39 N − (4.0 kg)a] − (3.0 kg)a = (5.6 kg)a
39 N = (5.6 kg)a + (7.0 kg)a = (12.6 kg)a
39 N
a =  = 3.1 m/s2
12.6 kg
b. F1 = 39 N − (4.0 kg)(3.1 m/s2) = 39 N − 12 N = 27 N
F2 = m2 a = (3.0 kg)(3.1 m/s2) = 9.3 N
M = 500.0 kg
w = 1000.0 rev/min
a. KE = 2Iw 2 = 2 2MR2w 2
1
1 1
KE = 5.48 × 106 J
P = 7457 W
W KE
b. P =  = 
∆t
∆t
KE 5.48 × 106 J
∆t =  =  = 735 s
P
7457 W
71. w1 = 2.0 rev/s
Li = Lf
r2 = 0.50 m
Mr12w1 = Mr22w 2
r1 = 1.0 m
r12w1

w2 = 
r22
M = 4m
(1.0 m)2(2.0 rev/s)
w2 = 
= 8.0 rev/s
(0.50 m)2
I Ch. 8–18
Holt Physics Solution Manual
2
1 min 2p rad
1
KE = 4(500.0 kg)(2.00 m)2 (1000.0 rev/min)  
60 s
1 rev
Copyright © by Holt, Rinehart and Winston. All rights reserved.
70. R = 2.00 m
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Givens
Solutions
72. eff = 0.64
W ut Fout dout
eff = o
= 
Win
Findin
m = 78 kg
dout = 4.0 m
din = 24 m
g = 9.81 m/s2
73. d = 2.0 m
Foutdout mgdout
 =  = (78 kg)(9.81 m/s2)(4.0 m)\(24 m)(0.64)
Fin = 
din(eff ) din(eff )
I
Fin = 2.0 × 102 N
Wout = Fg d(sin q)
q = 15°
Win = (Ff + Fg,x )d = [mkFg (cos q) + Fg (sin q)]d
mk = 0.160
Win = Fg d[mk(cos q) + (sin q )]
W ut
sin q
Fg d(sin q)
eff = o
= 
= 
Win
mk(cos q) + (sin q)
Fg d[mk(cos q) + (sin q )]
sin 15°
sin 15°
eff =  = 
0.15 + 0.26
(0.160)(cos 15°) + (sin 15°)
sin 15°
eff =  = 0.63 = 63%
0.41
74. eff = 0.875
Fin = 648 N
m = 150 kg
dout = 2.46 m
W ut Fout dout mgdout
eff = o
=  = 
Win
Findin
Findin
mgdout
(150 kg)(9.81 m/s2)(2.46 m)
din = 
=  = 6.4 m
Fin(eff )
(648 N)(0.875)
g = 9.81 m/s2
75. dout = 3.0 m
Fin = 2200 N
din = 14 m
Copyright © by Holt, Rinehart and Winston. All rights reserved.
m = 750 kg
g = 9.81 m/s2
76. Fg = 250 N
l
= 6.0 m
q = 60.0°
W ut Fout dout mgdout
eff = o
=  = 
Win
Findin
Findin
(750 kg)(9.81 m/s2)(3.0 m)
eff =  = 0.72 = 72%
(2200 N)(14 m)
Apply the first condition of equilibrium.
Fx = Ff − Fwall = 0
Fwall = Ff = Fn ms
Fy = Fn − Fg = 0
Fn = Fg
Fwall = Fg m s
Choose the base of the ladder as the pivot point and apply the second condition of
equilibrium.
l
−F  (cos q) + F m l (sin q) = 0
2l −Fg  (cos q) + Fwall l (sin q) = 0
2
g
g s
1
m s (sin q) = 2(cos q)
cos q
cos 60.0°
m s =  =  = 0.289
2(sin q)
(2)(sin 60.0°)
Section One—Pupil’s Edition Solutions
I Ch. 8–19
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Givens
77.
l
Solutions
= 15.0 m
Fg,l = 520.0 N
q = 60.0°
I
Fg,f = 800.0 N
df = 4.00 m
Apply the first condition of equilibrium in the x and y directions.
Fx = Fx,E − Fx,wall = 0
a. Fy = Fy,E − Fg,f − Fg,l = 0
Fy,E = Fg,f + Fg,l = 800.0 N + 520.0 N = 1320.0 N
Choose the base of the ladder as the pivot point and apply the second condition of
equilibrium.
l −Fg,l  (cos q) − Fg,f df (cos q) + Fx,wall l (sin q) = 0
2
F 2l + F d (cos q)
g,l
g,f f
Fx,wall =  = Fx,E
l (sin q)
[(520.0 N)(7.50 m) + (800.0 N)(4.00 m)](cos 60.0°)
Fx,E = 
(15.0 m)(sin 60.0°)
(3.90 × 103 N • m + 3.20 × 103 N • m)(cos 60.0°)
Fx,E = 
(15.0 m)(sin 60.0°)
(7.10 × 103 N • m)(cos 60.0°)
Fx,E =  = 273 N
(15.0 m)(sin 60.0°)
df = 9.00 m
b. Fx,E = Fs = msFn = msFy,E
F 2l + F d (cos q)
g,l
g,f f
F ,E
ms = x
= 
Fy,E
Fy,E l (sin q)
(3.90 × 103 N • m + 7.20 × 103 N • m)(cos 60.0°)
ms = 
(1320.0 N)(15.0 m)(sin 60.0°)
(11.1 × 103 N • m)(cos 60.0°)
ms =  = 0.324
(1320.0 N)(15.0 m)(sin 60.0°)
I Ch. 8–20
Holt Physics Solution Manual
Copyright © by Holt, Rinehart and Winston. All rights reserved.
[(520.0 N)(7.50 m) + (800.0 N)(9.00 m)](cos 60.0°)
ms = 
(1320.0 N)(15.0 m)(sin 60.0°)