SAMPLE PROBLEM 4/1 The system of four particles has the indicated particle masses, positions, velocities, and external forces. Determine F, F, F, T, G, Ho, Ho, HG, and HG. Solution The position of the mass center of the system is m (2di 2dj) + 2m (dk) + 3m(-2di) + 4m (dj) m + 2m +3m + 4m r = = r = Emiri Emi Emiri Emi ΣΕ Σmi =d(-0.4i + 0.2j + 0.2k) m(-vi + vj) + 2m (vj) + 3m(vk) + 4m (vi) 10m = v(0.3i + 0.3j + 0.3k) Fi + Fj 10m T= [ {/m₁v₁² - = 1 2 = F 10m (i+j) 11 [m(√2v)² + 2mv² + 3mv² + 4mv²] = 2 G = (Em)r = Ho = Er; x m₂r; = 0 - 2mvdi + 3mv (2d)j - 4mvdk Ⓡ = mvd(-2i+ 6j - 4k) Ho = EMo = -2dFk+ Fdj = Fd(j - 2k) For HG, we use Eq. 4/10: [HG = Ho + px mv] HG = mud(-2i+ 6j - 4k) - d(-0.4i + 0.2j + 0.2k) x = 10m (v)(0.3i + 0.3j + 0.3k) = mv (3i + 3j + 3k) HG = Fd(j - 2k) - d(-0.4i +0.2j + 0.2k) x 10m = Fd(0.2i + 0.8j - 1.4k) Ans. -mv² F mm) (i 10m m 1 2d Ans. 10mv (0.3i+ 0.3j + 0.3k) = mvd(-2i+ 4.2j - 2.2k) Ans. For HG, we could use Eq. 4/9 or Eq. 4/11 with P replaced by O. Using the latter, we have [HG = EMo - pxma] Ans. Ans. Ans. Ans. Ans. (i+j) Ⓒ Ans. √2v 2d x F 2m dl V V 3m 2d 3 4m F HELPFUL HINTS All summations are from i = 1 to 4, and all are performed in order of the mass numbers in the given figure. Because of the simple geometry, the cross products are performed by inspection. Using Eq. 4/10 with P replaced by O is more expedient than using Eq. 4/8 or 4/8a. The m in Eq. 4/10 is the total mass, which is 10m in this example. The quantity p in Eq. 4/10, with P replaced by O, is r. We again recognize that p = r here and that the mass of this system is 10m.
SAMPLE PROBLEM 4/1 The system of four particles has the indicated particle masses, positions, velocities, and external forces. Determine F, F, F, T, G, Ho, Ho, HG, and HG. Solution The position of the mass center of the system is m (2di 2dj) + 2m (dk) + 3m(-2di) + 4m (dj) m + 2m +3m + 4m r = = r = Emiri Emi Emiri Emi ΣΕ Σmi =d(-0.4i + 0.2j + 0.2k) m(-vi + vj) + 2m (vj) + 3m(vk) + 4m (vi) 10m = v(0.3i + 0.3j + 0.3k) Fi + Fj 10m T= [ {/m₁v₁² - = 1 2 = F 10m (i+j) 11 [m(√2v)² + 2mv² + 3mv² + 4mv²] = 2 G = (Em)r = Ho = Er; x m₂r; = 0 - 2mvdi + 3mv (2d)j - 4mvdk Ⓡ = mvd(-2i+ 6j - 4k) Ho = EMo = -2dFk+ Fdj = Fd(j - 2k) For HG, we use Eq. 4/10: [HG = Ho + px mv] HG = mud(-2i+ 6j - 4k) - d(-0.4i + 0.2j + 0.2k) x = 10m (v)(0.3i + 0.3j + 0.3k) = mv (3i + 3j + 3k) HG = Fd(j - 2k) - d(-0.4i +0.2j + 0.2k) x 10m = Fd(0.2i + 0.8j - 1.4k) Ans. -mv² F mm) (i 10m m 1 2d Ans. 10mv (0.3i+ 0.3j + 0.3k) = mvd(-2i+ 4.2j - 2.2k) Ans. For HG, we could use Eq. 4/9 or Eq. 4/11 with P replaced by O. Using the latter, we have [HG = EMo - pxma] Ans. Ans. Ans. Ans. Ans. (i+j) Ⓒ Ans. √2v 2d x F 2m dl V V 3m 2d 3 4m F HELPFUL HINTS All summations are from i = 1 to 4, and all are performed in order of the mass numbers in the given figure. Because of the simple geometry, the cross products are performed by inspection. Using Eq. 4/10 with P replaced by O is more expedient than using Eq. 4/8 or 4/8a. The m in Eq. 4/10 is the total mass, which is 10m in this example. The quantity p in Eq. 4/10, with P replaced by O, is r. We again recognize that p = r here and that the mass of this system is 10m.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Is there a typo is this solution? Explain.
![SAMPLE PROBLEM 4/1
The system of four particles has the indicated particle masses,
positions, velocities, and external forces. Determine r, r, r, T, G,
Ho, Ho, HG, and HG.
Solution The position of the mass center of the system is
m(2di - 2dj) + 2m(dk) + 3m(−2di) + 4m(dj)
m + 2m +3m 4m
=d(-0.4i + 0.2j + 0.2k)
r =
=
Emiri
Emi
Emir
Emi
EF
Emi
=
T E
= [ {}/{m/v₁²
m(-vi + vj) + 2m(vj) + 3m(vk) + 4m (vi)
10m
= v(0.3i + 0.3j + 0.3k)
Fi + Fj
10m
=
1
2
F
10m
(i+j)
[m(√√2v)² + 2mv² + 3mv² + 4mv²]
= mvd(-2i+ 6j − 4k)
Ho = M₁ = -2dFk + Fdj = Fd(j – 2k)
For HG, we use Eq. 4/10:
[HG = Ho+px mv]
HG = mvd(-2i+ 6j - 4k) - d(-0.4i + 0.2j + 0.2k) x
=
HG = Fd(j - 2k) – d(−0.4i + 0.2j + 0.2k) x 10m
= Fd(0.2i + 0.8j
-
G = (Em)r = 10m (v) (0.3i + 0.3j + 0.3k) = mv (3i + 3j + 3k)
Ho = Σr; × m₂r; = 0 − 2mvdi + 3mv (2d)j – 4mvdk 2
11
2
1.4k)
Ans.
-mv²
F
10m
m
1
(i+j)
2d
Ans.
10mv (0.3i+ 0.3j + 0.3k) = mvd(-2i+ 4.2j - 2.2k)
For HG, we could use Eq. 4/9 or Eq. 4/11 with P replaced by O. Using
the latter, we have
[HG = EMo - px ma]
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
√2v
2d
x
F
2m
2
dl
V
U
3m
2d
4
3
4m
F
y
HELPFUL HINTS
All summations are from i = 1 to 4, and
all are performed in order of the mass
numbers in the given figure.
Because of the simple geometry, the cross
products are performed by inspection.
® Using Eq. 4/10 with P replaced by O is
more expedient than using Eq. 4/8 or
4/8a. The m in Eq. 4/10 is the total
mass, which is 10m in this example. The
quantity p in Eq. 4/10, with P replaced
by O, is r.
We again recognize that p = r here and
that the mass of this system is 10m.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F01c923ca-9bc5-4831-bbcd-a65502aca82d%2Ff7d412e6-8c28-47af-b178-9c3586dcafd2%2Feyb3rvi_processed.jpeg&w=3840&q=75)
Transcribed Image Text:SAMPLE PROBLEM 4/1
The system of four particles has the indicated particle masses,
positions, velocities, and external forces. Determine r, r, r, T, G,
Ho, Ho, HG, and HG.
Solution The position of the mass center of the system is
m(2di - 2dj) + 2m(dk) + 3m(−2di) + 4m(dj)
m + 2m +3m 4m
=d(-0.4i + 0.2j + 0.2k)
r =
=
Emiri
Emi
Emir
Emi
EF
Emi
=
T E
= [ {}/{m/v₁²
m(-vi + vj) + 2m(vj) + 3m(vk) + 4m (vi)
10m
= v(0.3i + 0.3j + 0.3k)
Fi + Fj
10m
=
1
2
F
10m
(i+j)
[m(√√2v)² + 2mv² + 3mv² + 4mv²]
= mvd(-2i+ 6j − 4k)
Ho = M₁ = -2dFk + Fdj = Fd(j – 2k)
For HG, we use Eq. 4/10:
[HG = Ho+px mv]
HG = mvd(-2i+ 6j - 4k) - d(-0.4i + 0.2j + 0.2k) x
=
HG = Fd(j - 2k) – d(−0.4i + 0.2j + 0.2k) x 10m
= Fd(0.2i + 0.8j
-
G = (Em)r = 10m (v) (0.3i + 0.3j + 0.3k) = mv (3i + 3j + 3k)
Ho = Σr; × m₂r; = 0 − 2mvdi + 3mv (2d)j – 4mvdk 2
11
2
1.4k)
Ans.
-mv²
F
10m
m
1
(i+j)
2d
Ans.
10mv (0.3i+ 0.3j + 0.3k) = mvd(-2i+ 4.2j - 2.2k)
For HG, we could use Eq. 4/9 or Eq. 4/11 with P replaced by O. Using
the latter, we have
[HG = EMo - px ma]
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
Ans.
√2v
2d
x
F
2m
2
dl
V
U
3m
2d
4
3
4m
F
y
HELPFUL HINTS
All summations are from i = 1 to 4, and
all are performed in order of the mass
numbers in the given figure.
Because of the simple geometry, the cross
products are performed by inspection.
® Using Eq. 4/10 with P replaced by O is
more expedient than using Eq. 4/8 or
4/8a. The m in Eq. 4/10 is the total
mass, which is 10m in this example. The
quantity p in Eq. 4/10, with P replaced
by O, is r.
We again recognize that p = r here and
that the mass of this system is 10m.
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