Problem 4.128 Two systems of forces act on the beam. Are they equivalent? Strategy: Check the two conditions for equivalence. The sums of the forces must be equal, and the sums of the moments about an arbitrary point must be equal. System 1 100 N SON T

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Problem 4.128 Two systems of forces act on the beam.
Are they equivalent?
Strategy: Check the two conditions for equivalence.
The sums of the forces must be equal, and the sums of
the moments about an arbitrary point must be equal.
System 1
100 N
SO N
Im
Ans. Yes
System 2
50N
Solution: The strategy is to check the two conditions for equiv-
alence: (a) the sums of the forces must be equal and (b) the sums of
the moments about an arbitrary point must be equal. The sums of the
forces of the two systems: EFx 0, (both systems) and
System 1
100 N
= -100 +50 =-50 (N)
50 N
=-50 (N).
The sums of the forces are equal. The sums of the moments about the
left end are:
y
System 2
50 N
M1 = -(1)(100) = -100k (N-m) C.W
%3D
2 m
M2 =-(2)(50) =-100k (N-m).
C.W
Thus the sums of the moments about any point on the beam are equal
for the two sets of forces; the systems are equivalent. Yes
Transcribed Image Text:Problem 4.128 Two systems of forces act on the beam. Are they equivalent? Strategy: Check the two conditions for equivalence. The sums of the forces must be equal, and the sums of the moments about an arbitrary point must be equal. System 1 100 N SO N Im Ans. Yes System 2 50N Solution: The strategy is to check the two conditions for equiv- alence: (a) the sums of the forces must be equal and (b) the sums of the moments about an arbitrary point must be equal. The sums of the forces of the two systems: EFx 0, (both systems) and System 1 100 N = -100 +50 =-50 (N) 50 N =-50 (N). The sums of the forces are equal. The sums of the moments about the left end are: y System 2 50 N M1 = -(1)(100) = -100k (N-m) C.W %3D 2 m M2 =-(2)(50) =-100k (N-m). C.W Thus the sums of the moments about any point on the beam are equal for the two sets of forces; the systems are equivalent. Yes
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