A force F = 10 sin at N acts on a displacement of x = 2 sin (ư – T16)m. Determine (a) the work done during the first 6 s; (b) the work done during the first 12s. Solution: Rewriting Eq. (3.7-1) as W=[Fx°dt and substituting F = Fo sin t and x = X sin ( t- ) gives the work done per cycle of W=nF0X0 sin ở For the force and displacement given in this problem, Fo = 10 N, X = 2 m, = 1/6and the period = 2s. Thus, in the 6 s specified in (a), three complete cycles take place, and the work done is W=3(nF0X0sin o)=3rx10x2 × sin 30°=94.2 N •m The work done in part (b) is determined by integrating the expression for work between the limits 0 and 12s. W=@F0X0[cos 30°J01/2sin at cos at dt+sin 30°J01/2sin at dt] =Tx10x2[-0.8664rcos 2rt+0.50(12–sin 2nt4r)]01/2 =16.51N:1
A force F = 10 sin at N acts on a displacement of x = 2 sin (ư – T16)m. Determine (a) the work done during the first 6 s; (b) the work done during the first 12s. Solution: Rewriting Eq. (3.7-1) as W=[Fx°dt and substituting F = Fo sin t and x = X sin ( t- ) gives the work done per cycle of W=nF0X0 sin ở For the force and displacement given in this problem, Fo = 10 N, X = 2 m, = 1/6and the period = 2s. Thus, in the 6 s specified in (a), three complete cycles take place, and the work done is W=3(nF0X0sin o)=3rx10x2 × sin 30°=94.2 N •m The work done in part (b) is determined by integrating the expression for work between the limits 0 and 12s. W=@F0X0[cos 30°J01/2sin at cos at dt+sin 30°J01/2sin at dt] =Tx10x2[-0.8664rcos 2rt+0.50(12–sin 2nt4r)]01/2 =16.51N:1
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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! plz do this problem in steps it is solved already
![A force F = 10 sin at N acts on a displacement of x = 2 sin (nt – n16)m. Determine (a) the work done during the first 6 s; (b) the
work done during the first 12s.
Solution: Rewriting Eq. (3.7-1) as W=[Fx°dt and substituting F = Fo sin
t and x = X sin (
t- ) gives the work done per cycle of
W=nF0X0 sin ở
For the force and displacement given in this problem, Fo = 10 N, X = 2 m,
= n/6and the period
= 2s. Thus, in the 6 s specified in (a),
three complete cycles take place, and the work done is
W=3(nF0X0sin o)=3rx10x2 × sin 30°=94.2 N •m
The work done in part (b) is determined by integrating the expression for work between the limits 0 and 12s.
W=WF0X0[cos 30°f01/2sin nt cos at di+sin 30°f01/2sin nt dt]
=Tx10x2[-0.8664rcos 2rt+0.50(12–sin 2rt47)]01/2
=16.51N:1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3bee9a1b-b284-4a8d-a430-6be5e5a0e777%2Fcd1a1221-408f-459a-9986-e9de9fa90842%2Fzhfulwy_processed.png&w=3840&q=75)
Transcribed Image Text:A force F = 10 sin at N acts on a displacement of x = 2 sin (nt – n16)m. Determine (a) the work done during the first 6 s; (b) the
work done during the first 12s.
Solution: Rewriting Eq. (3.7-1) as W=[Fx°dt and substituting F = Fo sin
t and x = X sin (
t- ) gives the work done per cycle of
W=nF0X0 sin ở
For the force and displacement given in this problem, Fo = 10 N, X = 2 m,
= n/6and the period
= 2s. Thus, in the 6 s specified in (a),
three complete cycles take place, and the work done is
W=3(nF0X0sin o)=3rx10x2 × sin 30°=94.2 N •m
The work done in part (b) is determined by integrating the expression for work between the limits 0 and 12s.
W=WF0X0[cos 30°f01/2sin nt cos at di+sin 30°f01/2sin nt dt]
=Tx10x2[-0.8664rcos 2rt+0.50(12–sin 2rt47)]01/2
=16.51N:1
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