the belt forces shown with an equivalent force-couple system at A. = Fig. P3.120 225 mm 145 N ? 215 N 225 180 mm 10 Pulley B FB (0,215 145cos (20°), 145sin(20°)) N H MB (0.075) (215+145) 10 Pulley C Fc (0,- 155sin (10°) - 240sin (10°), ?) N 155 N 240 N m.N
the belt forces shown with an equivalent force-couple system at A. = Fig. P3.120 225 mm 145 N ? 215 N 225 180 mm 10 Pulley B FB (0,215 145cos (20°), 145sin(20°)) N H MB (0.075) (215+145) 10 Pulley C Fc (0,- 155sin (10°) - 240sin (10°), ?) N 155 N 240 N m.N
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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
Transcribed Image Text:1
3.120 Two 150-mm-diameter pulleys are mounted on line shaft AD. The
belts at B and Clie in vertical planes parallel to the yz plane. Replace
the belt forces shown with an equivalent force-couple system at A.
Fig. P3.120
235 mm
145 N
215 N
Mc = (?)(?) 2
F
225 mm
180 mm
10⁰
Pulley B
FB = (0,215 - 145cos (20°), 145sin(20°)) N
(0.075) (-215 +145)
MB
10⁰
Pulley C
Fc = (0,- 155sin(10°) - 240sin(10°), ?) N
=....?
MA = Mg + Mc +TAB × FB +raỞ × Fo
(?, ?, ?)
155 N
240 N
m.N
m.N
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