Under the man's pushing force P, the uniform cabinet is sliding on the ground with aconstant acceleration of a. Based on the given FBD, select all the correct equationsof motion.1 ft 1 ft→|-1 ft-1 ft→|WP4 ft4 ft3.5 ft3.5 ftBBF,NBOF = P – FA – FB = m · aΟΣ N+ N-W= m· aO F = P – FA – FB = 0O F, = NA+ NB – W = 0OE MA =— — Р.4— W .1+ NB - 2 3 —т. а:3.5OE MA = -P.4 – W .1+ NB · 2 = 0O MG = -P. 0. 5 – (FA+ FB) · 3. 5 – NA ·1+ NB ·1= 0O MB = -P ·4+W•1– NA · 2 = 0О Мв 3 — Р.4+ W:1-NA 2 — —т а 3.5

To choose: The correct equation for the motion. Given: The pushing force is P. The uniform cabinet…

Under the man's pushing force P, the uniform cabinet is sliding on the ground with a constant acceleration of a. Based on the given FBD, select all the correct equations of motion. 1 ft-1 ft- P i 4 ft == G 3.5 ft - B y = m.a P 4 ft 1 ft-1 ft- W G OF₂=P-FA - FB = OFy=NA + NB - W= = m.a OF₂=P-FA - FB = 0 OFy=NA + NB - W=0 OMA=-P.4-W.1+ NB 2 = -m.a.3.5 OMA=-P.4-W.1+ NB 2 = 0 OMG=-P.0.5 (FA+FB) 3.5 NA 1+ NB. 1 = 0 . ΟΣ ΜΒ: -P.4+ W.1- NA 2-0 OMB=-P.4+ W.1-N₁·2=-m.a.3.5 3.5 ft · ma

Under the man's pushing force P, the uniform cabinet is sliding on the ground with a constant acceleration of a. Based on the given FBD, select all the correct equations of motion. 1 ft-1 ft- P i 4 ft == G 3.5 ft - B y = m.a P 4 ft 1 ft-1 ft- W G OF₂=P-FA - FB = OFy=NA + NB - W= = m.a OF₂=P-FA - FB = 0 OFy=NA + NB - W=0 OMA=-P.4-W.1+ NB 2 = -m.a.3.5 OMA=-P.4-W.1+ NB 2 = 0 OMG=-P.0.5 (FA+FB) 3.5 NA 1+ NB. 1 = 0 . ΟΣ ΜΒ: -P.4+ W.1- NA 2-0 OMB=-P.4+ W.1-N₁·2=-m.a.3.5 3.5 ft · ma

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

Under the man's pushing force P, the uniform cabinet is sliding on the ground with a
constant acceleration of a. Based on the given FBD, select all the correct equations
of motion.
1 ft 1 ft→|
-1 ft-1 ft→|
W
P
4 ft
4 ft
3.5 ft
3.5 ft
B
B
F,
NB
OF = P – FA – FB = m · a
ΟΣ N+ N-W
= m· a
O F = P – FA – FB = 0
O F, = NA+ NB – W = 0
OE MA =
— — Р.4— W .1+ NB - 2 3 —т. а:3.5
OE MA = -P.4 – W .1+ NB · 2 = 0
O MG = -P. 0. 5 – (FA+ FB) · 3. 5 – NA ·1+ NB ·1= 0
O MB = -P ·4+W•1– NA · 2 = 0
О Мв 3 — Р.4+ W:1-NA 2 — —т а 3.5

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