a.ag2x = -15,2 & ag2y = -32,6b.ag2x = -9,3 & ag2y = -34,8c.aG2x = -20,7 & ag2y = -29,5d.ag2x = -3,1 & ag2y = -35,9e.aG2x = -32,6 & ac2y = -15,2f.ag2x = -35,9 & ag2y = -3,1g.aG2x = -29,5 & aG2y = -20,7h.ag2x = -34,8 & aczy = -9,3i.aG2x = -25,5 & agzy = -25,5j.a62x = 3,1 & ag2y = -35,9 (3)3(4)04>X|(1)References:a1 = 160 mm; a2 = 80 mm; a3 = 170 mm; a4 = 120 mm;02 = 15 degrees, W2= 30 rad / s, 02 = 0 rad/s2According to the above four-bar mechanism, the center of gravity of the limb (2) is horizontal andlocate the vertical linear acceleration components ( ag2x and ag2y) in unit m/s2. (Centers of gravitythe midpoints of the bars are)

∗a1=160mm=0.16m∗a2=80mm=0.08m∗a3=170mm=0.17m∗a4=120mm=0.12m To find:∗For centre of…

According to the above four-bar mechanism, the center of gravity of the limb (2) is horizontal locate the vertical linear acceleration components ( ag2x and ag2y) in unit m/s2. (Centers of gravity the midpoints of the bars are)

According to the above four-bar mechanism, the center of gravity of the limb (2) is horizontal locate the vertical linear acceleration components ( ag2x and ag2y) in unit m/s2. (Centers of gravity the midpoints of the bars are)

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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Concept explainers

Basic Terminology in Mechanics

The area of science and mathematics concerned the motion of the objects, the relation between the force matter, and motion if specified. Forces exerted over the objects result in displacements or changes in the position of an object concerning its environment. This branch of physics keeps its roots in Ancient Greece in the form of the scripts of Aristotle and Archimedes.

Question

a.ag2x = -15,2 & ag2y = -32,6
b.ag2x = -9,3 & ag2y = -34,8
c.aG2x = -20,7 & ag2y = -29,5
d.ag2x = -3,1 & ag2y = -35,9
e.aG2x = -32,6 & ac2y = -15,2
f.ag2x = -35,9 & ag2y = -3,1
g.aG2x = -29,5 & aG2y = -20,7
h.ag2x = -34,8 & aczy = -9,3
i.aG2x = -25,5 & agzy = -25,5
j.a62x = 3,1 & ag2y = -35,9

(3)
3
(4)
04
>X
|(1)
References:
a1 = 160 mm; a2 = 80 mm; a3 = 170 mm; a4 = 120 mm;
02 = 15 degrees, W2= 30 rad / s, 02 = 0 rad/s2
According to the above four-bar mechanism, the center of gravity of the limb (2) is horizontal and
locate the vertical linear acceleration components ( ag2x and ag2y) in unit m/s2. (Centers of gravity
the midpoints of the bars are)

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