Why are there no reaction forces of Fx and Fx at the joint F? 1,14 An aireraft tow bar is positioned by means of a single hydraulic cylinderconnected by a 25-mm-diameter steel rod to two identical arm-and-wheel units DEF.The mass of the entire tow bar is200 kg, and its centerof gravityis located at G For-the position shown, determine the normal stress in the rod.Problem 1.14Dimensions in mm1150100450B|E250850500675825FREE BODYENTRE Tow BAR :(200 kg)(4.81 m/s?) = 1962.0O N1150 mmAx+5EMA = 0 :A85OR1150(1962.00 N)R850 mmR = 2654,5 NFREE BODY - BOTH ARMWHEELUNITS:tan d675メニ8.4270°+5 EMe = 0 :100 mm(FeDcos x)(sso) - R(5o0) :450 mmEx500FeD550 cos 8.4270 (2654.5 N)= 2439.5 N(COMP.)2439.5 NTI (0.0125 m)?S00mm ' 675mmFeRAcoR= 2654.S KN4.9697 * 1o Pa4.97 MPa

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Why are there no reaction forces of Fx and Fx at the joint F?

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.

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Why are there no reaction forces of Fx and Fx at the joint F?

1,14 An aireraft tow bar is positioned by means of a single hydraulic cylinder
connected by a 25-mm-diameter steel rod to two identical arm-and-wheel units DEF.
The mass of the entire tow bar is200 kg, and its centerof gravityis located at G For-
the position shown, determine the normal stress in the rod.
Problem 1.14
Dimensions in mm
1150
100
450
B
|E
250
850
500
675
825
FREE BODY
ENTRE Tow BAR :
(200 kg)(4.81 m/s?) = 1962.0O N
1150 mm
Ax
+5EMA = 0 :
A
85OR
1150(1962.00 N)
R
850 mm
R = 2654,5 N
FREE BODY - BOTH ARM
WHEEL
UNITS:
tan d
675
メニ8.4270°
+5 EMe = 0 :
100 mm
(FeDcos x)(sso) - R(5o0) :
450 mm
Ex
500
FeD
550 cos 8.4270 (2654.5 N)
= 2439.5 N
(COMP.)
2439.5 N
TI (0.0125 m)?
S00mm ' 675mm
FeR
Aco
R= 2654.S KN
4.9697 * 1o Pa
4.97 MPa

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