A car engine with a given weight of w, is suspended in air by a cable that is connected to a MASSLESS ring. Whilethe ring is supported by two cables attached to the ceiling and the wall as shown in the figure below.(a) Engine, chains, and ring60°T3T2 Using the cartesian coordinate system, Which of the following is the correct NET FORCE on the y-axis for the ring?O EFy= T3cos60 + T1= mao EFy=T3cos60 + T1 = 0o EFy= T3sin60 - T1=0o EFy=T3cos60 – T1= maο Σ-Τι- TSsin60 = 0o EFy=T1+T35in60 = 0ο Σ-Τ- T; 0

The free-body diagram of the ring will be as follows,

Using the cartesian coordinate system, Which of the following is the correct NET FORCE on the y-axis for the ring? o EFy=T3cos60 + T1 = ma o EFy=T3cos60+ T1 =0 o EFy=T3sin60 – T1=0 o EFy=T3cos60-T1=ma o EFy=T1-T3sin60 = 0 O EFy=T1+T3sin60 = 0

Using the cartesian coordinate system, Which of the following is the correct NET FORCE on the y-axis for the ring? o EFy=T3cos60 + T1 = ma o EFy=T3cos60+ T1 =0 o EFy=T3sin60 – T1=0 o EFy=T3cos60-T1=ma o EFy=T1-T3sin60 = 0 O EFy=T1+T3sin60 = 0

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

Center of Gravity and Centroid

The term center of gravity indicates the position or location of a point about which the whole weight of a body or system is supposed to be concentrated, which means the whole weight of the body or system works through this point. It is closely related to mass distribution. For all positions of the body, the center of gravity would be a single point.

Question

A car engine with a given weight of w, is suspended in air by a cable that is connected to a MASSLESS ring. While
the ring is supported by two cables attached to the ceiling and the wall as shown in the figure below.
(a) Engine, chains, and ring
60°
T3
T2

Using the cartesian coordinate system, Which of the following is the correct NET FORCE on the y-axis for the ring?
O EFy= T3cos60 + T1= ma
o EFy=T3cos60 + T1 = 0
o EFy= T3sin60 - T1=0
o EFy=T3cos60 – T1= ma
ο Σ-Τι- TSsin60 = 0
o EFy=T1+T35in60 = 0
ο Σ-Τ- T; 0

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