Please answer nos. 2, 4, & 5. The 7-kg slender bar AB is released from rest in the position shown. Neglect friction in all contact surfaces. The acceleration of the mass center G immediatelyafter it was released from rest in terms of the angular acceleration of the bar a AB is (3a ABî – 2a AB).BNg6 m6 mAA4 m4 mNA2. Which of the following equations of motion is CORRECT and can be used to solve for the angular acceleration of rod AB?> Mc: 2.0W = (7)(3.0)(aAB)(3.0) + (7)(2.0)(aAB)(2.0) + (IAB)(CAB)A.ссw +ΣMc: -2.0W = -(7)(3.0)(ªAB)(2.0) + (7)(2.0)(aAB)(3.0) + (IAB)(ªAB)В.ccw +: > Mc:-2.0W = -(7)(3.0)(aAB)(3.0) + (7)(2.0)(aAB)(2.0) + (IAB)(aAB)C.ссw +ccw + > Mc: 2.0W = (7)(3.0)(aAB)(2.0) + (7)(2.0)(ªAB)(3.0) + (IaB)(ªaB)D.4. Which of the following best approximates the magnitude of the reaction at A?33.8 N52.8 N71.2 N23.8 N5. Which of the following gives the magnitude and direction of the acceleration of end A in m/s^2?5.30 m/s2 t4.53 m/s2 14.53 m/s2 t5.30 m/s2 1

Answered: Image /qna-images/answer/94598751-ab3d-410e-8838-f63b086f308f.jpg

The 7-kg slender bar AB is released from rest in the position shown. Neglect friction in all contact surfaces. The acceleration of the mass center G immediately after it was released from rest in terms of the angular acceleration of the bar a AB is (3a ABi – 2a AB). B B Ng 6 m 6 m

The 7-kg slender bar AB is released from rest in the position shown. Neglect friction in all contact surfaces. The acceleration of the mass center G immediately after it was released from rest in terms of the angular acceleration of the bar a AB is (3a ABi – 2a AB). B B Ng 6 m 6 m

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

Please answer nos. 2, 4, & 5.

The 7-kg slender bar AB is released from rest in the position shown. Neglect friction in all contact surfaces. The acceleration of the mass center G immediately
after it was released from rest in terms of the angular acceleration of the bar a AB is (3a ABî – 2a AB).
B
Ng
6 m
6 m
A
A
4 m
4 m
NA
2. Which of the following equations of motion is CORRECT and can be used to solve for the angular acceleration of rod AB?
> Mc: 2.0W = (7)(3.0)(aAB)(3.0) + (7)(2.0)(aAB)(2.0) + (IAB)(CAB)
A.
ссw +
Σ
Mc: -2.0W = -(7)(3.0)(ªAB)(2.0) + (7)(2.0)(aAB)(3.0) + (IAB)(ªAB)
В.
ccw +
: > Mc:-2.0W = -(7)(3.0)(aAB)(3.0) + (7)(2.0)(aAB)(2.0) + (IAB)(aAB)
C.
ссw +
ccw + > Mc: 2.0W = (7)(3.0)(aAB)(2.0) + (7)(2.0)(ªAB)(3.0) + (IaB)(ªaB)
D.
4. Which of the following best approximates the magnitude of the reaction at A?
33.8 N
52.8 N
71.2 N
23.8 N
5. Which of the following gives the magnitude and direction of the acceleration of end A in m/s^2?
5.30 m/s2 t
4.53 m/s2 1
4.53 m/s2 t
5.30 m/s2 1

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