Engineering Mechanics: Statics Plus Mastering Engineering with Pearson eText -- Access Card Package (14th Edition) (Hibbeler, The Engineering Mechanics: Statics & Dynamics Series, 14th Edition)
14th Edition
ISBN: 9780134160689
Author: Russell C. Hibbeler
Publisher: PEARSON
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Textbook Question
Chapter 1.6, Problem 11P
Represent each of the following with SI units having an appropriate prefix: (a) 8653 ms, (b) 8368 N, (c) 0.893 kg.
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Select the valid option from the list below.
E
F
G
20 kN
RAX = ?? KN
30°
30°
30°
30°
30°
30°
A
B
D
RAY = ?? KN
A The solution to the problem is found to be -10.0 kN.
B. The solution to the problem is found to be -20.0 KN.
○ C. The solution to the problem is found to be +11.5 kN.
D. The solution to the problem is found to be +23.1 kN.
E. No Valid Answer
Roy = ?? KN
Please do not rely too much on chatgpt, because its answer may be wrong. Please consider it carefully and give your own answer. You can borrow ideas from gpt, but please do not believe its answer.Very very grateful!
Please do not copy other's work,i will be very very grateful!!
Answer by selecting the correct options from the following multichoice selection.
ப
4m
B
A
C
D
3m
3 m
Figure Q17
FL
12 kN
E
16 KN
A. We should resolve forces in the horizontal direction to easily identify the internal force DF.
B. The solution to the problem is found to be -16 kN (C).
C. We should resolve forces in the vertical direction to first identify the internal force DF.
D. We should use Method of Joints at node F to find the internal force in member DF.
E. We should Method of Sections by cutting through members DF, DE and CE.
F. The starting point to solve this problem is to find all reactions at nodes A and B as they will be required for DF calculations.
G. The solution to the problem is found to be 16 kN (T).
H. The most appropriate method to find DF use is Method of Joints.
I. The most appropriate method to use is Method of Sections.
J. A good starting point to solve this problem is to find the horizontal reaction at node B but this is not required to the internal force
Chapter 1 Solutions
Engineering Mechanics: Statics Plus Mastering Engineering with Pearson eText -- Access Card Package (14th Edition) (Hibbeler, The Engineering Mechanics: Statics & Dynamics Series, 14th Edition)
Ch. 1.6 - What is the weight in newtons of an object that...Ch. 1.6 - Represent each of the following combinations of...Ch. 1.6 - Represent each of the following combinations of...Ch. 1.6 - (b) 350 lb/ft3 to kN/m3, (c) 8 ft/h to mm/s....Ch. 1.6 - Represent each of the following as a number...Ch. 1.6 - Round off the following numbers to three...Ch. 1.6 - Represent each of the following quantities in the...Ch. 1.6 - Represent each of the following combinations of...Ch. 1.6 - Represent each of the following combinations of...Ch. 1.6 - Represent each of the following combinations of...
Ch. 1.6 - Represent each of the following with SI units...Ch. 1.6 - Evaluate each of the following to three...Ch. 1.6 - Determine its density in SI units. Use an...Ch. 1.6 - (212 mN)2, (52 800 ms)2, and [548(105)]1/2 ms.Ch. 1.6 - is a dimensionality homogeneous equation which...Ch. 1.6 - To show this, convert 1 Pa = 1 N/m2 to lb/ft2....Ch. 1.6 - What is the density expressed in SI units? Express...Ch. 1.6 - Evaluate each of the following to three...Ch. 1.6 - If the density (mass/Volume) of concrete is 2.45...Ch. 1.6 - If the man is on the moon, where the acceleration...Ch. 1.6 - If they are 800 mm apart, determine the force of...
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- Please do not rely too much on chatgpt, because its answer may be wrong. Please consider it carefully and give your own answer. You can borrow ideas from gpt, but please do not believe its answer.Very very grateful! Please do not copy other's work,i will be very very grateful!!arrow_forward(19) Figure Q19 shows a framework consisting of horizontal members 3 m long and vertical members 4 m long. The framework is loaded at joints J and L with downward load forces of 2 kN. The applied forces cause a vertical reaction forces at A and G and no horizontal reaction force. You are asked to find the internal force in member JK - what would be your approach to solve this problem? Explain your solution process and some of your results by filling in the blanks below. 2 kN 2 kN H RAY RAX A K M N B C D E F 3 m 1 RGY 4m Fill in the multiple blanks. Figure Q19 Finding the vertical reactions is the starting point which can be done by taking moments at A and G but since this is symmetrical loading case the vertical reactions can simply be calculated by halving the total loading 4 kN. Ideally, we can solve the problem using the Method of cutting through the members JK, DJ and It would be sensible to select the left-hand side of the diagram as there are less full members and only one force…arrow_forward4m A 72 kN C E B D F 144 kN 3 m 3 m 3 m Figure Q16 Fill in the multiple blanks below. To find the reactions the starting point is to take moments at a suitable node location. Since node unknowns it is the ideal location to first take moments. By taking moments in a clockwise orientation we find a moment of there is an additional moment of 288 kNm from the load at C. From combining all moments together, we can then find the vertical reaction at F which is RFy= place. For best practice, it is a good approach to take moments at has two kNm due to the force load at node B and KN to 1 decimal in order to the find the vertical reaction RAY- Finally, we can sum forces in the horizontal direction to find the reaction RAX = -72 kN, assuming the reaction at A acts left-to-right. After which we can then sum forces in the vertical direction to verify the sum of RAY plus Rgy is the same as the total downwards force which should be KN.arrow_forward
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