calculated using a matrix determinant:Part C - Moment due to two forcesijkM =r x F =| rzryAs shown, a member is fixed at the origin, point O, and has two applied forces, F1 and F2, applied at the free end,rzF, F, Fpoint B.(Figure 3)Notice that the order of the elements of the matrixdeterminant is important; switching rows 2 and 3 of thedeterminant would change the sign of the moment frompositive to negative (or vice versa.)The forces are given by F1 = 115 Ni- 110 Nj+ 55 N k and F2 has magnitude 195 N and direction anglesa = 143.0°, B= 57.0°, and y= 75.2°. The dimensions are x1 = 1.30 m, y1 = 1.85 m, and z1 = 1.10 m.What is the moment about the origin due to the applied forces?Figure3 of 3Express the individual components of the Cartesian vector to three significant figures, separated by commas.• View Available Hint(s)vec?Mo =[i, j, k] N. mSubmitВ< Return to AssignmentProvide Feedback

Given data, The force F→1 = 115 i^-110 j^+55k^ N For force F2:- Force, F2 = 195 N α=143°, β=57° and…

As shown, a member is fixed at the origin, point O, and has two applied forces, F1 and F2, applied at the free end, point B. (Figure 3) The forces are given by F1 115 N i- 110 Nj+55 N k and F2 has magnitude 195 N and direction angles a = 143.0°, B= 57.0°, and y 75.2°. The dimensions are a1 = 1.30 m, y1 = 1.85 m, and z1 = 1.10 m. What is the moment about the origin due to the applied forces? Express the individual components of the Cartesian vector to three significant figures, separated by commas.

As shown, a member is fixed at the origin, point O, and has two applied forces, F1 and F2, applied at the free end, point B. (Figure 3) The forces are given by F1 115 N i- 110 Nj+55 N k and F2 has magnitude 195 N and direction angles a = 143.0°, B= 57.0°, and y 75.2°. The dimensions are a1 = 1.30 m, y1 = 1.85 m, and z1 = 1.10 m. What is the moment about the origin due to the applied forces? Express the individual components of the Cartesian vector to three significant figures, separated by commas.

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

Axial Load

When a bar of the uniform cross-section is acted by a load, such that the load acts along the longitudinal axis of the bar, such kinds of loadings are known as axial loadings. In general terms, a load is an external force acting on a component. An axial load acts perpendicular to the geometric cross-section of the component. Based on the direction of the application of the load, an axial load can be tensile or compressive. The analysis of bodies under the action of axial loads is generally studied under the branch of mechanics, known as statics.

Question

calculated using a matrix determinant:
Part C - Moment due to two forces
i
j
k
M =r x F =| rz
ry
As shown, a member is fixed at the origin, point O, and has two applied forces, F1 and F2, applied at the free end,
rz
F, F, F
point B.
(Figure 3)
Notice that the order of the elements of the matrix
determinant is important; switching rows 2 and 3 of the
determinant would change the sign of the moment from
positive to negative (or vice versa.)
The forces are given by F1 = 115 Ni- 110 Nj+ 55 N k and F2 has magnitude 195 N and direction angles
a = 143.0°, B= 57.0°, and y= 75.2°. The dimensions are x1 = 1.30 m, y1 = 1.85 m, and z1 = 1.10 m.
What is the moment about the origin due to the applied forces?
Figure
3 of 3
Express the individual components of the Cartesian vector to three significant figures, separated by commas.
• View Available Hint(s)
vec
?
Mo =[
i, j, k] N. m
Submit
В
< Return to Assignment
Provide Feedback

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