0-27Sm275 mm30°FHEA) d=0.275300:5mE500 mmD0:5mB) d=0.5500 mm300 mmC) d=0.5*cos30friction coefficient on the crate surface (at A and at B)The system can hold the crate as shown in the figure.The system is in equilibrium. Ignore weight of the system.D) d=0.275*sin30the weight of the crate is 200 N.N =normal force at A=normal force at BNothe moment about C for the system of ECAE) d=0.3EMc = F *(0.5)**cos 30°+ Fey * sin 30° *C0.275) – N,*(0.5)- UN*(d) = 0%3DWhat is 'd' value?d=?

Answered: 275 mm 30 30° E 0.5m FHE A) d=0.275 500…

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

0-27Sm
275 mm
30°
FHE
A) d=0.275
30
0:5m
E
500 mm
D
0:5m
B) d=0.5
500 mm
300 mm
C) d=0.5*cos30
friction coefficient on the crate surface (at A and at B)
The system can hold the crate as shown in the figure.
The system is in equilibrium. Ignore weight of the system.
D) d=0.275*sin30
the weight of the crate is 200 N.
N =normal force at A
=normal force at B
No
the moment about C for the system of ECA
E) d=0.3
EMc = F *(0.5)**cos 30°+ Fey * sin 30° *C0.275) – N,*(0.5)- UN*(d) = 0
%3D
What is 'd' value?
d=?

Expert Solution

Step 1

If a force F is inclined at an angle $θ$ with the horizontal, then the component of force along the horizontal direction is the product of F and $\cos θ$ . Similarly, the component of force along the vertical direction is product of F and $\sin θ$ .

Moment of force about a point is determined as the product of force and its perpendicular distance from the point considered. It can be expressed as,

$M = F d$

Here, moment is M, force is F and the perpendicular distance is d.

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