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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The name of the Course subject is Statics of Rigid Bodies

Friction Problem
350kN
A 350 kN block is resting on a rough
horizontal surface for which the
coefficient of friction is 0.45. Determine
the force P required to cause motion to
impend if applied to the block.
A. Horizontally
B. Downward at 30' with the horizontal.
C. What minimum force is required to
F
N
start motion?
Transcribed Image Text:Friction Problem 350kN A 350 kN block is resting on a rough horizontal surface for which the coefficient of friction is 0.45. Determine the force P required to cause motion to impend if applied to the block. A. Horizontally B. Downward at 30' with the horizontal. C. What minimum force is required to F N start motion?
O TWO FORCES PBQ PASS
COMPEETE SOLUTION IN JPEG FORMAT
SOLVE THE FOLLOWING PROBLEMS AND SUBMIT A PHOTO OF THE
TECHNOLOGICAL INSTITUTE OF THE PHILS;OC
STATIC OF RIGID BODIES (O01)
CVE THE FOLOWING PROBLEMS AND SUeMIT A PHOTO OF THE
P-200N
19
300
THROUGH A PONST A WHICH
IS Am TO THE RIGHT OF 3M
ABOVE A MOMENT CENTERO
FORCE P IS 200 N DIRECTED.
UP TO THE RIGHT AT 30° WITH
THE HORIZONTAL AND FORCE
Q IS 100N DIRECTED UP TO
THE LEFT AT 60° WITH THE
HORIZONTAL, DETERM INE
THE MOMENT OF THE
RESULTANT OF THESE TWO
FORCES TO o.
A
4m
M- ZFx= 200 Ki t 100 N
= 300
R= VrB00)²+(4*) + (3)
R= 300.091
Mz 200(4) †100 (3)
800 + 300
M=D1100 N
EQUILIBRIUM
%3D
DETERMINE
20N
40 N
M.100
300.9
= 3.67d
30N
ZFO)
>20cos (0) + 30 cos (0) + 40 cos(o)
Ide 3.67
%3D
tam- 40 N
30N
'A=30N
ZFly)-O
O = 20 sin(0) -30sın(0) + 40 cos(0) A - 30
Asre 40N
Acose 30N|
e-53-13
- 53. 13°
A 37.50
sin(53 B9)
Transcribed Image Text:O TWO FORCES PBQ PASS COMPEETE SOLUTION IN JPEG FORMAT SOLVE THE FOLLOWING PROBLEMS AND SUBMIT A PHOTO OF THE TECHNOLOGICAL INSTITUTE OF THE PHILS;OC STATIC OF RIGID BODIES (O01) CVE THE FOLOWING PROBLEMS AND SUeMIT A PHOTO OF THE P-200N 19 300 THROUGH A PONST A WHICH IS Am TO THE RIGHT OF 3M ABOVE A MOMENT CENTERO FORCE P IS 200 N DIRECTED. UP TO THE RIGHT AT 30° WITH THE HORIZONTAL AND FORCE Q IS 100N DIRECTED UP TO THE LEFT AT 60° WITH THE HORIZONTAL, DETERM INE THE MOMENT OF THE RESULTANT OF THESE TWO FORCES TO o. A 4m M- ZFx= 200 Ki t 100 N = 300 R= VrB00)²+(4*) + (3) R= 300.091 Mz 200(4) †100 (3) 800 + 300 M=D1100 N EQUILIBRIUM %3D DETERMINE 20N 40 N M.100 300.9 = 3.67d 30N ZFO) >20cos (0) + 30 cos (0) + 40 cos(o) Ide 3.67 %3D tam- 40 N 30N 'A=30N ZFly)-O O = 20 sin(0) -30sın(0) + 40 cos(0) A - 30 Asre 40N Acose 30N| e-53-13 - 53. 13° A 37.50 sin(53 B9)
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