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- OKAY. WE'RE NOW DOING MODULE 3, EXAMPLE PROBLEM 3.
AND BASICALLY, IT'S A PROBLEM SHOWING THREE FORCE VECTORS,
F1, F2 AND F3,
AND THEY'RE ASKING TO FIND THE RESULTANT
AND WHAT IS ITS DIRECTION MEASURED FROM THE +X-AXIS.
SO THE THREE FORCES WE HAVE--
LET'S DO OUR X, Y, Z AXIS, X, Y-AXIS.
THE FORCE F1--
F1 IS EQUAL TO 60 POUNDS.
AND THE WAY THEY GIVE THE DIRECTION IS A (1,1).
SO IT GOES 1 IN THE X AND 1 IN THE Y,
AND THE HYPOTENUSE OF THAT WOULD BE THEN SQUARE ROOT OF 2.
SO THAT GIVES YOU THE DIRECTION THAT'S ACTUALLY 45 DEGREES,
BUT WE'LL LEAVE IT THAT WAY FOR NOW.
AND THEN THE OTHER FORCE IS F2, AND THAT'S EQUAL TO 70 POUNDS,
AND ITS GIVEN DIRECTION IS 60 DEGREES FROM THERE.
AND THEN WE HAVE F3,
WHICH IS ACTING IN THE -Y DIRECTION,
SO F3 IS EQUAL TO 50 POUNDS MAGNITUDE
AND IT'S ACTING DOWN.
SO WE'RE GOING TO REPRESENT THESE FORCES
IN CARTESIAN FORMAT--
IN OTHER WORDS, THE "I," J AND K,
THE UNIT VECTOR FORMAT.
SO FORCE F1 THEN REPRESENTED THAT WAY
IS THE "I" DIRECTION, THE X COMPONENT
IS IN THE MINUS DIRECTION.
SO IT'S GOING TO BE -60 TIMES THE X COMPONENT,
WHICH IS REALLY 1 OVER SQUARE ROOT OF 2
IS THE RATIO.
AND THAT WOULD BE IN THE "I" DIRECTION
AS A UNIT VECTOR.
SO THE IBAR DECLINES ITS DIRECTION.
THEN IN THE Y DIRECTION, WHERE A +60,
AND ALSO IT'S 1 OVER SQUARE ROOT OF 2,
AND THAT'S IN THE JBAR DIRECTION.
SO CLEANING THAT UP,
WE GET A -42.4 IBAR PLUS 42.4 JBAR POUNDS.
THAT'S CARTESIAN REPRESENTATION OF FORCE F1.
OKAY. WE'LL DO THE SAME THING FOR F2.
NOW THE X COMPONENT OF F2 IS OVER HERE.
SO THAT'S GOING TO BE A -70 TIMES SINE OF 60 DEGREES JBAR.
EXCUSE ME. THAT'S IBAR, X DIRECTION.
CROSS THAT OUT, IT'S IBAR.
AND THEN -70 COSINE 60 DEGREES JBAR,
THE Y COMPONENT.
SO CLEANING THAT UP, WE GET A -60.6 IBAR
-35 JBAR POUNDS, THAT'S F2.
F3 IS GOING TO BE EASY.
THAT'S SIMPLY A -50 JBAR POUNDS.
OKAY.
SO THE RESULTANT FORCE THEN IS GOING TO BE THE SUM
OF THOSE THREE FORCES.
SO F SUB R RESULTANT IS GOING TO BE SUMMATION
OF ALL OF THOSE FORCES.
AND HERE'S WHAT WE GET.
WE GET OUR -42.4 -60.6 IBAR,
SO THAT'S ADDING ALL THE "I" COMPONENTS,
PLUS WE GET 42.4 - 35 - 50 J BAR,
ALL THOSE ARE POUNDS.
AND IF WE CLEAN THAT UP SOME, WE'RE GOING TO GET -103.0I
- 42.6 J BAR POUNDS.
THE MAGNITUDE NOW OF F SUB R--
THE MAGNITUDE OF F SUB R IS GOING TO BE SQUARE ROOT
OF THE COMPONENT SQUARED.
SO IT'S A -103.0 SQUARED PLUS A -42.6 SQUARED.
AND THAT COMES OUT TO 111 POUNDS,
THAT'S THE MAGNITUDE.
THEN OUR RESULTANT FORCE, IT'S MINUS-MINUS,
SO IT'S GOING TO BE SOME KIND OF A RESULTANT F SUB R HERE.
AND WE'LL LOOK AT THAT THETA PRIME,
SO WE'LL CALL THAT THETA PRIME IS EQUAL TO THE ARCTANGENT
OF OUR X/Y COMPONENTS, 42.57 DIVIDED BY 103.0,
AND IT COMES OUT TO 22.4 DEGREES.
AND SO THEN THE REAL THETA THAT THEY ASKED FOR
WAS FROM THE X-AXIS.
SO THAT WOULD BE FROM ALL THE WAY OVER HERE.
THAT'S THE THETA THEY WANT.
AND THAT WOULD JUST SIMPLY BE 22.4 PLUS 180 DEGREES
OR 202 DEGREES. �