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- OKAY. WE'RE NOW DOING MODULE 3, HOMEWORK PROBLEM 3.
THE PROBLEM STATES THAT ANGLE PHI IS 30,
F2 IS 3 KILONEWTON FORCE.
SO DETERMINE THE MAGNITUDE OF THE RESULTANT FORCE
ACTING ON THE PLANE AND ITS THETA DIRECTION
MEASURED CLOCKWISE FROM THE POSITIVE X-AXIS.
OKAY.
SO WE'RE GOING TO DRAW OUR X AND Y AXIS HERE
AND PUT ALL THE FORCES IN HERE.
SO WE HAVE A -40 KILONEWTON FORCE ACTING DOWN THIS WAY,
AND IT'S ACTING AT A 30-DEGREE ANGLE HERE.
AND WE HAVE A 3 KILONEWTON FORCE ACTING HERE,
AND IT'S A 30 DEGREES FROM THE X-AXIS.
AND WE HAVE A 5 KILONEWTON FORCE ACTING IN A DIRECTION DEFINED
BY A 3 DOWN AND 4 OVER DIRECTION.
AND SINCE THAT'S A STANDARD 3, 4, 5 TRIANGLE,
THEN WE KNOW THE HYPOTENUSE WILL BE 5.
SO NOW WE CAN TAKE--
ADD UP ALL OF THE X COMPONENTS OF THESE FORCES
AND THE Y COMPONENTS OF THESE FORCES
AND THEN WE'LL GET A TOTAL X AND Y COMPONENT
FOR THE RESULTANT FORCE OF THOSE.
SO LET'S ADD UP ALL OF THE X COMPONENTS
SO SOME OF THE FORCES IN THE X DIRECTION.
OKAY.
WE HAVE THE 4 KILONEWTON FORCE IS GOING TO BE IN THE MINUS--
X COMPONENT IS IN THE MINUS DIRECTION.
SO IT'LL BE A -4 KILONEWTONS TIMES THE SINE OF 30 DEGREES
TO GIVE THE X COMPONENT.
THEN WE HAVE 3--
WE HAVE A +3 KILONEWTONS TIMES THE COSINE OF 30 DEGREES
TO GIVE YOU THE X COMPONENT OF THAT FORCE.
AND THEN WE HAVE THE 5 KILONEWTONS
PLUS THE X COMPONENT IS GOING TO BE--
THE X IS 4, SO THAT'S 4/5 TIMES THE 5.
SO WE HAVE 4/5 TIMES 5 KILONEWTONS.
AND SO ALL OF THIS ADDS UP.
YOU ADD THIS ALL UP, YOU GET 4.60 KILONEWTONS ACTING
IN THE +X DIRECTION.
THEN SAME THING FOR SUMMATION OF FORCES IN THE Y DIRECTION.
OKAY.
IN THE Y, THE 4 KILONEWTONS IS THEN ACTING--
THE Y COMPONENT WILL BE DOWN.
SO IT'S GOING TO BE -4 KILONEWTONS TIMES
NOW IT'S THE COSINE OF 30 DEGREES PLUS--
THEN THE 3 KILONEWTON FORCE,
THE Y COMPONENT WOULD BE THE SINE OF 30 DEGREES.
AND THEN THE Y COMPONENT WOULD BE THE 3/5 RATIO.
SO IT'S A 3/5 TIMES THE 5 KILONEWTONS.
AND SO WE ADD ALL OF THIS ALL UP AND WE GET -4.96 KILONEWTONS,
AND THAT MEANS IT'S ACTING IN THE DOWN DIRECTION.
OKAY.
SO NOW WE HAVE THE X AND Y COMPONENTS
OF THE RESULTANT FORCE.
SO THEN WHAT IS THE MAGNITUDE OF THE RESULTANT FORCE?
SO THE MAGNITUDE OF THE RESULTANT FORCE
WOULD TAKE A SQUARE ROOT OF 4.60 SQUARED
PLUS THE -4.96 SQUARED,
AND WE GET A MAGNITUDE OF 6.77 KILONEWTONS.
AND THEN WE NEED TO FIND ITS DIRECTION
TO MAKE IT A VECTOR QUANTITY.
SO THEN THETA IS GOING TO BE EQUAL
TO THE ARCTANGENT OF F SUB RY DIVIDED BY F SUB RX,
THE MAGNITUDES IN THE Y AND X OF THE RESULTANT.
AND THIS WOULD BE THE ARCTANGENT OF 4.96
DIVIDED BY 4.60,
AND WE GET THETA EQUAL TO 47.2 DEGREES. �