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Title: Resolving of Vectors
Description: Resolving the x, y and z components of three dimensional vectors/forces.
Description: Resolving the x, y and z components of three dimensional vectors/forces.
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Ref: Hibbeler § 2
...
6, Bedford & Fowler: Statics § 2
...
5
Resolving forces refers to the process of finding two or more forces which, when combined, will
produce a force with the same magnitude and direction as the original
...
A resultant force is the force (magnitude and direction) obtained when two or more forces are
combined (i
...
, added as vectors)
...
g
...
These will be demonstrated here using a two-dimensional problem involving coplanar forces
...
The bigger one pulls on
the rope with a force of 270 N (about 60 lbf) at an angle of 55° from horizontal
...
a
...
What is the net force (magnitude and direction) on the hook – that is, calculate the
resultant force
...
That is, each angle
has been measured from the
positive x axis
...
The x- and y-components of force are
indicated schematically, as
Fx
55°
Fy
27
0N
The x- and y-components of the first force (270 N) can be calculated using a little trigonometry
involving the included angle, 55°:
cos(55°) =
Fx1
, or
270 N
Fx1 = (270 N ) cos(55°)
and
sin(55°) =
Fy1
270 N
, or
Fy1 = (270 N ) sin(55°)
...
The factor pi/180 is used
to convert the angle from degrees to radians
...
» F_x1 = 270 * cos( 55 * pi/180 )
F_x1 =
154
...
1711
Your Turn
180
N
110°
Answer, part a)
The larger boy exerts the greatest vertical force (221 N) on the hook
...
Fx
20°
Fy
Show that the x- and y-components of the second force (180 N acting at
110° from the x-axis) are 61
...
Note that trigonometry relationships are based on the
included angle of the triangle (20°, as shown at the right), not the
coordinate angle (-110° from the x-axis)
...
3157
The y-component of the resultant force is 390 N (directed down, or in the –y direction)
...
Then add the two x-components to determine the combined x-directed force, FRx, on the hook
...
FRx
27
18 0
N
77°
0N
62 N
FRy
FR
155 N
» F_Rx = F_x1 + ( -F_x2 )
F_Rx =
93
...
The result is an xcomponent of the resultant force of 93 N in the +x direction
...
» F_R = sqrt( F_Rx ^ 2 + F_Ry ^ 2 )
F_R =
401
...
FRx
θ
77°
» F_Rx = 93
...
316;
» theta = 180/pi * atan2( F_Ry, F_Rx )
theta =
The net force (magnitude and direction) on the hook is now known:
FR = 401 N (about 90 lbf) acting at an angle 76
...
FRy
Answer, part b)
FR
-76
...
Calculations use the included angles
%
(55° & 20°), with minus signs added to both y-component
%
equations to indicate the forces act in the -y direction,
%
and the F_x2 equation to show that this force acts in
%
the -x direction
...
3f N\t F_y1 = %+9
...
3f N\t F_y2 = %9
...
3f N\t F_Ry = %9
...
3f N\n',F_R);
% Calculate the angle of the resultant force
% (in degrees from the x-axis)
theta = atan2( F_Ry, F_Rx ) * 180/pi;
fprintf('theta = %7
Title: Resolving of Vectors
Description: Resolving the x, y and z components of three dimensional vectors/forces.
Description: Resolving the x, y and z components of three dimensional vectors/forces.