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Resolving Forces, Calculating Resultants
Ref: Hibbeler § 2.4-2.6, Bedford & Fowler: Statics § 2.3-2.
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. The most common use of the process is finding the components of the original force in the Cartesian coordinate directions: x, y, and z.
A resultant force is the force (magnitude and direction) obtained when two or more forces are combined (i.e., added as vectors).
Breaking down a force into its Cartesian coordinate components (e.g., F (^) x, F (^) y) and using Cartesian components to determine the force and direction of a resultant force are common tasks when solving statics problems. These will be demonstrated here using a two-dimensional problem involving co- planar forces.
Example: Co-Planar Forces
Two boys are playing by pulling on ropes connected to a hook in a rafter. The bigger one pulls on the rope with a force of 270 N (about 60 lb (^) f) at an angle of 55° from horizontal. The smaller boy pulls with a force of 180 N (about 40 lb (^) f) at an angle of 110° from horizontal.
a. Which boy is exerting the greatest vertical force (downward) on the hook?
b. What is the net force (magnitude and direction) on the hook – that is, calculate the resultant force.
270 N
180 N
-110° (^) -55°
Solution
First, consider the 270 N force acting at 55° from horizontal. The x- and y-components of force are indicated schematically, as
Note: The angles in this figure have been indicated as coordinate direction angles. That is, each angle has been measured from the positive x axis.
55°
270 N
F x
F
y
The x- and y-components of the first force (270 N) can be calculated using a little trigonometry involving the included angle, 55°:
270 N
F
cos( 55 °) = x^1 , or Fx 1 =( 270 N) cos( 55 °)
and
270 N
F
sin( 55 )
y 1 ° = , or Fy 1 = ( 270 N) sin( 55 °).
MATLAB can be used to solve for Fx1 and F (^) y1 using its built-in sin() and cos() functions, but these functions assume that the angle will be expressed as radians, not degrees. The factor pi/180 is used to convert the angle from degrees to radians. Note that pi is a predefined variable in MATLAB.
» F_x1 = 270 * cos( 55 * pi/180 )
F_x1 =
» F_x1 = 270 * sin( 55 * pi/180 )
F_x1 =
Your Turn
Show that the x- and y-components of the second force (180 N acting at 110° from the x-axis) are 61.5 N (-x direction) and 169 N (-y direction), respectively. 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).
Answer, part a)
The larger boy exerts the greatest vertical force (221 N) on the hook. The vertical force exerted by the smaller boy is only 169 N.
20°
180 N
110°
F
y
F x
270 N 180 N
77°
F
R^
F Rx
F
Ry
62 N
155 N
» F_Rx = F_x1 + ( -F_x2 )
F_Rx =
The minus sign was included before Fx2 because it is directed in the –x direction. The result is an x- component of the resultant force of 93 N in the +x direction.
Once the x- and y-components of the resultant force have been determined, the magnitude can be calculated using
2 Ry
2 FR = FRx +F
The MATLAB calculation uses the built-in square-root function sqrt().
» F_R = sqrt( F_Rx ^ 2 + F_Ry ^ 2 )
F_R =
The angle of the resultant force can be calculated using any of three functions in MATLAB:
Function Argument(s) Notes atan(abs(Fx / Fy)) one argument: abs(F (^) x / F (^) y) Returns the included angle
atan2(F (^) y, F (^) x) two arguments: F (^) x and F (^) y Returns the coordinate direction angle Angle value is always between 0 and π radians (0 and 180°) A negative sign on the angle indicates a result in one of the lower quadrants of the Cartesian coordinate system
cart2pol (F (^) x, F (^) y) two arguments: F (^) x and F (^) y Returns the positive angle from the positive x-axis to the vector Angle value always between 0 and 2π radians (0 and 360°) An angle value greater than 180° (π radians) indicates a result in one of the lower quadrants of the Cartesian coordinate system
The atan2() function is used here, and F (^) Ry is negative because it is acting in the –y direction.
» F_Rx = 93.302;
» F_Ry = -390.316;
» theta = 180/pi * atan2( F_Ry, F_Rx )
theta =
-76.
Answer, part b)
The net force (magnitude and direction) on the hook is now known:
FR = 401 N (about 90 lb (^) f) acting at an angle 76.6° below the x-axis.
F R
F (^) Rx
RyF
77°
θ
