Using Mathcad For Statics And Dynamics.pdf

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Using Mathcad for Statics and Dynamics
by Ron Larsen and Steve Hunt
Hibbeler
Bedford
and Fowler
2.4 - 2.6
2.3 - 2.5
2. Dot Products
2.9
2.5
3.3
3.2 - 3.3
4.2 - 4.3
2.6, 4.3
5. Moment of a Couple
4.6
4.4
4.10
7.3
5.1 - 5.3
5.1 - 5.2
8. Dry Friction
8.2
9.1
9.2
7.4
9.5
7.4
10.1 - 10.2
8.1 - 8.2
12.3
2.2
12.5
2.3
12.6
2.3
12.7
2.3
12.9
ch. 2
13.4
3.1 - 3.4
13.5
3.4
14.3
8.1 - 8.2
16.3
9.1
77071170.012.png
1
Resolving Forces, Calculating Resultants
Ref: Hibbeler § 2.4-2.6, Bedford & Fowler: Statics § 2.3-2.5
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. 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.
-110°
-55°
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.
Solution
First, consider the 270 N force acting at 55° from horizontal. The x- and y-components of force are
indicated schematically, as
77071170.013.png
F x
55°
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
°
)
=
F
x
, or
F x
=
(
270
N
)
cos(
55
°
)
270
N
and
sin(
55
°
)
=
F
y
, or
F y
=
(
270
N
)
sin(
55
°
)
.
270
N
Mathcad can be used to solve for F x1 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. You can use the deg unit
to explicitly tell Mathcad that the angle is in degrees and must be converted (by Mathcad) to radians.
F x1
:=
(
270 N
) cos 55 deg
(
)
F x1
=
154.866N
F y1
:=
(
270 N
) sin 55 deg
(
)
F y1
=
221.171N
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).
110°
20°
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.
F x
77071170.014.png 77071170.015.png 77071170.001.png 77071170.002.png 77071170.003.png 77071170.004.png
Solution, continued
To determine the combined force on the hook, F R , first add the two y-components calculated above,
to determine the combined y-directed force, F Ry , on the hook:
F Rx
77°
F Ry
:=
F y1
+
F y2
=
The y-component of the resultant force is 390 N (directed down, or in the –y direction). Note that the
direction has not been accounted for in this calculation.
F Ry
390.316N
Then add the two x-components to determine the combined x-directed force, F Rx , on the hook. Note
that the two x-component forces are acting in opposite directions, so the combined x-directed force,
F Rx , is smaller than either of the components, and directed in the +x direction.
77071170.005.png 77071170.006.png
 
F Rx
77°
62 N
155 N
F Rx
:=
F x1
+
( )
F x2
The minus sign was included before F x2 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.
F Rx
=
93.302N
Once the x- and y-components of the resultant force have been determined, the magnitude can be
calculated using
F
=
F
2
+
F
2
R
Rx
Ry
The Mathcad equation is basically the same…
F R
:=
F Rx
2
+
F Ry
2
F R
=
401.312N
77071170.007.png 77071170.008.png 77071170.009.png 77071170.010.png 77071170.011.png

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