The Wrench: Let’s suppose that we have already reduced a complicated force system to single (resultant ) force, , and a single couple with the moment, , as shown in the figure:

R C

N)22(10 kjiR

900( 2 3 )N m C i j k

x

y

z

And, we want to reduce this force system to the simplest one possible, a wrench.

Step 1: The unit vector parallel to :R )22(3

1

441

22kji

kjiu

Step 2: The component of parallel to :RC

900(1 4 6) 2 2100 2 2 300

3 3pa

i j kC C u u i j k u

Step 3: The component of perpendicular to :C R

900 100 1800 200 2700 200 500 2 4 5pe pa C C C i j k i j k

Continued on the next page

We note that

500 2 4 5 10( 2 2 ) 5000(2 8 10) 0pe C R i j k i j k

Hence, and are perpendicular.peC R

x

paC

Continued on the next page

N)22(10 kjiR

900( 2 3 )N m C i j k

y

z

peC

pe paC C and are perpendicular.

The moment of the couple has been resolved into a perpendicular component, , and a parallel component, . Theses vector could shown anywhere in space because a couple produces the same moment about all points in space. Here they are shown at the origin to make it easier to see that they are perpendicular and parallel to .

peC paC

R

N)22(10 kjiR

900( 2 3 )N m C i j k

x

y

z

R

R

couple

r

P (x,y,z)

peCpaC

Step 4: in a new position, denoted by P, such that the new resultant moment of all the couples, , is given by:

R

resC

res pa pe C C r R C C r R

Because is perpendicular to both and , there is no way that can be ; we can only eliminate . We put

r R r R resC0

peC

500 2 4 5 10( 2 2 )pe x y z C r R i j k i j k i j k 0

0 1 1 50

2 0 1 200

2 1 0 250

x

y

z

500 2 4 5 10

1 2 2

1000 20( ) 0, 2000 10( 2 ) 0, 2500 10(2 ) 0

x y z

y z z x x y

i j k

i j k 0

Continued on the next page

B

A

0 1 1

2 0 1 0

2 1 0

We note that

which indicates that there are not three independent equations. In fact, we see that the sum of the first two equations in B is equal to the third. As in the two-dimensional case, we have the equation of a straight line.

To see which line, we arbitrarily put x = 0 and then find y = - 250 and z = 200. This is one point on the line. Then we arbitrarily put y = 0 and find x = 125 and z = - 50 to find another.

The vector to the first point from the second is given by

0 125 250 0 200 ( 50) 125( 2 2 ) 375 v i j k i j k u

To show that this vector is parallel to , we note that .

Now we have reduced the force system to a single force acting along the line defined by B (point P is on this line) and a couple with a moment parallel to the force. Such a force system is called a wrench. Because the moment is in the direction opposite to that of the force, the wrench is said to be negative.

R v R 0

An Alternative Approach:We know that the final moment in the simplest force system is parallel to the resultant force, so we can go back to A and set the resultant moment equal to a vector parallel to the force:

900( 2 3 ) 10 ( 2 2 ) 3

1 2 2

2 2900 20( ) , 1800 10( 2 ) , 2700 10(2 )

3 3 3

res

Mx y z M

M M My z z x x y

i j k

C C r R i j k u i j k

where M is the magnitude of the moment parallel to the force (unknown at this point) and u is the unit vector parallel to R. Recalling that we can arbitrarily assign a value to one of the coordinates, we can put x = 0, y = 0, z = 0 in the three equations above and obtain the following three sets of equations with three equations in each set:

0 60 1 2700

60 0 2 5400

60 30 2 8100

x

y

M

We can solve these three sets of equations using, for example, MATLAB or MATHEMATICA:

0 (arbitrarily specified), 250 , 200 , 300

125 , 0 (arbitrarily specified), 50 , 300

100 , 50 , 0 (arbitrarily specified), 300

x m y m z m M N m

x m y m z m M N m

x m y m z m M N m

We note that . This solution agrees with the one obtained earlier. The solution for M is the same for all three arbitrary choices of the coordinates (indeed for all choices).

60 60 1 2700

0 30 2 5400

30 0 2 8100

y

z

M

0 60 1 2700

60 30 2 5400

60 0 2 8100

x

z

M

300 300peM C u