Chapter 2 Review of Forces and Moments - Brown University

Example 3. The structure shown is subjected to
a force T acting at E along the line EF.
Calculate the moment of T about points A and D.
[( /2) ( /2)tanθ
] [ ]
{ θ}
k
MB = r× F= −L i− L j × Pi −Wj
= ( L/2) W + Ptan
This example requires a lot more work. First we
need to write down the force as a vector. We
know the magnitude of the force is T, so we only
need to work out its direction. Since the force
acts along EF, the direction must be a unit vector
pointing along EF. It’s not hard to see that the
vector EF is
EF = ai − 3aj + 2ak
We can divide by the length of EF ( a 14 ) to
find a unit vector pointing in the correct
direction
e = ( i− EF
3 j+
2 k )/ 14
The force vector is
F= T ( i− 3j+
2 k )/ 14
Next, we need to write down the necessary position vectors
Force: r = 2ai+
3aj
Point A: rA
=−2ai
Point D: rD
= 4aj
Finally, we can work through the necessary cross products
MA
= ( r− rA)
× F
MD
= ( r− rD)
× F
= (4ai+ 3 aj) × T( i− 3j+
2 k) / 14 = (2 ai− aj) × T( i− 3j+
2 k) / 14
i j k
i j k
=
( Ta )
/ 14 4 3 0
1 −3 2
( i j k)
= Ta 6 −8 −15 / 14
Clearly, vector notation is very helpful when solving 3D problems!
( Ta )
= / 14 2 −1 0
1 −3 2
( i j k)
= Ta −2 −4 −5 / 14
Example 4. Finally, we work through a simple problem
involving distributed loading. Calculate expressions for the
moments exerted by the pressure acting on the beam about
points A and B.
j
A
x
dF = p dx
dx
p (per unit
length)
B
i
L