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where in two dimensions:
Σ F = ΣF x i + ΣF y j
ΣF = ( ΣF x ) 2 + ( ΣF y ) 2
Going from two-dimensional particle problems to two-dimensional rigid-body prob‐
lems involves only the addition of one more equation. This equation is, of course, the
moment equation relating the sum of all moments acting on the body to the body's
moment of inertia and its angular acceleration. In plane motion, the axis of rotation of
the rigid body is always perpendicular to the coordinate plane. And since there is only
one axis of rotation, there is only one inertia term and one angular acceleration term to
consider. Thus, you can write:
M cg = I α
where M cg is the total moment and is calculated with the formulas discussed in the
section “Force and Torque” on page 80 in Chapter 3 , and I is calculated about the axis
of rotation using the techniques discussed in the section “Mass, Center of Mass, and
Moment of Inertia” on page 9 in Chapter 1 .
In their component forms, the set of equations of motion for two-dimensional kinetics
problems are:
ΣF x = m a x
ΣF y = m a y
ΣM cg = I α
Since these equations indicate linear motion on the xy-plane, the angular acceleration
will be about the z-axis perpendicular to the xy-plane. Likewise, the moment of inertia,
I , will be taken about the z-axis.
Recall from Chapter 3 that we calculate moment by taking the cross product of the
position vector for the force under consideration and the force vector. This means that,
unlike with particle kinetics, you now have to keep track of exactly where on the body
each force is applied. This is best illustrated with an example.
Consider the box of uniform density shown in Figure 4-6 . Uniform density means that
its center of gravity is at the box's geometric center. Find the value of the minimum force,
F p , applied at the upper edge of the box, required to start tipping the box over.
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