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Figure 4-6. Box-free body diagram
In Figure 4-6 , F p is the applied force, R 1 and R 2 are the reaction forces at supports one
and two, F f1 and F f2 are the forces due to friction at points one and two, and mg is the
weight of the box.
This is an example of the type of problem where you know something about the motion
of the object and have to find the value of one or more forces acting on it. To find the
value of the force that will be just enough to start tipping the box, you need to look at
the instant when the reaction force at support two is 0. This implies that all of the weight
of the box is now supported at point one and the box is starting to rotate over. At this
instant, just before it starts to rotate, the angular acceleration of the box is 0. Note that
the box's linear acceleration isn't necessarily 0—that is, you can push on the box and it
may slide without actually tipping over.
The equations of motion for this problem are:
ΣF x = F p − F f1 − F f2 = m a x
ΣF y = R 1 + R 2 − m g = m a y = 0
ΣM cg = F p (h/2) + R 2 (w/2) − R 1 (w/2) + F f2 (h/2) + F f1 (h/2) = I
α = 0
Rewriting the second preceding equation when R 2 is 0 shows that R 1 is equal to the
weight of the box. Further, when R 2 is 0, the R 2 ( w / 2 ) term drops out of the moment
equation, which can be rewritten by solving for F p in terms of R 1 . Note that when R 2
goes to 0, so does F f2 . After some algebra, the equation looks as follows:
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