Dynamics II:
Fnet = ma and Components of Force
Click here for an overview of problem-solving
strategies.
Through a sequence of about a dozen example problems in lecture, you
have seen how to combine forces in situations of increasing complexity
to determine the net force and the resultant motion. The problem
described in the diagram to the
right uses all of the skills you have been practicing: resolving forces
into components, computing forces along two different axes, equilibrium
vs. non-equilibrium, and two-body interaction. The block labeled
m1 is initially stuck due to static friction, but can be
freed to slide uphill by increasing the hanging mass m2. After
the block begins to slide, kinetic friction takes over and the block
accelerates slowly uphill.
Let q=50o, m1=200g,
ms=0.54, mk=0.11.
Let's begin by assuming that we have added mass m2 until just
before the block would begin sliding uphill (the system is still at rest).
To calculate the amount of mass used, we first calculate the normal force
on the block:
N = m1g cosq = (0.2kg)(9.8m/s2)(0.643) = 1.26 N.
The force of friction fs = msN just before
slippage, so fs = (0.54)(1.26 N) = 0.68 N.
Now we need the downhill component of the weight:
wx = m1g sinq =
(0.2kg)(9.8m/s2)(0.766) = 1.50 N.
With the system at rest, the uphill tension force T = m2g so therefore:
m2g = wx + fs = 2.18 N, and m2
= 0.222 kg = 222 g.
Now, if the block is given a slight uphill "nudge", the mass
m2 will be sufficient to accelerate the mass m1 due
to the reduced friction force. The friction becomes:
fk = mk N = (0.11)(1.26 N) = 0.139 N.
But now that the system is accelerating, uphill and downhill forces
are no longer equal. For the block, the net force uphill is given by:
Fnet = m1a = T - (wx + fk) = T - 1.64 N.
For the hanging mass, the net force downward is given by:
Fnet = m2a = m2g - T = (0.222kg)(9.8m/s2) - T = 2.18 N - T.
Rearranging the above equation gives T = 2.18 N - a(0.222kg). This can now be substituted
for T in the equation for the block's motion to eliminate T:
(0.2kg)a = {2.18 N - a(0.222kg)} - 1.64 N, or
(0.422kg)a = 0.54 N, a = 1.28 m/s2.
The tension in the string can then be obtained:
T = 2.18 N - (1.28 m/s2)(0.222kg) = 1.90 N.
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