Conquering the Physics GRE

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1.1.1 Blocks on ramps
Here’s a basic scenario: a block of mass m is on a ramp inclined at an angle ✓, and suppose we
want to know the coecient of static friction µ required to keep it in place. The usual solution
method is to resolve any forces F into components along the ramp (Fk) and perpendicular
to the ramp (F?). Rather than fuss with trigonometry or similar triangles, we can just do
this by considering limiting cases, a theme that we’ll return to throughout this book. In this
case, we have to resolve the gravitational force Fg. If the ramp is flat (✓ = 0), then there is
no force in the direction of the ramp, so gravity acts entirely perpendicularly, and Fg,k = 0.
On the other hand, if the ramp is sheer vertical (✓ = ⇡/2), then gravity acts entirely parallel
to the ramp (Fg,? = 0), and the block falls straight down. Knowing that there must be sines
and cosines involved, and the magnitude of Fg is mg, this uniquely fixes
Fg,k = mg sin ✓, Fg,? = mg cos ✓.
For the block not to accelerate perpendicular to the ramp, we need the perpendicular forces
to balance, which fixes the normal force to be N = mg cos ✓. Then the frictional force
is Ff = µmg cos ✓, which must balance the component of gravity parallel to the ramp,
Fg,k = mg sin ✓. Setting these equal gives
µmg cos ✓ = mg sin ✓ =) µ = tan ✓.
Again, we can check this by limiting cases. If ✓ = 0, then we don’t need any friction to hold
the block in place, and µ = 0. If ✓ = ⇡/2, we need an infinite amount of friction to glue the
block to the ramp and keep if from falling vertically, so µ = 1. Both of these check out.
Standard variants on this problem include applied forces and blocks attached to pulleys
which hang over the side of the ramp, but surprisingly, neither the basic problem nor its
variants have shown up on recent exams. Perhaps it is considered too standard by the GRE,
such that most students will have memorized the problem and its variants so completely
that it’s not worth testing. In any case, consider it a simple review of how to resolve forces
into components by using a limiting-cases argument, as this can potentially save you a lot
of time on the exam.
1.1.2 Falling and hanging blocks
The next step up in complexity is to have two or more blocks interacting – for example,
two blocks tied together with a rope, falling under the influence of gravity, or the same
blocks hanging from a ceiling. These kinds of questions test your ability to identify precisely
which forces are acting on which blocks. A foolproof, though time-consuming, method is
to use free-body diagrams, where you draw each individual block and only the forces acting
on it. This avoids the pitfall of double-counting, or applying the same force twice to two
di↵erent objects, and ensures that you take into careful account the action/reaction balance
of Newton’s third law.
Sometimes, though, simple physical reasoning will suce, especially in situations where
the blocks aren’t really distinct objects. For example, consider placing one block on top of

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