If you don’t know where we got those vector components, refer back to Chapter 9 .
Step 3 : Write equations for the net force in each direction.
Note that the block is in equilibrium in the direction perpendicular to the plane, so the equation for F (^) net,
perpendicular^ (but not the equation for F^ net, down the plane ) can be set equal to 0.
Step 4 : Solve.
We can rewrite F (^) f , because
Ff = μFN = μ mg (cos θ )
Plugging this expression for F (^) f into the “F (^) net, down the plane ” equation, we have
It always pays to check the reasonability of the answer. First, the answer doesn’t include any variables
that weren’t given. Next, the units work out: g has acceleration units; neither the sine or cosine of an angle
nor the coefficient of friction has any units.
Second, compare the answer to something familiar. Note that if the plane were vertical, θ = 90°, so
the acceleration would be g —yes, the block would then be in free fall! Also, note that friction tends to
make the acceleration smaller, as you might expect.
For this particular incline, what coefficient of friction would cause the block to slide with constant
speed?
Constant speed means a = 0. The solution for FN in the perpendicular direction is the same as before: F (^) N
= mg (cosθ ). But in the down-the-plane direction, no acceleration means that F (^) f = mg (sinθ ). Because μ
= F (^) f /F (^) N ,
Canceling terms and remembering that sin/cos = tan, you find that μ = tanθ when acceleration is zero.
You might note that neither this answer nor the previous one includes the mass of the block, so on the
same plane, both heavy and light masses move the same way!