University Physics I - Classical Mechanics, 2019

168 CHAPTER 8. MOTION IN TWO DIMENSIONS
Newton’s second law, as given by equations (8.4) applied to this system, then reads:
for the motion along the plane, and
F g x + F k x = ma x = F g sin θ − F k (8.15)
F g y + F n y = ma y = −F g cos θ + F n (8.16)
for the direction perpendicular to the plane. Of course, since there is no motion in this direction,
a y is zero. This gives us immediately the value of the normal force:
F n = F g cos θ = mg cos θ (8.17)
since F g = mg. We can also use the result (8.17), together with Eq. (6.30), to get the magnitude
of the friction force, assuming we know the coefficient of kinetic (or sliding) friction, μ k :
Substituting this and F g = mg in Eq. (8.15), we get
We can eliminate the mass to obtain finally
F k = μ k F n = μ k mg cos θ (8.18)
ma x = mg sin θ − μ k mg cos θ (8.19)
a x = g (sin θ − μ k cos θ) (8.20)
which is the desired result. In the absence of friction (μ k =0)thisgivesa = g sin θ, aswehadin
Chapter 2. Note that, if you reduce the tilt of the surface (that is, make θ smaller), the cos θ term
in Eq. (8.20) growsandthesinθ term gets smaller, so we must make sure that we do not use this
equation when θ is too small or we would get the absurd result that a x < 0, that is, that the force
of kinetic friction has overcome gravity and is accelerating the object upwards!
Of course, we know from experience that what happens when θ is very small is that the block does
not slide: it is held in place by the force of static friction. The diagram for such a situation looks
thesameasFig.8.4, except that ⃗a = 0, the force of friction is F s instead of F k , and of course its
magnitude must match that of the x component of gravity. Equation (8.15) then becomes
ma x =0=F g sin θ − F s (8.21)
Recall from Chapter 6 that the force of static friction does not have a fixed value: rather, it will
match the applied force up to a maximum value given by Eq. (6.29):
F s max = μ s F n = μ s mg cos θ (8.22)
where I have used Eq. (8.17), since clearly the equation (8.16) still applies along the vertical
direction. So, on the one hand we have the requirement that F s = mg sin θ to keep the block from