Examples

1. Inclined plane of length L with friction

   a) At what angle does the block barely overcome the force of static friction? b) How fast is the block going when it reaches the bottom.

   

   Static friction exerts as much force as necessary to keep the block at rest up to the maximum it can exert, $f_s({\rm max}) = \mu_s N$. If we decompose the force into components parallel to and perpendicular to the plane, and note that the components perpendicular to the plane must cancel, we get:
   

   $$\sum F_{\vert\vert} = mg\sin\theta - \mu_s N = 0$$ (2.35)

   
   
   (for $\theta \le \theta_{\rm max}$) and
   

   $$\sum F_\perp = N - mg\cos\theta = 0$$ (2.36)

   
   

   From the latter,
   

   $$N = mg\cos\theta$$ (2.37)

   
   
   so that the first equation becomes:
   

   $$\sum F_{\vert\vert} = mg\sin\theta_{\rm max} - \mu_s mg\cos\theta_{\rm max} = 0$$ (2.38)

   
   
   at the maximum possible angle. Solving for the angle, we get:
   

   $$\theta_{\rm max} = \tan^{-1}\mu_s$$ (2.39)

   
   

   Once it is moving, $F_{\vert\vert} > 0$ (down the plane) so we get:
   

   $$\sum F_{\vert\vert} = mg\sin\theta - \mu_s mg\cos\theta = ma_{\vert\vert}$$ (2.40)

   
   
   (for $theta > theta_{\rm max}$) or
   

   $$a_{\vert\vert} = g\sin\theta - \mu_s g \cos\theta$$ (2.41)

   
   
   from which we can find whatever we like from the methods of chapters 1-3.

2. A block of mass $m_1 = 4$ kg sits on a table with $\mu_s = 0.5$, $\mu_k = 0.4$. It is attached by a massless unstretchable string running over a pulley to a block of mass $m$ hanging off of a table. a) What is the largest mass that $m$ can be before the system starts to move? b) Suppose that $m = 4$ kg. Describe the subsequent motion (find $a$).
   

   The system is effectively one dimensional, with the string and pulley serving to ``bend'' the force around corners without loss. Just a smattering of what you need to solve the problem (I'm leaving some of it undone):
   

   $$\sum F_{y1} = N - m_1g = 0$$ (2.42)

   $$\sum F_{x1} = T - \mu_s N = 0$$ (2.43)

   $${\rm or } \sum F_{x1} = T - \mu_k N = m_1 a$$ (2.44)

   $$\sum F_{y2} = T - m_2g = m_2a$$ (2.45)

   
   

   One finds $N$, puts it in one of the next two equations (depending on what you are finding), and solves it and the last equation simultaneously (eliminating $T$) to find either $m_2$ or $a$ and $T$. Once you have $a$, you own the solution...

3. Find the minimum braking distance of a car travelling at 30 m/sec running on tires with $\mu_s = 0.5$ and $\mu_k = 0.3$ a) equipped with ABS such that the tires do not skid, but rather roll (so that they exert the maximum static friction only); b) the same car, but without ABS and with the wheels locked in a skid (kinetic friction only)
   

   It is simplest to use $2a_x\Delta x = v_f^2 - v_0^2$, with $v_f = 0$ and $\Delta x = D$, the stopping distance.

   The only forces exerted are gravity, the normal force, and the force of static (ABS) or kinetic (no-ABS) friction. To find $a_x$, we use Newton's Law:
   

   $$\sum_x F_x = -\mu_{(s,k)} N = m a_x$$ (2.46)

   $$\sum_y F_y = N - mg = m a_y = 0$$ (2.47)

   
   
   (where $\mu_s$ is for static friction and $\mu_k$ is for kinetic friction), or:
   

   $$m a_x = - \mu_{(s,k)}mg$$ (2.48)

   
   
   so
   

   $$a_x = -\mu_{(s,k)} g$$ (2.49)

   
   
   Thus
   

   $$-2 \mu_{(s,k)} g D = -v_0^2$$ (2.50)

   
   
   or
   

   $$D = \frac{v_0^2}{2 \mu_{(s,k)} g}$$ (2.51)

   
   
   Numerically, $D_{\rm ABS} = 90$ m, and $D_{\rm normal} = 150$ m. Another nice feature of ABS is that when fully applied they will always stop your car in ``exactly'' the minimum distance. I'm spending $800 for ABS for my next car...

4. A car is rounding a curve of $R = 50$m banked at an angle of 20$\circ$. The car is travelling at $v = 30$ m/sec. Find: a) the normal force exerted by the road on the car; b) the force of static friction exerted by the road on the tires.
   

   Note that we don't know $f_s$; it is presumably less than $\mu_s N$, but we have to solve for $N$ to be sure. We pick a direction for it down the plane, although if $v$ is less than a certain value it will point up the plane (and our answer should reflect that). We use coordinates lined up with the eventual direction of $\vec{F}_{tot}$: +x parallel to the ground (and $R$). We write Newton's second law:
   

   $$\sum_x F_x = N\sin\theta + f_s\cos\theta = ma_x = \frac{mv^2}{R}$$ (2.52)

   $$\sum_y F_y = N\cos\theta - mg - f_s\sin\theta = ma_y = 0$$ (2.53)

   
   
   and solve the y equation for $N$:
   

   $$N = \frac{mg+f_s\sin\theta}{\cos\theta}$$ (2.54)

   
   
   substitute into the x equation:
   

   $$(mg+f_s\sin\theta)\tan\theta + f_s\cos\theta = \frac{mv^2}{R}$$ (2.55)

   
   
   and finally solve for $f_s$:
   

   $$f_s = \frac{\frac{mv^2}{R} - mg\tan\theta}{\sin\theta\tan\theta + \cos\theta}$$ (2.56)

   
   

   From this we see that if
   

   $$\frac{mv^2}{R} > mg\tan\theta$$ (2.57)

   
   
   or
   

   $$\frac{v^2}{Rg} > \tan\theta$$ (2.58)

   
   
   then $f_s$ is positive (down the incline), otherwise it is negative (up the incline). When $\frac{v^2}{Rg} = \tan\theta$, $f_s = 0$ and the car would round the curve even on ice.

   Other questions I might have asked: Within what range of speeds will the car be able to round the curve, given $\mu_s = 0.3$? If $\mu_s = 0.5$, what is the smallest (largest) angle for which the car can round the curve?