1.7 Friction

Friction is a very complicated process which occurs on a miroscopic scale, so in order to model in on a macroscopic scale we must use simplified empirical laws. In general, friction is a force which opposes change in motion. Hence if a force is applied parallel to a surface, then the frictional force will be antiparallel to this, perpendicular to the normal force.

1.7.1 Static Friction

Static friction appears when two objects are motionless with respect to one another. If a force is applied between the two objects and they don’t move, there must be a frictional force opposing the motion.

\boldsymbol{f}_{s}=-\boldsymbol{F}_{app}.

(1.54)

As the applied force gets larger, the static friction must get larger to preserve equilibrium, until a maximum limit is reached and the object starts moving.

Definition 1.11.

The maximum magnitude of static friction is given by

f_{s,max}=\mu_{s}\lvert\boldsymbol{N}\rvert.

(1.55)

Hence,

0\leq\lvert\boldsymbol{f}_{s}\rvert\leq f_{s,max}.

(1.56)

1.7.2 Kinetic Friction

Kinetic friction opposes the motion of two surfaces sliding against each other.

Definition 1.12.

Kinetic friction is given by

\boldsymbol{f}_{k}=-\mu_{k}\lvert\boldsymbol{N}\rvert\hat{\boldsymbol{v}}.

(1.57)

Example 1.6.

Find the stopping distance of a block sliding down a slope.

Example 1.7.

Three blocks — A, B, and C — are connected over a slope by massless inextensible ropes going through frictionless pulleys. Blocks A and B both have a weight of $25\text{\,}\mathrm{N}$ a coefficient of kinetic friction of $\mu_{k}=0.35$ . The angle of the slope is $30 °$ and block C is falling with constant velocity. What is the weight of block C?

We will start to solve this problem by drawing free-body diagrams for all the blocks. In each diagram, we will orient the $x$ -axis along the direction of motion.

Since the ropes are inextensible, the tension throughout them must be uniform. Therefore all the blocks must be moving together at the same speed. This speed is constant, so Newton’s second law tells us that the resultant forces along both axes, $\Sigma F_{x}$ and $\Sigma F_{y}$ , must be zero for all three blocks. Looking at the diagram for block C, this tells us that $W_{C}$ , the weight of block C, is equal in magnitude to $T_{BC}$ , the tension in the rope connecting B and C.

From the diagram for block B we have

	$\displaystyle\Sigma F_{y}$	$\displaystyle=N_{B}-W_{B}\cos\theta=0$		(1.58)
	$\displaystyle\implies N_{B}$	$\displaystyle=W_{B}\cos\theta$		(1.59)

and

$\displaystyle\Sigma F_{x}$	$\displaystyle=T_{BC}-(f_{B}+T_{AB})-W_{B}\sin\theta=0$	(1.60)
$\displaystyle\implies T_{BC}$	$\displaystyle=f_{B}+T_{AB}+W_{B}\sin\theta$	(1.61)
	$\displaystyle=\mu_{k}W_{B}\cos\theta+T_{AB}+W_{B}\sin\theta,$	(1.62)

from block A we get

	$\displaystyle\Sigma F_{y}$	$\displaystyle=N_{A}-W_{A}=0$		(1.63)
	$\displaystyle\implies N_{A}$	$\displaystyle=W_{A}$		(1.64)

and

$\displaystyle\Sigma F_{x}$	$\displaystyle=T_{AB}-f_{A}=0$	(1.65)
$\displaystyle\implies T_{AB}$	$\displaystyle=f_{A}=\mu_{k}N_{A}$	(1.66)
	$\displaystyle=\mu_{k}W_{A}.$	(1.67)

Combining all these results, we get

$\displaystyle W_{C}$	$\displaystyle=T_{BC}$	(1.68)
	$\displaystyle=\mu_{k}W_{B}\cos\theta+W_{B}\sin\theta+\mu_{k}W_{A}$	(1.69)
	$\displaystyle=$7.58\text{\,}\mathrm{N}$+$12.5\text{\,}\mathrm{N}$+$8.75\text{% \,}\mathrm{N}$$	(1.70)
	$\displaystyle=$28.83\text{\,}\mathrm{N}$.$	(1.71)

Now, imagine if the rope between blocks A and B is cut. What will happen to block C?

With both blocks A and B providing block C with enough friction to balance gravity, it seems that if we remove block A then there will no longer be enough resistance and block C will have to accelerate downwards. We can try calculating the acceleration and find out if our hypothesis is correct. The forces on blocks B and C are all the same except that we no longer have $T_{AB}$ . Thus for block C we get

\displaystyle\Sigma F_{x}

\displaystyle=W_{C}-T_{BC}=\frac{W_{C}}{g}a,

(1.72)

and for block B we get

	$\displaystyle\Sigma F_{x}$	$\displaystyle=T_{BC}-f_{B}-W_{B}\sin\theta=\frac{W_{B}}{g}a$		(1.74)
	$\displaystyle\implies T_{BC}$	$\displaystyle=\mu_{k}W_{B}\cos\theta+W_{B}\sin\theta+\frac{W_{B}}{g}a.$		(1.75)

Combining these, we get

$\displaystyle W_{C}-\frac{W_{C}}{g}a$	$\displaystyle=\mu_{k}W_{B}\cos\theta+W_{B}\sin\theta+\frac{W_{B}}{g}a$	(1.76)
$\displaystyle\frac{(W_{C}+W_{B})}{g}a$	$\displaystyle=W_{C}-W_{B}\sin\theta-\mu_{k}W_{B}\cos\theta$	(1.77)
$\displaystyle a$	$\displaystyle=\frac{W_{C}-W_{B}(\sin\theta+\mu_{k}\cos\theta)}{W_{C}+W_{B}}g$	(1.78)
	$\displaystyle=$1.59\text{\,}\mathrm{m}\text{\,}{\mathrm{s}}^{-2}$.$	(1.79)

This acceleration is nonzero and positive, which indicates that block C does indeed accelerate downwards as we thought.