6.2 Friction  (Page 3/12)

[link] and [link] include the dependence of friction on materials and the normal force. The direction of friction is always opposite that of motion, parallel to the surface between objects, and perpendicular to the normal force. For example, if the crate you try to push (with a force parallel to the floor) has a mass of 100 kg, then the normal force is equal to its weight,

perpendicular to the floor. If the coefficient of static friction is 0.45, you would have to exert a force parallel to the floor greater than

to move the crate. Once there is motion, friction is less and the coefficient of kinetic friction might be 0.30, so that a force of only

keeps it moving at a constant speed. If the floor is lubricated, both coefficients are considerably less than they would be without lubrication. Coefficient of friction is a unitless quantity with a magnitude usually between 0 and 1.0. The actual value depends on the two surfaces that are in contact.

Many people have experienced the slipperiness of walking on ice. However, many parts of the body, especially the joints, have much smaller coefficients of friction—often three or four times less than ice. A joint is formed by the ends of two bones, which are connected by thick tissues. The knee joint is formed by the lower leg bone (the tibia) and the thighbone (the femur). The hip is a ball (at the end of the femur) and socket (part of the pelvis) joint. The ends of the bones in the joint are covered by cartilage, which provides a smooth, almost-glassy surface. The joints also produce a fluid (synovial fluid) that reduces friction and wear. A damaged or arthritic joint can be replaced by an artificial joint ( [link] ). These replacements can be made of metals (stainless steel or titanium) or plastic (polyethylene), also with very small coefficients of friction.

Natural lubricants include saliva produced in our mouths to aid in the swallowing process, and the slippery mucus found between organs in the body, allowing them to move freely past each other during heartbeats, during breathing, and when a person moves. Hospitals and doctor’s clinics commonly use artificial lubricants, such as gels, to reduce friction.

The equations given for static and kinetic friction are empirical laws that describe the behavior of the forces of friction. While these formulas are very useful for practical purposes, they do not have the status of mathematical statements that represent general principles (e.g., Newton’s second law). In fact, there are cases for which these equations are not even good approximations. For instance, neither formula is accurate for lubricated surfaces or for two surfaces siding across each other at high speeds. Unless specified, we will not be concerned with these exceptions.

Static and kinetic friction

A 20.0-kg crate is at rest on a floor as shown in [link] . The coefficient of static friction between the crate and floor is 0.700 and the coefficient of kinetic friction is 0.600. A horizontal force $\overrightarrow{P}$ is applied to the crate. Find the force of friction if (a) $\overrightarrow{P}=20.0\text{N,}$ (b) $\overrightarrow{P}=30.0\text{N,}$ (c) $\overrightarrow{P}=120.0\text{N,}$ and (d) $\overrightarrow{P}=180.0\text{N}\text{.}$

Strategy

The free-body diagram of the crate is shown in [link] (b). We apply Newton’s second law in the horizontal and vertical directions, including the friction force in opposition to the direction of motion of the box.

Solution

Newton’s second law gives

$\begin{array}{cccc}{\displaystyle \sum F_x=m{a}_x}\hfill & & & {\displaystyle \sum F_y=m{a}_y}\hfill \\ P-f=m{a}_x\hfill & & & N-w=0.\hfill \end{array}$

Here we are using the symbol f to represent the frictional force since we have not yet determined whether the crate is subject to station friction or kinetic friction. We do this whenever we are unsure what type of friction is acting. Now the weight of the crate is

$w=(20.0\text{kg})(9.80{\text{m/s}}^2)=196\text{N,}$

which is also equal to N . The maximum force of static friction is therefore $\left(0.700\right)\left(196\text{N}\right)=137\text{N}\text{.}$ As long as $\overrightarrow{P}$ is less than 137 N, the force of static friction keeps the crate stationary and $f_{\text{s}}=\overrightarrow{P}.$ Thus, (a) $f_s=20.0\text{N,}$ (b) $f_s=30.0\text{N,}$ and (c) $f_s=120.0\text{N}\text{.}$

(d) If $\overrightarrow{P}=180.0\text{N,}$ the applied force is greater than the maximum force of static friction (137 N), so the crate can no longer remain at rest. Once the crate is in motion, kinetic friction acts. Then

$f_{\text{k}}={\mu }_{\text{k}}N=(0.600)(196\text{N})=118\text{N,}$

and the acceleration is

$a_x=\frac{\overrightarrow{P}-f_{\text{k}}}{m}=\frac{180.0\text{N}-118\text{N}}{20.0\text{kg}}=3.10{\text{m/s}}^2\text{.}$

Significance

This example illustrates how we consider friction in a dynamics problem. Notice that static friction has a value that matches the applied force, until we reach the maximum value of static friction. Also, no motion can occur until the applied force equals the force of static friction, but the force of kinetic friction will then become smaller.

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