Question

In order to determine the coefficients of friction between rubber and various surfaces, a student uses a rubber eraser and an incline. In one experiment, the eraser begins to slip down the incline when the angle of inclination is 35.6° and then moves down the incline with constant speed when the angle is reduced to 30.8°. From these data, determine the coefficients of static and kinetic friction for this experiment.

Answer 1

The coefficients of static and kinetic friction for this experiment are 0.716 and 0.596, respectively.

Step-by-step explanation:

The Free Body Diagram associated with the experiment is presented as attachment included below.

Friction is a contact force that occurs as a reaction against any change in state of motion, which is fostered by gravity.

Normal force is another contact force that appears as a reaction to the component of weight perpendicular to the direction of motion. Let consider a framework of reference consisting in two orthogonal axes, one being parallel to the direction of motion (x-axis) and the other one normal to it (y-axis). Equations of motion are described herein:

`\Sigma F_(x) = W \cdot \sin \theta - f = 0`

`\Sigma F_(y) = N - W \cdot \cos \theta = 0`

Where:

`W` - Weight of the eraser, measured in newtons.

`f` - Friction force, measured in newtons.

`N` - Normal force, measured in newtons.

`\theta` - Angle of the incline, measured in degrees.

The maximum allowable static friction force is:

`f = \mu_(s) \cdot N`

Where:

`\mu_(s)` - Coefficient of static friction, dimensionless.

`N` - Normal force, measured in newtons.

Likewise, the kinetic friction force is described by the following model:

`f = \mu_(k) \cdot N`

Where:

`\mu_(k)` - Coefficient of static friction, dimensionless.

`N` - Normal force, measured in newtons.

And weight is equal to the product of the mass of eraser and gravitational constant (
`g = 9.807\,(m)/(s^(2))`)

In this exercise, coefficients of static and kinetic friction must be determined. First equation of equilibrium has to be expanded and coefficient of friction cleared:

`m\cdot g \cdot \sin \theta - \mu\cdot N = 0`

`\mu = (m\cdot g \cdot \sin \theta)/(N)`

But
`N = m\cdot g \cos \theta`, so that:

`\mu = \tan \theta`

Now, coefficients of static and kinetic friction are, respectively:

`\mu_(s) = \tan 35.6^(\circ)`

`\mu_(s) \approx 0.716`

`\mu_(k) \approx \tan 30.8^(\circ)`

`\mu_(k) \approx 0.596`

The coefficients of static and kinetic friction for this experiment are 0.716 and 0.596, respectively.