Physics, 25.06.2020 05:01 cailynrachael

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 35.6Â° and then moves down the incline with constant speed when the angle is reduced to 30.8Â°. From these data, determine the coefficients of static and kinetic friction for this experiment.

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Answer from: aletadaboss

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

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:

- Weight of the eraser, measured in newtons.

- Friction force, measured in newtons.

- 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.

- 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.

- Normal force, measured in newtons.

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

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:

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

$\mu = \frac{m\cdot g \cdot \sin \theta}{N}$

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.

In order to determine the coefficients of friction between rubber and various surfaces, a student us

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