Measure of the coefficient of static friction

The angle θm is the largest angle that the inclined plane can make without the body slipping.

There are countless situations in which we observe the movement of objects sliding on inclined planes. An inclined plane is a surface inclined with respect to the horizontal, on which objects can slide by the action of the force of gravity. Examples: slides, slopes, toboggans and truck buckets unloading.

In some situations, related to the study of the inclined plane, we will see the adoption of a frictionless plane; in other situations, we will consider friction. For an inclined plane, in which friction is taken into account, we have the possibility to determine the value of the coefficient of friction between the body and the contact surface.

Let’s look at the figure above: if we place a body on an inclined plane with friction and slowly increase the slope of the plane, we will see that the body will begin to slide from a certain angle. This angle is called the limit angle or maximum angle. Here we will represent the limiting angle by θ m . Therefore, we can conclude that this is the maximum angle that the inclined plane can have without the object slipping (see the figure above).

At this point, the friction force is μ and .N . Putting a x =0 and the value of θ m in the expression for x calculated above, we get:

g.sinθ m -μ and .g.cosθ m =0

Isolating the coefficient of friction μ and , we get:

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Let’s look at an example:

A certain toboggan, without water lubrication, has a friction coefficient μ = 0.3. Determine what acceleration will be experienced by the buoy on the 45° segment.

Resolution:

As we know, the toboggan is an inclined plane with friction, so there is a need to include the friction force in the resolution scheme. The schema is as follows:

According to Newton’s second law for the y components:

total y = ma y

Nm.g.cos θ=ma y =0

N=mgcos θ

Finding the value of normal force

The components on the x axis result in:

total x = ma x

mgsen θ-μ.N=ma x

mgsen θ-μ.mgcosθ=ma x

Isolating the acceleration x , we get:

x =g.sin θ-μ.g.cosθ

For an inclination of 45º, we have:

sin45°≅0.71
cos45°≅0.71

and using the coefficient of friction given in the statement μ = 0.3 and g = 10 m/s 2 , we calculate the numerical value of x :

x =10 x 0.71-0.3 x 10 x 0.71

ax = 4.97 m/s 2

Therefore, the acceleration of the dry toboggan is 4.97 m/s 2 .

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