Question

A box of mass 50 kg is placed on an inclined plane. When the angle of the plane is increased to 30º, the box begins to slide downwards. Calculate the coefficient of static friction between the plane and the box. Draw the free body diagram.

Answer 1

Answer on Question 65995, Physics, Mechanics, Relativity

Question:

A box of mass $50~kg$ is placed on an inclined plane. When the angle of the plane is increased to $30{}^{\circ}$ , the box begins to slide downwards. Calculate the coefficient of static friction between the plane and the box. Draw the free body diagram.

Solution:

There are three forces that act on the sliding box: the force of gravity $mg$ directed downward and can be resolved into two perpendicular components ( $F_{\parallel} = mgsin\theta$ and $F_{\perp} = mgcos\theta$ ), the force of reaction directed perpendicular to the surface and friction force $F_{fr}$ directed opposite to the motion of the box. Let's draw a free-body diagram and write all forces that act on the box:

$$m \vec {g} + \vec {N} + \overrightarrow {F _ {f r}} = m \vec {a} = 0.$$

Then projected the forces on axis $x$ and $y$ we get:

$$m g \sin \theta - F _ {f r} = 0, (1)$$ $$N - m g c o s \theta = 0. (2)$$

Let's find the static friction force that acts on the box:

$$F _ {f r} = \mu_ {s} N = \mu_ {s} m g c o s \theta .$$

Substituting the friction force into the first equation we get:

$$mgsin\theta - \mu_s mgcos\theta = 0.$$

From this formula we can find the coefficient of static friction between the plane and the box:

$$mgsin\theta = \mu_s mgcos\theta,$$ $$\mu_s = \frac{sin\theta}{cos\theta} = tan\theta = tan30{}^\circ = 0.58.$$

Answer:

$$\mu_s = 0.58.$$

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