# A block weighing 14 kg is on a plane with an incline of pi/6 and friction coefficient of 1/5. How much force, if any, is necessary to keep the block from sliding down?

Jun 10, 2018

In order for the object to not slide down, a force of ${F}_{0} = 45.75$ Newtons must be applied upon it.

#### Explanation:

Let us visualise the situation:

To analyze the motion of the block, we must know about the forces which act on it.

First of all, the gravity force $G = m g$ which we can break down into ${G}_{x}$ and ${G}_{y}$, where ${G}_{x}$ is parallel to the plane and ${G}_{y}$ is perpendicular to it. Second of all, the normal force $N$ and finally the force of friction ${F}_{f}$.

We ask: what force ${F}_{0}$ must we apply on the block such that it remains motionless? We can assume this force to be on the same direction as ${G}_{x}$ and ${F}_{f}$, parallel to the plane, and then apply Newton's Second Law:

$F = m a$

Where $F$ is the sum of all the forces applied on a certain direction.

If the plane has an inclined $\theta$ and since the triangle formed by the vectors $\vec{G}$, ${\vec{G}}_{y}$ and ${\vec{G}}_{x}$ is similar to the original triangle, we get the following relations:

$\left\{\begin{matrix}{G}_{x} = m g \sin \theta \\ {G}_{y} = m g \cos \theta\end{matrix}\right.$

Now, on the parallel direction to the plane, we have:

${F}_{\text{parallel}} = - {F}_{0} + {G}_{x} - {F}_{f} = m a$

The sign of the force is determined by whether it acts for or against motion. In our case, ${F}_{0}$ must be opposing motion.

In order for the object to not move, the accerelation $a$ must be zero, hence

${F}_{0} = {G}_{x} - {F}_{f}$

Finding ${F}_{f}$; The friction force is calculated by the formula ${F}_{f} = \mu N$, where $\mu$ is the coefficient of friction. Since the object is not moving perpendiculary, it means that

${F}_{\text{perpendicular}} = 0 = N - {G}_{y} \implies N = {G}_{y} = m g \cos \theta$

If we take $g$ to be $10$:

$N = 140 \cos \left(\frac{\pi}{6}\right) = 70 \sqrt{3} \approx 121.24 \text{ N}$

$\implies {F}_{f} = \frac{1}{5} \cdot 121.24 \approx 24.25 \text{ N}$

Finding ${G}_{x}$;

${G}_{x} = m g \sin \theta = 140 \sin \left(\frac{\pi}{6}\right) = 70 \text{ N}$

Therefore, for the block to not move, we must apply a force of:

${F}_{0} = 70 - 24.25 = 45.75 \text{ N}$