Question

A 5kg box rest on an inclined plane with angle of inclination 30degree. calculate the force required to keep, if there is no sliding down the plane given that the coefficient of friction is 0.30

Answer 1

Note: the direction of 'extra' force ${\vec F}$ is not specified in the problem so we shall assume it is directed along the slope

Let $Ox$ axis be along the inclined plane (positive direction in direction of possible sliding) and $Oy$ axis is perpendicular to the inclined plane. There are 4 power actiong on the box: gravitational force $m\vec g$ , force of friction ${{\vec F}_{{\rm{frict}}}}$ , normal force ${\vec N}$ and the 'extra' force ${\vec F}$ we need to find.

Let's write down the Newton's second law assume there's no sliding

$$\vec F + m\vec g + \vec N + {{\vec F}_{{\rm{frict}}}} = 0$$

Assume that the force ${\vec F}$ has component only along the $Ox$ - axis let's project this equations to the $Oy$ axis. The componenst of vectors are $\vec g = g(\cos \alpha ,\sin \alpha )$ , $\vec N = (0,N)$ , ${{\vec F}_{{\rm{frict}}}} = ( - {F_{{\rm{frict}}}},0)$ and $F = ( - F,0)$ (whe choose minus sign for x-component of extra force because it should be directed against the direction of possible sliding). Then the equation for the $Oy$ axis is

$$- mg\cos \alpha + N = 0$$

thus

$$N = mg\cos \alpha$$

Now we use ${F_{{\rm{frict}}}} = \mu N$ (Amontons-Coulomb friction law) and write down the equation for the $Ox$ axis

$$- {F_x} + mg\sin \alpha - \mu mg\cos \alpha = 0$$

thus

$${F_x} = mg(\sin \alpha - \mu \cos \alpha )$$

$${F_x} \approx 5{\rm{[kg}}] \cdot 9.8[{{\rm{m}} \over {{{\rm{s}}^{\rm{2}}}}}] \cdot (\sin {\pi \over 6} - 0.3\cos {\pi \over 6}) \approx 11.8[{\rm{N}}]$$

So $\left| {\vec F} \right| \approx 11.8[{\rm{N}}]$

This is the answer.