Question

Question from friction.​

Answer 1

Mass of the object, $\sf{m=200\ kg.}$

Angle of inclination, $\sf{\theta=30^o}.$

Coefficient of limiting friction, $\sf{\mu=\dfrac{1}{\sqrt3}}.$

Let $\sf{g=10\ m\,s^{-2}.}$

We know the component of weight of the object parallel to the plane is,

$\sf{\longrightarrow mg\sin\theta=200\times10\sin30^o}$

$\sf{\longrightarrow mg\sin\theta=200\times10\times\dfrac{1}{2}}$

$\sf{\longrightarrow mg\sin\theta=1000\ N}$

This component allows the object to slide down along the plane.

And the frictional force acting on the object is,

$\sf{\longrightarrow \mu\,mg\cos\theta=\dfrac{1}{\sqrt3}\times200\times10\cos30^o}$

$\sf{\longrightarrow \mu\,mg\cos\theta=\dfrac{1}{\sqrt3}\times200\times10\times\dfrac{\sqrt3}{2}}$

$\sf{\longrightarrow \mu\,mg\cos\theta=1000\ N}$

We can keep the mass in equilibrium by applying an external force in two ways.

1. In order to avoid it from sliding down only.

• The frictional force is acting upward along the plane since the object was sliding down.

• The applied force is also in the direction of this friction.

• The weight is in downward direction so it's opposite to the applied force.

As the net force acting on it is zero,

$\sf{\longrightarrow F+\mu\,mg\cos\theta-mg\sin\theta=0}$

$\sf{\longrightarrow F+1000-1000=0}$

$\sf{\longrightarrow F=0}$

So this is the least force.

$\sf{\longrightarrow\underline{\underline{F_{min}=0}}}$

2. In order to move upward with constant speed.

• As the object is getting pushed upward, the frictional force is acting downward, opposite to applied force.

• The weight is in downward direction, along with frictional force, opposite to applied force.

As the net force acting on it is zero,

$\sf{\longrightarrow F-\mu\,mg\cos\theta-mg\sin\theta=0}$

$\sf{\longrightarrow F-1000-1000=0}$

$\sf{\longrightarrow F-2000=0}$

$\sf{\longrightarrow F=2000\ N}$

This is the greatest force.

$\sf{\longrightarrow\underline{\underline{F_{max}=2000\ N}}}$