Question

Find the least pulling force which acting at an angle of ${45}^0$ with the horizontal, will slide a body weighing 5Kg along a rough horizontal surface. The coefficient of friction${\mu }_s={\mu }_k=1/3$. If a force of double this is applied along the same direction, find the resulting acceleration of the block.

Answer 1

When the block is about to start sliding, the frictional force acting on block reaches to its limiting value $f={\mu }_{s}N$
In vertical direction: $N+Psin{45}^0=mg$
$\Rightarrow N=mg=Psin{45}^0$

In horizontal direction:
$Pcos{45}^0={\mu }_{s}N$

or $Pcos{45}^0={\mu }_s(mg-Psin{45}^0)$

or $P=\frac{{\mu }_{s}mg}{cos{45}^0+\mu sin{45}^0}=\frac{25}{\sqrt{2}}N$

If applied force is 2P: Friction will be kinetic nature. The block will move

$f={\mu }_{k}N$.

$N=mg-2Psin{45}^0$ and

$2Pcos{45}^0-{\mu }_{k}N=ma$
$2Pcos{45}^0-{\mu }_k(mg-2Psin{45}^0)=ma$
$a=\frac{2P}{m\sqrt{2}}({\mu }_k+1)-{\mu }_{k}g=\frac{10}{3}m{s}^{-2}$