Question

A solid sphere of mass 0.50$kg$ is kept on a horizontal surface. The coefficient of static friction between the surfaces in contact is 2$/7$ . What maximum force can be applied at the highest point in the horizontal direction so that the sphere does not slip on the surface?

Answer 1

Given: $\mu =\frac{2}{7}$ and $m=0.50kg$

According to the question,

$\Sigma T=0$

$\Rightarrow$ $F\times R=l_a\times F\times R$ ( $a_c=R\alpha$)

$\Rightarrow$ $F=\frac{2}{5}mR\alpha +\mu mg$ ...........$\left(i\right)$

Now, $F=m{a}_c-\mu mg$

$\Rightarrow$ $a_c=\frac{F+\mu mg}{m}$ ..............$\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$, we get

$\Rightarrow$ $F=\frac{2}{5}m\left(\frac{F+\mu mg}{m}\right)+\mu mg$

$\Rightarrow$ $F=\frac{2}{5}\left(F+\mu mg\right)+\mu mg$

Putting all the values, we have

$\Rightarrow$ $F=\frac{5\times 2}{3}=\frac{10}{3}=3.3N$