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 $\frac{2}{7}$. 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

If we take moment about the center than

$F\times R=I\alpha +f\times R$

$\Rightarrow F=\frac{2}{5}mR\alpha +\mu mg$ ...(1)

$\text{Again,}F=m{a}_c-mmg$ ...(2)

$[because{a}_c=r\alpha ]$

$\Rightarrow a=\frac{(F+\mu mg)}{m}$

\(\text{Putting the value of}a_c \text{in equation}(1),

$\text{we get,}$

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

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

$F=\frac{2}{5}F+\frac{2}{5}+\frac{2}{5}\times \times \frac{2}{7}\times 0.5\times 10+2.7\times 0.5\times 10$

$\Rightarrow \frac{3F}{5}=\frac{4}{7}+\frac{10}{7}=2$

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