A solid sphere of mass 10 kg is placed on a rough surface having coefficient of friction $μ$ = 0.1. A constant force F = 7 N is applied along a line passing through the center of the sphere as shown in the figure. The value of frictional force on the sphere is

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7 N

1 N

3 N

2 N

answer is B.

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Detailed Solution

Let f the be frictional force towards left for pure rolling.

$\begin{array}{l} a = R \times α \\ where a = \frac{F - f}{m} = \frac{7 - f}{10} - - - - (1) \\ α = \frac{τ}{I} = \frac{f R}{I} = \frac{f R}{\frac{2}{5} m R^{2}} = \frac{5 f}{2 m R} \end{array}$

$Hence, \frac{7 - f}{10} = \frac{5 f R}{20 R} f r o m (1) \Rightarrow f = 2 N$

Limiting friction = $f_{L} = μ m g = (0.1) (10) (10) = 10 N$

Since, $f < f_{L}$ sphere will keep on doing pure rolling.

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