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

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. 3.33 N is the maximum force can be applied at the highest point in the horizontal direction so that the sphere does not slip on the surface.

• Momentum is given by,
• F × R = lα × f × R
• ⇒ F = 2/5 mRα +βmg
• and
• F = mac - βmg
• ∴ac = (F + βmg) / m
• Equating above equations, we get,
• ⇒$\dfrac{2}{5} * m * (\dfrac{F + \beta mg}{m})+\beta mg\\\\\dfrac{2}{5} (F + \beta mg) +\beta mg$
• F = 2/5 F +2/5 * 0.5 * 10 + 2/7 * 0.5 * 10
• 3F/5 = 4/7 + 10/7 =2
• F = (5 * 2) / 3 = 10/3
• ∴F = 3.33N

Answer 2

Force of 3.33 N can be applied in the horizontal direction at the highest point, so there will not be any slip on the surface

Explanation:

Mass of the solid sphere = 0.50 Kg

$\text {Coefficient of static friction} = \frac{2}{7}$

About the center, if we take a moment  

F X R = lα × f × R  

F = 2/5 mRα + μmg    ------- (1)

Then, $\mathrm{F}=\mathrm{ma}_{\mathrm{c}}-\mu \mathrm{mg}$ ------- (2)

$a_c = F + \frac{ \mu mg}{m}$

Applying the value of ac in equation 1, we get

= $\frac{2}{5} \times m \times[\frac {F+\mu m g}{ m} ]+\mu m g$

$\frac{2}{5} \times[F+\mu m g]+\mu m g$

$\frac{2}{5} \times F+\frac{2}{5}\times 0.5 \times 10+ \frac{2}{7} \times0.5 \times 10$

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

$F = \frac{5\times 2}{3}$

F = 3.33 N

The amount of force required at the highest point in the horizontal direction so that the sphere does not slip on the surface is 3.333 N