1 Answer | Add Yours
The particle of mass m is placed on the sphere. The gravitational force of attraction on the particle due to the Earth is equal to m*g in a direction vertically downwards.
This can be divided into two components, one tangential to the sphere and the other along the radius at the point of contact towards the center of the sphere. The former component is m*g*cos 60 and the latter is m*g*sin 60. The normal force between the particle and the sphere is equal to m*g*sin 60. If the coefficient of friction between the two is `mu` , the frictional force on the particle due to the sphere is `m*g*(sqrt3/2)*mu` . It is given that this is equal to `(m*g)/2` . Equating the two expressions gives:
`m*g*(sqrt3/2)*mu = (m*g)/2`
=> `mu = 1/sqrt 3`
The coefficient of friction between the particle and the sphere is `sqrt 3/3`.
We’ve answered 333,385 questions. We can answer yours, too.Ask a question