A small ring of mass 100 g slides on a smooth straight wire in the direction i + 2j + 2k. Apart from the normal reaction of the wire, forces on the ring are its weight - k N and a constant force due to the wind of 1/2i + 1/2j N. The ring has an initial speed of 2 m s^-1. i) Find the unit vector in

Question

A small ring of mass 100 g slides on a smooth straight wire in the direction i + 2j + 2k.
Apart from the normal reaction of the wire, forces on the ring are its weight - k N and a constant force due to the wind of [latex]\frac{1}{2}i + \frac{1}{2}j N.[/latex] The ring has an initial speed of 2 m [latex]s^{-1}[/latex].
i) Find the unit vector in the direction of the wire.
ii) Find the resolved part of the total forces on the ring in the direction of motion.
iii) Hence find the acceleration of the ring and the distance travelled before coming to rest.

Accepted Answer

i) |i + 2j + 2k| = √(1² + 2² + 2²) = 3.

Hence the unit vector in the direction of motion is

[latex]\hat{n} = \frac{1}{3}(i + 2j + 2k) = \frac{1}{3}i + \frac{2}{3}j + \frac{2}{3}k. [/latex]

ii) The normal reaction R of the wire is perpendicular to the wire and has no component parallel to it. The resultant of the other forces is

[latex]F = \frac{1}{2}i + \frac{1}{2}j - k.[/latex]

The resolved part in the direction of motion is

F • [latex]\hat{n} = \left(\frac{1}{6} + \frac{1}{3} - \frac{2}{3}\right) = - \frac{1}{6} N.[/latex]

iii) Applying F = ma in the direction of motion gives

[latex]- \frac{1}{6} = \frac{1}{10} a [/latex] (mass is [latex]\frac{1}{10} kg [/latex])

⇒ a = [latex]- \frac{5}{3}.[/latex]

The ring decelerates at [latex]\frac{5}{3} m s^{-2}.[/latex]

To find when the ring comes to rest, apply v² = u² + 2as :

0 = 4 - 2 × [latex]\frac{5}{3}s [/latex]

⇒ s = [latex] \frac{6}{5}.[/latex]

The ring comes to rest after [latex] \frac{6}{5}[/latex] m.

Question 17.5: A small ring of mass 100 g slides on a smooth straight wire ...

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