Question

A ring of mass 4 kg is threaded on a fixed rough vertical rod. A light string is attached to the ring,

and is pulled with a force of magnitude T N acting at an angle of 60◦ to the downward vertical (see

diagram). The ring is in equilibrium.

(i) The normal and frictional components of the contact force exerted on the ring by the rod are R N

and F N respectively. Find R and F in terms of T.

Answer 1

Explanations & Calculations

• By resolving the pulling force into orthogonal components we can find the verticle & horizontal applications of it.
• The vertical component of the pulling force that acts downwards on the ring is

$\qquad\qquad \begin{aligned} \small P_v &=\small T\times\cos60=\frac{T}{2} \end{aligned}$

• And it's horizontal component which applies horizontally on the ring (pulling away) is

$\qquad\qquad \begin{aligned} \small P_h &=\small T\times \sin60=\frac{\sqrt3 T}{2} \end{aligned}$

• Now, the ring's weight is $\small 4g$ downwards and the frictional force is holding the ring sliding downwards the rod hence it applies on the rod upwards.
• Since the ring is in equilibrium, all the vertical forces should be balanced or the net vertical force on the ring should be zero. Then,

$\qquad\qquad \begin{aligned} \small \uparrow\Sigma F&=\small F-4g-\frac{T}{2}=0\\ \small F&=\small 4g+\frac{T}{2} \end{aligned}$

• Frictional force is found then.

• Treating the ring's horizontal equilibrium the same way, we can find the normal force that acts on the ring.

$\qquad\qquad \begin{aligned} \small \to \frac{\sqrt3 T}{2}-R&=\small 0\\ \small R&=\small \frac{\sqrt 3 T}{2} \end{aligned}$

• The contact force on the ring is obtained by $\small \sqrt{R^2+F^2}$