Answer to Question #231301 in Mechanics | Relativity for Roshi

Question #231301

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.


1
Expert's answer
2021-09-01T11:20:50-0400

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

Pv=T×cos60=T2\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

Ph=T×sin60=3T2\qquad\qquad \begin{aligned} \small P_h &=\small T\times \sin60=\frac{\sqrt3 T}{2} \end{aligned}

  • Now, the ring's weight is 4g\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,

ΣF=F4gT2=0F=4g+T2\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.

3T2R=0R=3T2\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 R2+F2\small \sqrt{R^2+F^2}

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