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I am learning about the 'bending' of a cyclist

enter image description here

I was told that the reaction force also 'tilts' when the cyclist is tilting (like in the figure), How is this possible? Isn't the reaction force always normal to the cyclist?

Qmechanic
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Hayden Soares
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3 Answers3

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I think $R$ is meant to be the force exerted by the cycle on the cyclist. The confusing thing about the diagram is that it draws the cyclist and cycle as one object, but then shows $R$, which is an internal force as far as this combined object is concerned. The diagram also omits the equal and opposite force $-R$ which the cyclist exerts on the cycle, the normal force that the road exerts on the cycle, and the cycle's weight.

It would be much clearer to draw two separate free-body diagrams - one for the cyclist, with forces $R$ from the cycle and the cyclist's weight $mg$; and one for the cycle with forces $-R$ from the cyclist, the cycle's weight $Mg$, friction from the road surface (horizontal) and the normal force from the road surface (vertical).

gandalf61
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well it is just that reaction force is not always normal to the plane.Instead the normal force is always normal to the plane. The reaction force is composed of two parts

  1. The normal force which is normal to the surface i.e, the $ R\cos(\theta)$
  2. And the friction force which is horizontal i.e, the $ R\sin(\theta)$
Niladri Sarkar
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Most probably I didnt get your question. What do you mean by Reaction force is it Normal force. If yes then the normal always acts perpendicular it is just that we have spitted it into components of force (R cos-theta, R sin-theta) which can be compared to other components of force in the same direction(or same axis)

mr.anonymous
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  • $R$ cannot be the normal force exerted by the road on the cycle, because the normal force must be perpendicular to the road i.e. vertical. – gandalf61 Feb 20 '21 at 17:13
  • since the cycle is inclined to the road therefore the force normal is perpendicular to the the cycle tyre.. just imagine the the tyre as a big box on a road – mr.anonymous Feb 20 '21 at 17:18
  • If that were true then a tilted cycle on a smooth horizontal surface would experience an unbalanced horizontal force in the direction of its tilt, and it would move sideways. This is clearly not the case. – gandalf61 Feb 20 '21 at 17:48
  • @gandalf61 i didnt understand ur point. – mr.anonymous Feb 21 '21 at 06:27
  • The normal force must always be perpendicular to the road surface. See https://www.physicscentral.com/experiment/askaphysicist/physics-answer.cfm?uid=20100126081219. Otherwise a tilted cycle on a smooth road would be pushed sideways. Which it isn’t. – gandalf61 Feb 21 '21 at 07:46
  • if cycle is tilted and moving in circular orbit(which is why cyclist tilt cycles to avoid skidding ) so by tilting the outward force(centrifugal) breaks into components and is reduced . And it is pushed sideways(inwards(centripetal force + friction)and outwards(centrifugal)) and hence cyclist remains stable – mr.anonymous Feb 21 '21 at 08:05
  • see thishttps://imgur.com/a/MFJNhyl – mr.anonymous Feb 21 '21 at 08:36
  • I have moved this extended discussion to a chat room. We can continue it here. – gandalf61 Feb 21 '21 at 10:24