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I'm studying mechanics and we were always taught that for a particle on an inclined plane, the reaction force $R = mg\cos\theta$. However in this mechanics module, under circular motion, when a motorbike/other object is on a smooth bend in a track, and is moving, in the form of a circle inclined at an angle, the equation becomes $R\cos\theta = mg$.

I can see that if you resolve R through $\theta$ you get it in the same direction as $mg$, but if you resolve $mg$ through the angle you also get it in the same direction as R, but yet the two equations will give you different answers and I don't understand why it is $R\cos\theta = mg$ instead of $R = mg\cos\theta$, when finding the reaction force.

FreezingFire
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Tiernan Watson
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1 Answers1

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When have an object either stationary on an inclined plane or moving straight up or down the incline, the reactive force perpendicular to the plane does not have to "hold up" the entire weight of the object. The remaining part of the weight of the object is counteracted by friction, or is allowed to accelerate the object downward along the incline, or goes to some combination of those two effects.

In short, in an inclined plane exercise we desire the reactive force to be less than the object's weight. Multiplying the weight by $\cos\theta$, which is less than $1$, fits that desire.

In a typical "banked turn" exercise, such as with your motorbike, the reactive force has to hold up all the weight of the motorbike (since it is considered undesirable to require friction to supply any of that necessary force, and the motorbike is not supposed to "slide down"), and the reactive force also has to push the motorbike toward the center of the turn in order to make the motorbike follow a curved path.

In short, in a banked turn exercise we need the reactive force to be more than the object's weight; in other words, the weight $mg$ must be less than the reactive force $R$, which is accomplished when $mg = R \cos\theta$.

The reason the factor is $\cos\theta$ and not some other function of $\theta$ is a matter of the shapes of the force diagrams for the two exercises.

David K
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