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Gravity Components

Part of the Mechanics section of Coddy's Physics journey. Lesson 21 of 97.

On a ramp the weight mg\relax mg splits into a part down along the slope and a part into the slope. The angle between mg\relax mg and the part into the slope is the ramp angle θ\theta:

F∥=mgsin⁡θ,F⊥=mgcos⁡θF_{\parallel} = mg\sin\theta, \quad F_{\perp} = mg\cos\theta

The weight mg of a box on a ramp split into two dashed components: mg sin theta down along the slope and mg cos theta into the slope; the angle between mg and the component into the slope is the ramp angle theta

The angle between mg and the part into the slope equals the ramp angle θ.

Nothing accelerates into the slope, so N=mgcos⁡θN = mg\cos\theta. On a 30∘30^\circ ramp a 10 kg10\ \mathrm{kg} box is pulled down the slope with 49 N49\ \mathrm{N}, and N is about 84.9 N84.9\ \mathrm{N} instead of 98 N98\ \mathrm{N}.

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Challenge

A 10 kg10\ \mathrm{kg} box slides down a frictionless 60∘60^\circ ramp. Find the normal force NN the ramp pushes on it with.

Try it yourself

Physics diagram: ramp, crate, vec; arrows v

Solve for NN

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