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Lesson10 min read

Choose axes that do the work

A slope looks complicated until you choose a useful direction and resolve the weight.

Sample unit
BY THE END, YOU CAN

Resolve the weight of a block into components parallel and perpendicular to a slope.

Paper, pen & a little patience.

Start with the written lesson below. A handwritten video will follow.

Forces on a block on a frictionless slopeWeight acts vertically down. The normal force acts perpendicular to the slope, up and to the left. The block accelerates down the slope.30°mmgNstart with the forces.ONE OBJECT. TWO FORCES. A CLEARER PICTURE.
A frictionless slope. Draw the actual forces before resolving them.

Keep the same first step

A block is sliding on a fixed, straight, frictionless slope at angle θ\theta above the horizontal. Choose the block as your system.

Two forces act on it: its weight mgmg, vertically downward, and the normal force NN, perpendicular to the slope. The block’s motion is not another force.

Turn the axes, not gravity

Choose xx down the slope and yy perpendicular to the slope, pointing away from the surface.

Gravity still points vertically downward. We are changing the coordinate directions used to describe it, not the physical force.

This choice helps because the block does not accelerate into or away from the fixed straight surface. Its acceleration perpendicular to the slope is zero.

Resolve the weight

The angle between the weight and the direction into the slope is θ\theta. In the component triangle, the side parallel to the slope is opposite this angle:

Fg,x=mgsin⁡θ,Fg,y=−mgcos⁡θ.F_{g,x} = mg\sin\theta, \qquad F_{g,y} = -mg\cos\theta.

Try two checks before memorizing anything:

  • When θ=0\theta = 0, the surface is horizontal. The downhill component must vanish.
  • As the slope becomes steeper, more of the weight acts along it.

The two components are a way of representing the same weight. Do not count both components and the original weight as three forces.

Write one equation per direction

Perpendicular to the slope:

N−mgcos⁡θ=0.N - mg\cos\theta = 0.

Along the slope:

mgsin⁡θ=max.mg\sin\theta = ma_x.

So N=mgcos⁡θN = mg\cos\theta and ax=gsin⁡θa_x = g\sin\theta. The normal force is smaller than the weight for a nonzero slope in this particular model.

Make a prediction

Would doubling the mass double the acceleration on this frictionless slope?

Compare your prediction

No. In mgsin⁡θ=maxmg\sin\theta = ma_x, the mass cancels. Doubling the mass doubles the downhill force, but it also doubles the mass being accelerated. The acceleration is unchanged.

This result assumes the same slope, negligible air resistance, and no friction or other applied force.

Now try A block on an incline without referring to the equations above. Explain each step aloud or beside your drawing.

KEEP THE IDEAS CONNECTED

Connected learning.