The Inclined Plane

Block Acceleration

the leftover pull

With the into-slope press balanced, the unbalanced along-slope pull accelerates the block down the ramp, at a rate independent of its mass.

Example

With the into-slope press balanced, the unbalanced along-slope pull accelerates the block down the ramp — at a rate that does not depend on its mass. Exact arithmetic here means exact results for the stated model inputs; measured inputs still carry uncertainty and significant-figure limits.

highlighted = computed this step

Only the along-slope pull is left

Into the slope, the normal force cancels the press. Along the slope there is nothing to cancel the 6 newton pull (the ramp is frictionless), so that is the net force.

Fnet=W∥=6 N (down the slope)F_{\text{net}} = W_\parallel = \hl{6}\ \text{N (down the slope)}
The unbalanced along-slope pullA block on a ramp with one arrow pulling down along the slope.mdown-slope

Net force gives acceleration

Divide the net force by the mass: 6 newtons over 1 kilogram is 6 metres per second squared, down the slope.

a=Fnetm=6 N1 kg=6 m/s2a = \frac{F_{\text{net}}}{m} = \frac{6\ \text{N}}{1\ \text{kg}} = \hl{6}\ \text{m}/\text{s}^{2}
The block accelerates down the slopeA block on a ramp with an arrow showing it accelerating down along the surface.ma

The mass did not matter

Try a heavier block: 2 kilograms has twice the weight, 20 newtons, and twice the along-slope pull, 12 newtons, but also twice the mass. Its acceleration is 12 over 2, still 6 metres per second squared. The mass cancels, leaving acceleration equal to gravity times the along-slope fraction.

a=m gsin⁡θm=gsin⁡θ=10 m/s2⋅35=6 m/s2a = \frac{m\,g\sin\theta}{m} = g\sin\theta = 10\ \text{m}/\text{s}^{2} \cdot \frac{3}{5} = \hl{6}\ \text{m}/\text{s}^{2}

More mass gives more pull and the same acceleration

On a frictionless ramp, more mass gives more weight and more net down-slope force in the same ratio. Dividing by that same mass leaves the acceleration unchanged.

mFneta1 kg6 N6 m/s22 kg12 N6 m/s23 kg18 N6 m/s2\begin{array}{c|c|c}m & F_{\text{net}} & a \\ \hline 1\ \text{kg} & 6\ \text{N} & 6\ \text{m}/\text{s}^{2} \\ 2\ \text{kg} & 12\ \text{N} & 6\ \text{m}/\text{s}^{2} \\ 3\ \text{kg} & 18\ \text{N} & 6\ \text{m}/\text{s}^{2}\end{array}

Steeper ramp gives more acceleration

Mass cancels, but the ramp fraction does not. A larger rise fraction leaves more of gravity along the surface.

risehypga010 m/s20 m/s23510 m/s26 m/s24510 m/s28 m/s2\begin{array}{c|c|c}\frac{\text{rise}}{\text{hyp}} & g & a \\ \hline 0 & 10\ \text{m}/\text{s}^{2} & 0\ \text{m}/\text{s}^{2} \\ \tfrac{3}{5} & 10\ \text{m}/\text{s}^{2} & 6\ \text{m}/\text{s}^{2} \\ \tfrac{4}{5} & 10\ \text{m}/\text{s}^{2} & 8\ \text{m}/\text{s}^{2}\end{array}
mechanics The leftover 6 N over a 1 kg block gives a clean 6 m/s^2, and the mass cancels out of a = g sin theta.