Vector components for an object on a slope
In this case on an object on a slope, the weight is a single force that can be resolved into two independent components:
- one acting along the slope Fs
- the other acting perpendicular to the slope Fp
Expressed mathematically:
Component of weight parallel to slope, Fs equals:
\(W\sin\theta = mg\sin\theta\)
Component of weight perpendicular to slope, Fp equals:
\(W\cos\theta = mg\cos\theta\)
Question
A 10 kg box slides down a frictionless slope. The slope is at 30° to the horizontal.
Find the component of the weight acting parallel to the slope.
The component of weight parallel to the slope is:
\(W\sin \theta = mg\sin \theta\)
\(= 10 \times 9.8 \times \sin 30^\circ\)
\(= 49N\)
Question
Now find the acceleration of the box down the incline.
\(F = 49N\)
\(m = 10kg\)
\(a = ?\)
\(F = ma\)
\(49 = 10 \times a\)
\(a = 4.9m{s^{ - 2}}\)
Question
If the slope is steeper, how does this affect the two components?
\(mg\sin \theta\) parallel to slope will increase and \(mg\cos \theta\) will decrease.
Question
How will this affect the acceleration?
\(mg\sin \theta\) will cause increased force and \(mg\cos \theta\) will decrease causing less friction as there will be less force pushing the block onto the surface. So the acceleration will increase.
