Numford
Vectors

Components of weight on an inclined plane

Of a weight of 10 N, 5 N acts down a 30 degree slope and 8.66 N presses into it.

Vectors

Drag the glowing tips, or type values. Angles are measured from the x axis, and θ is the angle from A to B. Pick a result card to see it built in the picture.

Angle betweenn/acos θ = A·B / AB
Unit vectors-1jB = n/a

Vector A of size 10 and vector B of size 0. The resultant is 10, 0 degrees from A. The dot product is 0 and the cross product has size 0.

Show the formulas
Resultant
One of the vectors is zero, so R = 10.
Components
A·B = AxBx + AyBy + AzBz = 0 × 0 + -10 × 0 + 0 × 0 = 0
Determinant
A×B = (AyBz - AzBy) i + (AzBx - AxBz) j + (AxBy - AyBx) k = 0
Unit vector
 = A / |A| = -1j
Working

Step by step

  1. Given a weight W = 10 N (a 1 kg block with g = 10 m/s²) on a slope of α = 30°.
  2. Along the slope: W sin α = 10 × sin 30° = 5 N
  3. Perpendicular to the slope: W cos α = 10 × cos 30° = 8.66 N
  4. Check: √(25 + 75) = 10. The two parts rebuild the weight.
  5. On a smooth slope the block accelerates down it at g sin α. The normal force balances the perpendicular part.
Derivation

Why it works

  1. Take one axis along the slope and the other perpendicular to it.
  2. The weight W points straight down. The normal to the slope is tilted by the angle α of the incline from the vertical.
  3. So the angle between W and the perpendicular axis is also α.
  4. Along the slope: W sin α. Perpendicular to the slope: W cos α.
  5. The slope component pulls the block down the plane. The perpendicular one is balanced by the normal force.
Questions

Components on an inclined plane: quick answers

Why is the component along the slope W sin α?

The angle between the weight and the perpendicular to the slope is α. So the part perpendicular to the slope is W cos α, and the part along it is W sin α.

What are the components of 10 N on a 30 degree slope?

Along the slope 5 N and perpendicular 8.66 N.

What happens as the slope gets steeper?

W sin α grows and W cos α shrinks. At 90 degrees the whole weight acts along the "slope", and at 0 degrees none of it does.

Try your own vectors