Motion in One & Two Dimensions
- If a constant force acts upon an object, then it will experience a resultant acceleration, determined using F = ma
- This motion can be investigated in one or two-dimensional planes, such as along the ground or on a slope
- One-dimensional planes involve just up and down or left and right (on the ground)
- Two-dimensional planes involve both up and down and left and right (on a slope)
- On a slope, it is often simpler to resolve the forces into parallel and perpendicular components, rather than horizontally and vertically:
- The weight, mg of the object acts vertically down
- The frictional force, F between the slope and the object acts along the plane of the slope, in the direction opposing the motion
- The normal reaction force, R acts perpendicular to the plane of contact between the object and the slope
The normal reaction force R, weight W and friction F on a block moving down a slope
Worked example
The diagram below shows the forces acting on a water drop on the windscreen of a stationary car.
The windscreen makes an angle of 30° to the horizontal. The weight of the water drop is 8.0 × 10–5 N.The normal contact force on the water drop is R. There is also a force F acting on the water drop as shown. The water drop is stationary.Determine:
(a) The component of the weight of the water-drop perpendicular to the windscreen
(b) The component of the weight of the water-drop parallel to the windscreen
(c) The magnitude of R
(d) The magnitude of F
Part (a)
- Perpendicular component = W cos 30°
- Perpendicular component = (8.0 × 10–5) cos 30° = 6.9 × 10–5 N
Part (b)
- Parallel component = W sin 30°
- Parallel component = (8.0 × 10–5) sin 30° = 4.0 × 10–5 N
Part (c)
- R is equal to the perpendicular component of the weight
- Perpendicular forces must be equal and opposite
R = 6.9 × 10–5 N
Part (d)
- F is equal to the parallel component of the weight
- Parallel forces must be equal and opposite
F = 4.0 × 10–5 N
Worked example
The diagram below shows a rider on a sledge sliding down an icy slope. The frictional forces acting on the sledge and the rider are negligible. The normal contact force N and the total weight W of the sledge and rider are shown.
The acceleration of the sledge and rider down the slope is 2.0 m s–2.Determine the angle made by the slope to the horizontal.
Step 1: Write down the known quantities
- Acceleration, a = 2.0 m s–2
- Weight, W = mg
- Component of weight parallel the slope, F = W sin θ = mg sin θ
Step 2: Write down the equation relating force and acceleration
F = ma
Step 3: Substitute in the component of weight down the slope
mg sin θ = ma
Step 4: Rearrange for sin θ and calculate the angle
a = g sin θ
sin θ = 2.0 / 9.81
θ = 12°
