Topic 6 of 6
From resultant force to motion
Build the force model first. For a body of constant mass, the resultant force determines its acceleration.
Momentum is p = mv. When m is constant, its rate of change is dp/dt = m dv/dt = ma. Combining this with the second law gives:
Fresultant = ma
Use the resultant component along the chosen direction, not one selected applied force. The acceleration follows the resultant force; it need not point along the body's current velocity.
Estimate the scale before detailed calculation
Suppose a small cart has mass roughly 1 kg, moves at a speed around 1 m/s, and changes velocity by about 1 m/s in the same direction over about 2 s. These rough assumptions suggest momentum magnitude of order 1 kg m/s = 1 N s and acceleration magnitude about 0.5 m/s2.
If its speed stays around 1 m/s over those 2 s, expect travel of order 2 m. These are scale checks, not precise predictions. They help expose a result such as a kilometre of travel for this short, slow cart run.
A cart pulled along a level track
A 0.80 kg cart is moving right with initial velocity +0.40 m/s. A string pulls it right with 0.60 N and track resistance acts left with 0.20 N. These forces stay constant over the next 2.0 s, during which the cart continues moving right.
Find the resultant from the actual forces on the cart
Scale note: the horizontal arrows share one scale, and the two vertical arrows share another. Horizontally, 0.60 - 0.20 = +0.40 N. Vertically, support and weight balance. These are the four interactions; the resultant is their sum.
Worked force-to-motion model
Use the resultant, then justify constant acceleration
- Identify the forces: string on cart, track resistance on cart, track support and Earth's gravitational force. With g = 9.81 m/s2, weight and support each have magnitude 0.80(9.81) = 7.848 N, about 7.85 N.
- Resolve horizontally: Fx = +0.60 - 0.20 = +0.40 N. Vertical forces balance.
- Calculate acceleration: a = 0.40/0.80 = +0.50 m/s2. Constant resultant and constant mass justify constant acceleration.
- Predict the ending motion: v = 0.40 + 0.50(2.0) = +1.40 m/s. Displacement s = 0.40(2.0) + ½(0.50)(2.0)2 = +1.80 m.
- Check momentum: Δp = 0.80(1.40 - 0.40) = +0.80 kg m/s. Dividing by 2.0 s gives +0.40 N, the same resultant.
Resistance is leftward because the stated cart keeps moving right. If the cart reverses, reconsider the resistance direction. If the pull is removed, calculate a new resultant before using the motion equations.
Optional check A 0.80 kg cart is moving right when its pulling string goes slack. Track resistance remains 0.20 N leftward. What is its acceleration immediately afterwards, taking right as positive?
Infer motion from position data
One measurement method is to film a cart beside a length scale in its plane of motion. Fix the camera approximately perpendicular to that plane, establish the time base, calibrate length and identify the same point on the cart in each frame. A suitable position sensor provides another route.
A scale outside the cart's motion plane or a changed camera view can bias the distance conversion. Repeating a run does not repair that calibration. Keep the run within a supported, clear path with enough run-out.
The following are supplied rounded model values for the cart, starting at position +0.200 m. They are used to practise the analysis; they are not laboratory observations.
| Time / s | Position / m |
|---|---|
| 0.000 | 0.200 |
| 0.500 | 0.463 |
| 1.000 | 0.850 |
| 1.500 | 1.363 |
| 2.000 | 2.000 |
For each neighbouring pair, divide displacement by its own time interval. The first gives (0.463 - 0.200)/(0.500 - 0.000) = 0.526 m/s. This is an interval-average velocity.
Under a constant-acceleration model, associate that value with the interval's midpoint time: (0.000 + 0.500)/2 = 0.250 s. This follows because velocity is a straight-line function of time. For more general changing motion, such an assignment is an approximation.
Plotting the interval velocities against their midpoint times allows a straight-line fit. Its gradient estimates acceleration; its intercept estimates the initial velocity for the chosen t = 0. Compare mass × fitted acceleration with the resultant 0.40 N, not the string pull alone.
Choose time intervals short enough to resolve change but not so short that small position differences are dominated by reading noise. Differencing can magnify that noise. A fixed position-zero offset cancels in each difference; a length scale-factor error does not.
Do not plot an interval-average velocity at its later endpoint by default. Also, x against t2 is not automatically a straight line here: the initial velocity is nonzero, so position includes a term proportional to t.
Optional check In a constant-acceleration cart model, x changes from 0.200 m at 0.000 s to 0.463 m at 0.500 s. The interval-average velocity is 0.526 m/s. At what representative time should it be plotted?
Check the cart velocities and fitted motion
The four midpoint times are 0.250, 0.750, 1.250 and 1.750 s. Their interval velocities are 0.526, 0.774, 1.026 and 1.274 m/s.
A free-intercept straight-line fit gives gradient 0.4992 m/s2 and intercept 0.4008 m/s. These represent approximately a = 0.50 m/s2 and u = 0.40 m/s. The fitted relationship is velocity = initial velocity + acceleration × midpoint time.
Inferred resultant = 0.80 × 0.4992 = 0.39936 N, or about 0.40 N. The small difference from the ideal model follows from the deliberately rounded supplied positions. Extra spreadsheet digits do not establish the precision of a real camera measurement.