Topic 3 of 4
Resultant force and acceleration
The resultant force determines the acceleration. Calculate it from all the forces on the body before using F = ma.
Acceleration is the change in velocity per unit time. Use a free-body diagram to find the resultant and choose a positive direction.
Fresultant = maFor a body of constant mass: force in N, mass in kg and acceleration in m/s2. Acceleration is in the direction of the resultant force.
A resultant force of 1 N gives a 1 kg mass an acceleration of 1 m/s2. The equation concerns acceleration, not the velocity at that instant. An object may be moving quickly while its resultant force and acceleration are both zero.
Unbalanced forces on a moving trolley
Selected body: 2.0 kg trolley
The trolley is moving right along a horizontal track.
Horizontal resultant: 7 - 3 = 4 N right. The vertical forces balance.
Worked example
Subtract the opposing force first
The trolley shown is moving forwards. Find its acceleration. Take forwards as positive.
- Find the horizontal resultant: Fresultant = 7 - 3 = +4 N. The forces oppose each other, so subtract their magnitudes.
- Rearrange: a = Fresultant / m.
- Substitute: a = 4 / 2.0 = 2.0 m/s2 forwards.
The balanced vertical forces give no vertical acceleration. Using 7 / 2.0 would treat the pull as the entire resultant and ignore the track's resistance.
A forward-moving object can accelerate backwards
In a different interval, suppose the trolley is still moving forwards but its string pull is 3 N and track resistance is 7 N. Keeping forwards positive gives:
a = -4 / 2.0 = -2.0 m/s2The acceleration is backwards while the velocity is forwards, so the trolley slows during this interval.
The negative sign specifies direction. It does not mean the trolley is already moving backwards. If the trolley reaches a stop, reassess the forces before predicting what happens next; the same resistance direction cannot simply be assumed throughout a reversal.
How mass affects acceleration
With the same resultant force, a larger mass has a smaller acceleration. A 4 N resultant gives a 2 kg trolley an acceleration of 2 m/s2, but a 4 kg trolley an acceleration of 1 m/s2. Doubling mass halves acceleration when the resultant is unchanged.
Investigating the relationship
The following idealised data describe a trolley of fixed mass 0.50 kg. The forces listed are resultant forces, after accounting for resistance.
| Resultant force / N | Acceleration / (m/s2) |
|---|---|
| 0.20 | 0.40 |
| 0.40 | 0.80 |
| 0.60 | 1.20 |
Doubling the resultant from 0.20 to 0.40 N doubles acceleration from 0.40 to 0.80 m/s2. The ratio Fresultant / a is 0.50 kg in every row, consistent with the fixed mass. An acceleration-against-resultant-force graph would be a straight line through the origin for this model.
For a real investigation, keep the trolley's total mass unchanged and use a level track. Change the horizontal pull, estimate acceleration from motion readings, and account for resistance when finding the resultant. A motion sensor or timed velocity readings can provide the changes needed for a = (v - u) / t.
Repeat readings to judge variation. A slightly sloping track adds a component of weight along the track, while changing resistance can make a measured pull different from the assumed resultant. Address those causes instead of calling every disagreement "human error". Real data need not fall exactly on the model values.