Apply Newton's second law with friction to engineering problems such as pulling, braking and lifting.
Why it matters
Statics assumes nothing accelerates. As soon as something speeds up, slows down or turns, Newton's second law takes over: the net force equals mass times acceleration. Engineers use it to size brakes and motors, to find the force on a lift cable while it accelerates, to set conveyor speeds, and to estimate stopping distances. This lesson applies F = m·a to the kind of particle problems that appear in engineering dynamics courses.
Kinetics of a particle
A body can be treated as a particle when its rotation does not matter, only its overall motion. Then:
ΣF = m·a (as vectors)
ΣFx = m·ax ΣFy = m·ay
The method is the same as statics, with one change: the right-hand side is m·a instead of zero.
- Draw a free-body diagram showing every force on the body.
- Choose axes, ideally with one axis along the acceleration.
- Write ΣF = m·a for each axis.
- Solve, then use kinematics (v = u + a·t, v² = u² + 2·a·s) if you need speeds, times or distances.
Friction
Dry friction opposes sliding. When the body slides, the kinetic friction force is
f = μk · N
where N is the normal force and μk the coefficient of kinetic friction. Before sliding starts, static friction can resist up to μs·N, with μs usually a little larger than μk. Friction coefficients vary a lot with surfaces, contamination and wear, so treat textbook values as rough.
Worked example 1: pulling a crate
A 50 kg crate is pulled across a floor by a rope at 30° above horizontal with a tension of 200 N. The coefficient of kinetic friction is 0.30. Find the acceleration once the crate is moving.
y-axis (no vertical acceleration):
N + 200·sin30° − 50·9.81 = 0 → N = 490.5 − 100 = 390.5 N
x-axis:
200·cos30° − 0.30·N = 50·a
173.2 − 117.2 = 50·a → a ≈ 1.12 m/s²
The upward part of the rope pull reduces the normal force and therefore the friction. Pulling slightly upward is easier than pushing slightly downward, which increases N.
Worked example 2: stopping distance
A 1,500 kg car travels at 25 m/s (90 km/h) and brakes hard on dry pavement with an effective friction coefficient of 0.70. Assume the brakes hold the tyres at the edge of grip.
ΣFx = −μ·m·g = m·a → a = −μ·g = −0.70 · 9.81 = −6.87 m/s²
v² = u² + 2·a·s → 0 = 25² − 2·6.87·s → s ≈ 45.5 m
The mass cancels out: a heavier car needs more braking force but has the same deceleration for the same μ. Notice that stopping distance grows with the square of speed: at 35 m/s the same car needs about 89 m, plus the distance covered during the driver's reaction time.
Worked example 3: lifting a load
A crane lifts a 2,000 kg load, accelerating it upward at 0.5 m/s². The cable tension is
T − m·g = m·a → T = m·(g + a) = 2,000 · 10.31 = 20,620 N
That is about 5 % more than the static weight of 19,620 N. Sudden starts and stops (snatch loading) can raise cable forces far more, which is one reason lifting equipment carries rated capacities and design factors set by standards.
Work and energy (a shortcut)
When you only need speeds, not times, the work-energy principle avoids solving for acceleration:
Σ(work done by forces) = ½·m·v₂² − ½·m·v₁²
For the braking car, friction does work −μ·m·g·s and the kinetic energy drops from ½·m·25² to zero, giving the same 45.5 m.
Common mistakes
- Putting m·a on the free-body diagram as a force. It is the result of the forces, not one of them.
- Assuming N = m·g. Any force with a vertical component, or a slope, changes N.
- Using static friction for a sliding body. Once it slides, use μk.
- Forgetting that a can be negative. Choose a positive direction and stick with it.
Recap
- ΣF = m·a: statics with the right-hand side no longer zero.
- Friction while sliding is μk·N, and N depends on all the vertical forces.
- Combine F = m·a with kinematics for times and distances, or use work-energy.
- Stopping distance grows with speed squared.
Try it
On the Do tab, finish the acceleration line of a crate-pulling function and check the worked example.