Mechanics

Chapter 5: Force and Motion

Evan Luo

6 min read

1. What is a Force? Why does it matter?

  • A force is simply a push or a pull that one object (the agent) exerts on another.

  • Every force has magnitude (how strong) and direction → we treat forces as vectors.

  • Two broad families:

    1. Contact forces – need physical touch (e.g., friction, tension).
    2. Long-range forces – act at a distance (e.g., gravity, magnetism).

2. Net Force & Newton's Second Law

  • Net force = the vector sum of all forces acting on an object:

    F⃗net=∑iF⃗i\vec{F}_{\text{net}} = \sum_i \vec{F}_i

  • Experiments show:

    1. Acceleration a⃗\vec{a} grows in direct proportion to F⃗net\vec{F}_{\text{net}}.
    2. For a given F⃗net\vec{F}_{\text{net}}, heavier objects (more mass mm) accelerate less.

  • Combining both ideas gives Newton's 2nd Law:

    a⃗=F⃗netm\boxed{\vec{a} = \frac{\vec{F}_{\text{net}}}{m}}

    The acceleration points in the same direction as F⃗net\vec{F}_{\text{net}}.

  • Units:

    • Mass → kilograms (kg)
    • Force → newtons (N) – one newton is 1 kg⋅m/s21\ \text{kg} \cdot \text{m/s}^2.

3. Equilibrium & Newton's First Law

  • If no net force acts (∑F⃗=0\sum \vec{F} = 0), the object is in equilibrium:

    • At rest → it stays at rest.
    • Moving at constant velocity → it keeps that velocity.
    • Written as: a⃗=0\vec{a} = 0.

  • This is Newton's 1st Law ("law of inertia").

Key takeaway: motion itself needs no force to continue—only a change in motion needs force.

4. Everyday Forces to Recognise

SymbolForce nameDirection ruleUseful equationComment
F⃗G\vec{F}_GGravityStraight down (toward Earth's centre)FG=mgF_G = mgLong-range
F⃗sp\vec{F}_{\text{sp}}SpringToward spring's relaxed (equilibrium) length—Hooke's law appears later
T⃗\vec{T}TensionAlong (and pulling) the string/rope—Same at every point in an ideal rope
n⃗\vec{n}NormalPerpendicular to the surface—"Support" force
f⃗k\vec{f}_kKinetic frictionParallel & opposite to sliding motionfk=μknf_k = \mu_k nSurface must be sliding
f⃗s\vec{f}_sStatic frictionParallel & opposite to intended motionfs≤μsnf_s \leq \mu_s nHolds an object at rest

Free-body diagram: draw the object alone, then add each force with the correct direction and label. This makes the force equation easier to write correctly.

5. From Straight-Line Motion to Rotation – Introducing Torque

  • A force can also twist an object about a pivot.

  • We measure this twisting tendency with torque τ⃗\vec{\tau}:

    τ⃗=r⃗×F⃗\boxed{\vec{\tau} = \vec{r} \times \vec{F}}

    where:

    • r⃗\vec{r} = position vector from the pivot to where the force is applied,
    • ×\times = vector (cross) product.

  • Magnitude:

    τ=F rsin⁡ϕ=F d⊥\tau = F\,r\sin\phi = F\,d_\perp
    • ϕ\phi = angle between F⃗\vec{F} and r⃗\vec{r}.
    • d⊥d_\perp = shortest (perpendicular) distance from pivot to the force's line of action.

  • Sign convention (viewing +z coming out of the page):

    • Counter-clockwise (CCW) twist → +τ+\tau
    • Clockwise (CW) twist → −τ-\tau

  • Component trick: sometimes it is faster to resolve F⃗\vec{F} into xx- and yy-parts and add their torques:

    τ=Fx dy,⊥+Fy dx,⊥\tau = F_x\,d_{y,\perp} + F_y\,d_{x,\perp}

6. Net Torque

Just like forces, torques add:

τ⃗net=∑iτ⃗i\vec{\tau}_{\text{net}} = \sum_i \vec{\tau}_i

The net torque tells you the overall rotational effect of all forces on the object.

7. Quick Practice (try before peeking!)

A. The square plate shown on the left has a pivot point O at its centre. If F1=26 NF_1 = 26\ \text{N}, F2=14 NF_2 = 14\ \text{N}, and F3=18 NF_3 = 18\ \text{N}, what is the net torque?

Square plate with forces

Using the torque formula τ=Frsin⁡ϕ\tau = Fr\sin\phi for each force:

For F1F_1 (26 N):

  • Distance from pivot to force application point: r=0.127 mr = 0.127\ \text{m}
  • Angle between force and position vector: ϕ=135∘\phi = 135^\circ
  • Torque: τ1=−(26 N)sin⁡135∘(0.127 m)=−2.33 Nm\tau_1 = -(26\ \text{N})\sin135^\circ(0.127\ \text{m}) = -2.33\ \text{Nm} (CW)

For F2F_2 (14 N):

  • Distance from pivot to force application point: r=0.127 mr = 0.127\ \text{m}
  • Angle between force and position vector: ϕ=135∘\phi = 135^\circ
  • Torque: τ2=(14 N)sin⁡135∘(0.127 m)=1.26 Nm\tau_2 = (14\ \text{N})\sin135^\circ(0.127\ \text{m}) = 1.26\ \text{Nm} (CCW)

For F3F_3 (18 N):

  • Distance from pivot to force application point: r=0.127 mr = 0.127\ \text{m}
  • Angle between force and position vector: ϕ=90∘\phi = 90^\circ
  • Torque: τ3=(18 N)sin⁡90∘(0.127 m)=2.29 Nm\tau_3 = (18\ \text{N})\sin90^\circ(0.127\ \text{m}) = 2.29\ \text{Nm} (CCW)

Net torque: τnet=τ1+τ2+τ3=−2.33 Nm+1.26 Nm+2.29 Nm=1.22 Nm\tau_{\text{net}} = \tau_1 + \tau_2 + \tau_3 = -2.33\ \text{Nm} + 1.26\ \text{Nm} + 2.29\ \text{Nm} = 1.22\ \text{Nm} (CCW)

Click to reveal

B. The pulley shown on the right has a mass of 5.0 kg5.0\ \text{kg} and a diameter of 30 cm30\ \text{cm}. The pulley has a pivot point O located half way between its center and its rim. A rope holding two 15 kg15\ \text{kg} masses wraps around the pulley. What is the net torque?

Pulley with masses

We need to calculate the torque from the pulley's weight and the two hanging masses:

For the 15 kg15\ \text{kg} mass on the left:

  • Force: F=mg=15 kg×9.8 m/s2=147 NF = mg = 15\ \text{kg} \times 9.8\ \text{m/s}^2 = 147\ \text{N}
  • Distance from pivot to force line of action: r=0.075 mr = 0.075\ \text{m}
  • Angle between force and moment arm: ϕ=90∘\phi = 90^\circ
  • Torque: τ1=(147 N)(0.075 m)sin⁡90∘=11.025 Nm\tau_1 = (147\ \text{N})(0.075\ \text{m})\sin90^\circ = 11.025\ \text{Nm} (CCW)

For the pulley's weight (5.0 kg5.0\ \text{kg}):

  • Force: F=mg=5.0 kg×9.8 m/s2=49 NF = mg = 5.0\ \text{kg} \times 9.8\ \text{m/s}^2 = 49\ \text{N}
  • Distance from pivot to center of mass: r=0.075 mr = 0.075\ \text{m}
  • Angle between force and moment arm: ϕ=90∘\phi = 90^\circ
  • Torque: τ2=−(49 N)(0.075 m)sin⁡90∘=−3.675 Nm\tau_2 = -(49\ \text{N})(0.075\ \text{m})\sin90^\circ = -3.675\ \text{Nm} (CW)

For the 15 kg15\ \text{kg} mass on the right:

  • Force: F=mg=15 kg×9.8 m/s2=147 NF = mg = 15\ \text{kg} \times 9.8\ \text{m/s}^2 = 147\ \text{N}
  • Distance from pivot to force line of action: r=0.225 mr = 0.225\ \text{m}
  • Angle between force and moment arm: ϕ=90∘\phi = 90^\circ
  • Torque: τ3=−(147 N)(0.225 m)sin⁡90∘=−33.075 Nm\tau_3 = -(147\ \text{N})(0.225\ \text{m})\sin90^\circ = -33.075\ \text{Nm} (CW)

Net torque: τnet=τ1+τ2+τ3=11.025 Nm+(−3.675 Nm)+(−33.075 Nm)=−25.725 Nm\tau_{\text{net}} = \tau_1 + \tau_2 + \tau_3 = 11.025\ \text{Nm} + (-3.675\ \text{Nm}) + (-33.075\ \text{Nm}) = -25.725\ \text{Nm} (CW)

Click to reveal

8. Key Points to Remember

  1. Draw a free-body diagram first—it clarifies both linear and rotational problems.
  2. For straight-line motion, relate F⃗net\vec{F}_{\text{net}} to a⃗\vec{a} via Newton's 2nd law.
  3. For rotation, replace forces with torques about the chosen pivot, then sum them.
  4. Zero net force   ⟹  \implies no linear acceleration; zero net torque   ⟹  \implies no angular acceleration.
  5. Sign conventions (for torque directions) keep results consistent—pick one and stick to it.

Feel free to ask for more examples, deeper derivations, or practice problems!

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