4.1 Force

Definition

A force is a push or pull exerted on an object resulting from its interaction with another object. A force can cause an object to accelerate by changing its speed, altering its direction of velocity, or causing both. Force is a vector quantity, possessing both magnitude and direction, and follows the rules of vector addition.

SI Unit

The SI unit of force is the newton (N)(\text{N}). One newton is defined as the amount of net force required to accelerate a mass of one kilogram at a rate of one meter per second squared: 1 N=1 kg⋅m/s21 \text{ N} = 1 \text{ kg}\cdot\text{m/s}^2

Contact and Field Forces

Forces in classical mechanics are categorized into two fundamental classes based on how objects interact:

1. Contact Forces

Contact forces arise from direct physical contact between the interacting surfaces or bodies. Common examples include:

  • Normal force exerted perpendicular to a supporting surface
  • Tension transmitted through a taut string, cord, or cable
  • Frictional force acting parallel to contacting surfaces to oppose sliding motion
  • Applied forces exerted by direct pushing, pulling, or muscle action

2. Field Forces

Field forces act across empty space without requiring physical contact between bodies. An object creates a physical field throughout surrounding space that exerts a force on any other responsive body placed within that region. Common examples include:

  • Gravitational force between masses (such as Earth pulling downward on an object)
  • Electrostatic force between stationary electric charges
  • Magnetic force between magnetic poles or moving electric charges
Figure 4.1.1: Classification of forces showing representative contact interactions and non-contact field interactions.

Force as a Vector Quantity

Because force is a vector, a complete description of any force requires specifying its magnitude, direction, and point of application. A force F→\vec{F} acting in a two-dimensional Cartesian plane can be resolved into perpendicular components:

F→=Fxi^+Fyj^\vec{F} = F_x \hat{i} + F_y \hat{j}

Where:

  • FxF_x is the component of the force along the x-axis
  • FyF_y is the component of the force along the y-axis
  • i^\hat{i} and j^\hat{j} are standard unit vectors pointing along the positive x- and y-axes, respectively

The magnitude F of the force is given by:

F=Fx2+Fy2F = \sqrt{F_x^2 + F_y^2}

The angle \theta that the force vector makes relative to the positive x-axis is determined by:

tan⁡θ=FyFx\tan\theta = \frac{F_y}{F_x}

Figure 4.1.2: Resolution of an applied force vector into horizontal and vertical rectangular components.

Net Force and the Principle of Superposition

When multiple forces act simultaneously on a single body, their combined physical effect is identical to that of a single equivalent force called the net force (or resultant force) F→net\vec{F}{\text{net}}:

F→net=∑F→=F→1+F→2+⋯+F→n\vec{F}{\text{net}} = \sum \vec{F} = \vec{F}_1 + \vec{F}_2 + \dots + \vec{F}_n

The components of the net force along the coordinate axes are calculated by summing the corresponding components of all applied forces:

Fnet,x=∑Fx=F1x+F2x+⋯+FnxF_{\text{net},x} = \sum F_x = F_{1x} + F_{2x} + \dots + F_{nx}

Fnet,y=∑Fy=F1y+F2y+⋯+FnyF_{\text{net},y} = \sum F_y = F_{1y} + F_{2y} + \dots + F_{ny}

The overall magnitude of the net force is:

Fnet=(∑Fx)2+(∑Fy)2F_{\text{net}} = \sqrt{\left(\sum F_x\right)^2 + \left(\sum F_y\right)^2}

The direction \theta of the resultant force vector relative to the positive x-axis is:

tan⁡θ=∑Fy∑Fx\tan\theta = \frac{\sum F_y}{\sum F_x}

If the vector sum of all forces equals zero (∑F→=0)(\sum \vec{F} = 0), the forces are balanced, and the object maintains a constant state of motion. If ∑F→≠0\sum \vec{F} \neq 0, the forces are unbalanced, producing an acceleration in the direction of the net force vector.

Figure 4.1.3: Concurrent forces acting on an object combined into a single resultant net force vector using vector addition.

Example 1: Net Force Along a Straight Line

Two horizontal forces act concurrently on a crate resting on a smooth floor. Force F→1\vec{F}_1 has a magnitude of45 N 45 \text{ N} and acts toward the right (+x). Force F→2\vec{F}_2 has a magnitude of 20 N20 \text{ N} and acts toward the left (-x). Find the magnitude and direction of the net force acting on the crate.

Figure 4.1.4: Collinear horizontal forces acting in opposite directions on a resting crate.

Calculation:

Assigning positive to the right (+x) and negative to the left (-x):

F1=+45 NF_1 = +45 \text{ N}

F2=−20 NF_2 = -20 \text{ N}

Summing the collinear components yields:

Fnet,x=F1+F2F_{\text{net},x} = F_1 + F_2

Fnet,x=45+(−20)=+25 NF_{\text{net},x} = 45 + (-20) = +25 \text{ N}

The net force acting on the crate has a magnitude of 25 N25 \text{ N} and is directed to the right.

Example 2: Net Force in Two Dimensions

Two ropes pull simultaneously on an eye-bolt anchored to a wall. Rope A exerts a force F→A\vec{F}_A of magnitude 60.0 N60.0 \text{ N} directed at an angle of 30.0∘30.0^\circ above the horizontal (+x). Rope B exerts a force F→B\vec{F}_B of magnitude 80.0 N80.0 \text{ N} directed horizontally along the positive x-axis. Determine the magnitude and direction of the resultant force acting on the eye-bolt.

Figure 4.1.5: Vector addition diagram showing two pulling forces acting at an angle on an eye-bolt.

Vector Components:

For Rope A:

FAx=FAcos⁡(30.0∘)=(60.0)(0.8660)≈51.96 NF_{Ax} = F_A \cos(30.0^\circ) = (60.0)(0.8660) \approx 51.96 \text{ N}

FAy=FAsin⁡(30.0∘)=(60.0)(0.5000)=30.00 NF_{Ay} = F_A \sin(30.0^\circ) = (60.0)(0.5000) = 30.00 \text{ N}

For Rope B:

FBx=80.0 NF_{Bx} = 80.0 \text{ N}

FBy=0 NF_{By} = 0 \text{ N}

Net Force Components:

Summing along the x-axis:

Fnet,x=FAx+FBx=51.96+80.0=131.96 NF_{\text{net},x} = F_{Ax} + F_{Bx} = 51.96 + 80.0 = 131.96 \text{ N}

Summing along the y-axis:

Fnet,y=FAy+FBy=30.00+0=30.00 NF_{\text{net},y} = F_{Ay} + F_{By} = 30.00 + 0 = 30.00 \text{ N}

Magnitude of the Resultant Force:

Fnet=(Fnet,x)2+(Fnet,y)2F_{\text{net}} = \sqrt{(F_{\text{net},x})^2 + (F_{\text{net},y})^2}

Fnet=(131.96)2+(30.00)2=17413.4+900=18313.4≈135.3 NF_{\text{net}} = \sqrt{(131.96)^2 + (30.00)^2} = \sqrt{17413.4 + 900} = \sqrt{18313.4} \approx 135.3 \text{ N}

Direction of the Resultant Force:

tan⁡θ=Fnet,yFnet,x=30.00131.96≈0.2273\tan\theta = \frac{F_{\text{net},y}}{F_{\text{net},x}} = \frac{30.00}{131.96} \approx 0.2273

θ=arctan⁡(0.2273)≈12.8∘\theta = \arctan(0.2273) \approx 12.8^\circ

The net force acting on the eye-bolt has a magnitude of 135.3 N135.3 \text{ N} and acts at an angle of 12.8∘12.8^\circ above the horizontal.

Key Points

  • A force is a vector push or pull resulting from an interaction between two bodies.
  • The SI unit of force is the newton (N)(\text{N}), equivalent to 1 kg⋅m/s21 \text{ kg}\cdot\text{m/s}^2.
  • Contact forces require physical contact between objects; field forces act over distance through gravitational, electric, or magnetic fields.
  • Forces obey the principle of superposition, combining by vector addition into a single net force F→net=∑F→\vec{F}_{\text{net}} = \sum \vec{F}.
  • Balanced forces produce a net force of zero (∑F→=0)(\sum \vec{F} = 0), whereas unbalanced forces cause acceleration.