Force as a Vector Assigned to a Position
A force has both a size and a direction. A force field represents how that force varies from place to place: it assigns a force vector to each position in a region. It can therefore describe a position-dependent push or pull without requiring us to follow the motion of an object.
For example, a field might assign a force pointing toward a fixed center, or a force pointing in the same direction everywhere. The central question here is: what force vector does the field assign at this particular position?
The position is the input; the force vector is the output. If \(\mathbf r=(x,y)\) is a position in the field’s region, then \(\mathbf F(\mathbf r)\) is the force assigned there. The components \(F_1\) and \(F_2\) give the force’s horizontal and vertical components. A three-dimensional field has a third component as well.
In physical applications, position coordinates may be measured in metres and force components in newtons. The field formula must be interpreted with units that make its output a force. In the examples below, coefficients and coordinates are stated with units so that each evaluated force is in newtons.
Reading Magnitude and Direction
A field value is a vector, not just a number. Its magnitude gives the strength of the force, while its direction tells which way it points. In two dimensions, if \(\mathbf F(x,y)=\langle P,Q\rangle\), then its magnitude is \(\sqrt{P^2+Q^2}\). In three dimensions, for \(\langle P,Q,R\rangle\), it is \(\sqrt{P^2+Q^2+R^2}\).
When the force is nonzero, dividing it by its magnitude gives a unit vector in the force’s direction. A zero vector has magnitude zero, but it has no direction: dividing it by its magnitude would require division by zero.
Worked Example: An Inward Force at Two Positions
Worked Example: An Inward Force at Two Positions
Consider the field \(\mathbf F(x,y)=\langle -(2\,\mathrm{N/m})x,-(2\,\mathrm{N/m})y\rangle\). Find the force at \((3\,\mathrm m,4\,\mathrm m)\), its strength and direction, and then compare it with the force at \((0\,\mathrm m,5\,\mathrm m)\).
At the first position, substitute \(x=3\,\mathrm m\) and \(y=4\,\mathrm m\) into both components:
The magnitude is
Since the force is nonzero, its unit direction vector is
This direction points toward the origin. To compare, evaluate the same field at the second position:
Its magnitude is \(\sqrt{0^2+(-10)^2}\,\mathrm N=10\,\mathrm N\), and its direction is \(\langle0,-1\rangle\). Thus, the two forces have equal strength but point in different directions. Substituting each position into the original formula reproduces the stated components, and each magnitude calculation gives \(10\,\mathrm N\).
Why Magnitude and Direction Separate the Force
For a nonzero vector \(\mathbf v\), its magnitude scales the unit vector in its direction to recover the original vector. This fact explains how strength and direction together describe the field value completely.
So \(\mathbf u\) is a unit vector. Moreover, \(\|\mathbf v\|\mathbf u=\|\mathbf v\|(\mathbf v/\|\mathbf v\|)=\mathbf v\). Applying this to a force vector shows that its magnitude and unit direction recover its components. The condition \(\mathbf v\ne\mathbf 0\) matters: for the zero vector, the proposed division is undefined.
Worked Example: A Force Whose Direction Changes
Worked Example: A Force Whose Direction Changes
Let \(\mathbf F(x,y)=\langle(2\,\mathrm{N/m})y,(1\,\mathrm{N/m})(x-y)\rangle\). Evaluate the force at \(A=(1\,\mathrm m,3\,\mathrm m)\) and \(B=(2\,\mathrm m,0\,\mathrm m)\), then compare strength and direction.
At \(A\), the first component is \((2\,\mathrm{N/m})(3\,\mathrm m)=6\,\mathrm N\). The second component is \((1\,\mathrm{N/m})(1\,\mathrm m-3\,\mathrm m)=-2\,\mathrm N\). Therefore,
Its magnitude is \(\sqrt{6^2+(-2)^2}\,\mathrm N=\sqrt{40}\,\mathrm N=2\sqrt{10}\,\mathrm N\). Its unit direction vector is \(\langle6,-2\rangle/(2\sqrt{10})=\langle3/\sqrt{10},-1/\sqrt{10}\rangle\).
At \(B\), the first component is \((2\,\mathrm{N/m})(0\,\mathrm m)=0\,\mathrm N\). The second is \((1\,\mathrm{N/m})(2\,\mathrm m-0\,\mathrm m)=2\,\mathrm N\). Hence,
This force has magnitude \(2\,\mathrm N\) and unit direction \(\langle0,1\rangle\). The forces differ in both strength and direction. Checking the original components at each point gives \(\langle6,-2\rangle\) at \(A\) and \(\langle0,2\rangle\) at \(B\), as calculated.
Zero Force and Undefined Direction
A field can assign the zero vector at a position. At that point, the force has no strength and no defined direction. This is different from a force with small but nonzero magnitude: every nonzero vector has a direction, even when it is weak.
Worked Example: Finding a Position with Zero Force
Consider \(\mathbf F(x,y)=\langle(1\,\mathrm{N/m})x,-(1\,\mathrm{N/m})y\rangle\). At the origin,
The magnitude is \(\sqrt{0^2+0^2}\,\mathrm N=0\,\mathrm N\), so there is no direction vector to calculate. To contrast, at \((2\,\mathrm m,1\,\mathrm m)\), substitution gives \(\mathbf F=\langle2\,\mathrm N,-1\,\mathrm N\rangle\), with magnitude \(\sqrt{2^2+(-1)^2}\,\mathrm N=\sqrt5\,\mathrm N\). Its direction is \(\langle2/\sqrt5,-1/\sqrt5\rangle\), whose magnitude is \(\sqrt{4/5+1/5}=1\). Substitution confirms both field values, and the zero at the origin is not a point where a direction can be assigned.
A Reliable Evaluation Process
For a formula-defined force field, keep the position, the vector components, and the strength distinct. A reliable calculation follows these steps:
Record the coordinates at which the force is requested and check that the position belongs to the field’s domain.
Compute the components separately, keeping their order consistent with the field’s definition.
Use its components for direction and calculate its magnitude for strength. Normalize only if the vector is nonzero.
These steps help prevent a common confusion: evaluating a force field does not mean evaluating its magnitude. The field returns a vector; taking the magnitude is a separate calculation.
Common Mistake: Treating a Component as the Whole Force
Suppose a force at a point is \(\langle6\,\mathrm N,-2\,\mathrm N\rangle\). A mistaken calculation might report its strength as \(6\,\mathrm N\), using only the first component. That ignores the vertical component. The correct magnitude is
The first component alone does not give the force’s strength; it describes only its horizontal part. Another error is to normalize a zero force. For \(\langle0,0\rangle\), division by the magnitude would mean division by zero, so the force has no unit direction vector.
Why Position-Dependent Force Models Matter
A force field gives a compact way to compare forces at different locations without confusing a location with the force acting there. For example, two positions can have equal force magnitudes but different directions, as in the inward field above. Other fields can change both magnitude and direction, or assign zero force at particular points.
This viewpoint is useful whenever the force acting on an object depends on where the object is. The field describes the force available at each position; evaluating it answers the local question, “What force is assigned here?” It does not by itself describe the object’s path or motion. Those questions require additional information and models.
Check Your Understanding
Answer each question using the distinction between position, force vector, magnitude, and direction.
- For \(\mathbf F(x,y)=\langle x,-y\rangle\), what is the force vector at \((2,3)\)?
- What is the magnitude of \(\langle-5,12\rangle\), and what is its unit direction vector?
- Can a force vector have magnitude zero and still have a direction? Explain.
- If forces at two positions have equal magnitudes, must their vectors be equal? Why or why not?
- Why is it incorrect to use just one component of a two-dimensional force as its strength?