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A Mixed Workshop on Forces, Work, and Projections

Learn to combine force vectors, calculate work and angles, and resolve a force into components along and perpendicular to displacement.

Beginner 10 min read

What You'll Learn

  • Combine multiple constant forces into a resultant force before calculating total work.
  • Calculate work from force and displacement components using the dot product.
  • Find the angle between nonzero force and displacement vectors.
  • Decompose a force into components parallel and perpendicular to a chosen direction.
  • Check projection and work calculations using units, dot products, and vector addition.

Putting Forces, Work, and Projections Together

A force problem can ask for more than one quantity. You may need to combine several force vectors, find the work done over a displacement, calculate the angle between the resultant force and the displacement, and identify which parts of the force act along or across the direction of travel.

The calculations fit together through the dot product. The central choices are to use the displacement vector as the direction of motion, keep vector and scalar quantities distinct, and check that the resulting units and signs make sense.

Key idea: For a constant force \(\mathbf{F}\) acting through displacement \(\mathbf{d}\), the work is \(W=\mathbf{F}\cdot\mathbf{d}\). The component of \(\mathbf{F}\) parallel to \(\mathbf{d}\) determines the work; its perpendicular component contributes zero.

The Essential Relationships

For vectors \(\mathbf{F}=(F_x,F_y,F_z)\) and \(\mathbf{d}=(d_x,d_y,d_z)\), their dot product is \(F_xd_x+F_yd_y+F_zd_z\). If the force is measured in newtons and displacement in metres, the dot product has units of joules.

$$ W=\mathbf{F}\cdot\mathbf{d} =F_xd_x+F_yd_y+F_zd_z =|\mathbf{F}|\,|\mathbf{d}|\cos\theta $$

Here \(\theta\) is the angle between the two vectors. The angle formula applies when both vectors are nonzero; equivalently, \(\cos\theta=(\mathbf{F}\cdot\mathbf{d})/(|\mathbf{F}|\,|\mathbf{d}|)\). A positive dot product gives an acute angle and positive work, a zero dot product gives a right angle and zero work, and a negative dot product gives an obtuse angle and negative work.

For a nonzero direction vector \(\mathbf{d}\), the unit vector in its direction is \(\mathbf{u}=\mathbf{d}/|\mathbf{d}|\). The scalar component of \(\mathbf{F}\) along that direction is \(\mathbf{F}\cdot\mathbf{u}\). The corresponding vector component and the perpendicular remainder are:

$$ \mathbf{F}_{\parallel} =(\mathbf{F}\cdot\mathbf{u})\mathbf{u} =\frac{\mathbf{F}\cdot\mathbf{d}}{\mathbf{d}\cdot\mathbf{d}}\mathbf{d}, \qquad \mathbf{F}_{\perp}=\mathbf{F}-\mathbf{F}_{\parallel}. $$

The scalar component can be negative: that means the parallel part points opposite to the chosen direction. The vector \(\mathbf{F}_{\parallel}\) is not the same thing as the scalar component. In a force problem, the scalar component has units of newtons, while the vector component is a force vector in newtons.

Why the Parallel Component Determines Work

The projection formulas are useful because they explain which part of a force does work along a displacement. Suppose \(\mathbf{d}\ne\mathbf{0}\) and split \(\mathbf{F}\) into \(\mathbf{F}_{\parallel}+\mathbf{F}_{\perp}\). By construction, the perpendicular remainder has zero dot product with \(\mathbf{d}\). Therefore:

$$ \begin{aligned} \mathbf{F}\cdot\mathbf{d} &=(\mathbf{F}_{\parallel}+\mathbf{F}_{\perp})\cdot\mathbf{d}\\ &=\mathbf{F}_{\parallel}\cdot\mathbf{d} +\mathbf{F}_{\perp}\cdot\mathbf{d}\\ &=\mathbf{F}_{\parallel}\cdot\mathbf{d}. \end{aligned} $$

More explicitly, \(\mathbf{F}_{\parallel}=((\mathbf{F}\cdot\mathbf{d})/(\mathbf{d}\cdot\mathbf{d}))\mathbf{d}\), so its dot product with \(\mathbf{d}\) equals \((\mathbf{F}\cdot\mathbf{d})\). The perpendicular term vanishes because \(\mathbf{F}_{\perp}\cdot\mathbf{d}=0\). This proves that work comes entirely from the force component parallel to displacement. The perpendicular force may change the direction of motion, but in this constant-force work calculation it contributes no work along the given displacement.

Worked Examples

Worked Example: Work, Angle, and Force Components

A constant force \(\mathbf{F}=(8,6,0)\) N acts through displacement \(\mathbf{d}=(3,4,0)\) m. Find the work, the angle between force and displacement, and the parallel and perpendicular components of the force relative to the displacement.

1. Calculate the work. Multiply corresponding components and add:

$$ W=\mathbf{F}\cdot\mathbf{d} =(8)(3)+(6)(4)+(0)(0) =24+24+0 =48\text{ J}. $$

2. Calculate the angle. The magnitudes are \(|\mathbf{F}|=\sqrt{8^2+6^2+0^2}=\sqrt{100}=10\) N and \(|\mathbf{d}|=\sqrt{3^2+4^2+0^2}=\sqrt{25}=5\) m. Hence \(\cos\theta=48/(10\cdot5)=0.96\), so \(\theta=\cos^{-1}(0.96)\approx16.26^\circ\) to two decimal places.

3. Find the parallel component. Since \(\mathbf{d}\cdot\mathbf{d}=25\),

$$ \mathbf{F}_{\parallel} =\frac{48}{25}(3,4,0) =\left(\frac{144}{25},\frac{192}{25},0\right) =(5.76,7.68,0)\text{ N}. $$

4. Find the perpendicular component.

$$ \mathbf{F}_{\perp} =\mathbf{F}-\mathbf{F}_{\parallel} =(8-5.76,\,6-7.68,\,0) =(2.24,-1.68,0)\text{ N}. $$

Check. The components add to the original force: \((5.76,7.68,0)+(2.24,-1.68,0)=(8,6,0)\). Also, \(\mathbf{F}_{\perp}\cdot\mathbf{d}=(2.24)(3)+(-1.68)(4)=6.72-6.72=0\). Finally, \(\mathbf{F}_{\parallel}\cdot\mathbf{d}=(5.76)(3)+(7.68)(4)=17.28+30.72=48\) J, agreeing with the work.

Worked Example: Negative Work and an Opposing Component

A force \(\mathbf{F}=(-4,1,0)\) N acts through displacement \(\mathbf{d}=(1,2,2)\) m. Find the work, the angle between the vectors, and the decomposition of the force relative to the displacement.

1. Calculate the work.

$$ W=\mathbf{F}\cdot\mathbf{d} =(-4)(1)+(1)(2)+(0)(2) =-4+2+0 =-2\text{ J}. $$

2. Calculate the angle. The force magnitude is \(|\mathbf{F}|=\sqrt{(-4)^2+1^2+0^2}=\sqrt{17}\) N. The displacement magnitude is \(|\mathbf{d}|=\sqrt{1^2+2^2+2^2}=\sqrt{9}=3\) m. Thus \(\cos\theta=-2/(3\sqrt{17})\), and \(\theta=\cos^{-1}(-2/(3\sqrt{17}))\approx99.30^\circ\) to two decimal places. The obtuse angle is consistent with negative work.

3. Resolve the force. Here \(\mathbf{d}\cdot\mathbf{d}=9\), so

$$ \mathbf{F}_{\parallel} =\frac{-2}{9}(1,2,2) =\left(-\frac{2}{9},-\frac{4}{9},-\frac{4}{9}\right)\text{ N}. $$

The negative coefficient shows that this component points opposite to the displacement. Subtract it from the original force:

$$ \mathbf{F}_{\perp} =(-4,1,0)-\left(-\frac{2}{9},-\frac{4}{9},-\frac{4}{9}\right) =\left(-\frac{34}{9},\frac{13}{9},\frac{4}{9}\right)\text{ N}. $$

Check. Adding the components recovers \((-4,1,0)\) N. The perpendicular check gives \(\mathbf{F}_{\perp}\cdot\mathbf{d}=-34/9+26/9+8/9=0\). The parallel component gives \(\mathbf{F}_{\parallel}\cdot\mathbf{d}=-2/9-8/9-8/9=-2\) J, as required.

Worked Example: Combining Forces Before Resolving Them

Two constant forces act on an object as it moves through displacement \(\mathbf{d}=(2,1,2)\) m. The forces are \(\mathbf{F}_1=(5,0,12)\) N and \(\mathbf{F}_2=(-1,2,0)\) N. Find the total work, the angle between the resultant force and displacement, and the resultant force's parallel and perpendicular components.

1. Find the resultant force. Add corresponding components:

$$ \mathbf{F}_{\text{net}} =\mathbf{F}_1+\mathbf{F}_2 =(5-1,\,0+2,\,12+0) =(4,2,12)\text{ N}. $$

2. Calculate total work. The work of each force is \(\mathbf{F}_1\cdot\mathbf{d}=(5)(2)+(0)(1)+(12)(2)=10+0+24=34\) J and \(\mathbf{F}_2\cdot\mathbf{d}=(-1)(2)+(2)(1)+(0)(2)=-2+2+0=0\) J. Thus total work is \(34+0=34\) J. Directly, \(\mathbf{F}_{\text{net}}\cdot\mathbf{d}=(4)(2)+(2)(1)+(12)(2)=8+2+24=34\) J.

3. Find the angle. The magnitudes are \(|\mathbf{F}_{\text{net}}|=\sqrt{4^2+2^2+12^2}=\sqrt{164}=2\sqrt{41}\) N and \(|\mathbf{d}|=\sqrt{2^2+1^2+2^2}=3\) m. Therefore \(\cos\theta=34/(6\sqrt{41})=17/(3\sqrt{41})\), giving \(\theta\approx27.78^\circ\) to two decimal places.

4. Resolve the resultant force. Since \(\mathbf{d}\cdot\mathbf{d}=9\),

$$ \mathbf{F}_{\parallel} =\frac{34}{9}(2,1,2) =\left(\frac{68}{9},\frac{34}{9},\frac{68}{9}\right)\text{ N}. $$

Subtracting from the resultant gives

$$ \mathbf{F}_{\perp} =(4,2,12)-\left(\frac{68}{9},\frac{34}{9},\frac{68}{9}\right) =\left(-\frac{32}{9},-\frac{16}{9},\frac{40}{9}\right)\text{ N}. $$

Check. The two components sum to \((4,2,12)\) N. Their perpendicular condition is \(\mathbf{F}_{\perp}\cdot\mathbf{d}=-64/9-16/9+80/9=0\). Also, \(\mathbf{F}_{\parallel}\cdot\mathbf{d}=136/9+34/9+136/9=306/9=34\) J, matching the total work.

Common Pitfall: Projecting Onto a Nonunit Direction

A frequent mistake is to calculate the parallel component as \((\mathbf{F}\cdot\mathbf{d})\mathbf{d}\), omitting the denominator \(\mathbf{d}\cdot\mathbf{d}\). That expression generally has the wrong magnitude. For example, in the first worked example, it would give \(48(3,4,0)=(144,192,0)\) N. Its dot product with \(\mathbf{d}\) is \(432+768=1200\) J, not \(48\) J, and the proposed vector is not a component of the original force.

The correction is to divide by the squared length of the direction vector: \(\mathbf{F}_{\parallel}=((\mathbf{F}\cdot\mathbf{d})/(\mathbf{d}\cdot\mathbf{d}))\mathbf{d}\). In that example, \(\mathbf{d}\cdot\mathbf{d}=25\), so the correct projection is \((48/25)(3,4,0)=(5.76,7.68,0)\) N. Alternatively, first make a unit vector \(\mathbf{u}=\mathbf{d}/|\mathbf{d}|\), then use \((\mathbf{F}\cdot\mathbf{u})\mathbf{u}\).

A Reliable Order of Work

For a problem combining forces, work, angle, and projections, organize the calculation before substituting numbers:

1
Identify the displacement.
Use its direction as the reference direction for work and force components. The displacement must be nonzero to define this direction.
2
Combine forces if needed.
Add the force vectors component by component to obtain the net force.
3
Calculate work and angle.
Use the dot product for work; use the normalized dot product for the angle, provided both vectors are nonzero.
4
Resolve and check.
Compute the parallel projection, subtract it to get the perpendicular part, then check that the parts add back to the force and that the perpendicular part has zero dot product with displacement.

These checks catch different errors: vector addition tests whether the decomposition reconstructs the force, perpendicularity tests the direction of the remainder, and the work calculation confirms that the parallel component gives the same work as the full force.

Check Your Understanding

For each question, keep track of whether the answer is a scalar or a vector, and include units when the quantities represent forces, distances, or work.

  1. If a nonzero force has a negative dot product with a nonzero displacement, what does that tell you about the work and the angle between them?
  2. Why must the formula for projection onto \(\mathbf{d}\) include the denominator \(\mathbf{d}\cdot\mathbf{d}\)?
  3. A force has dot product zero with a nonzero displacement. What is its work, and what is its parallel component?
  4. Two forces act through the same displacement. Explain why their total work equals the work of their resultant force.
  5. What two checks can verify a proposed parallel-perpendicular decomposition of a force?

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