The Dependency of theta in the dot product

How the work done depends on the angle $θ$

The work $W$ done on an object acted upon by a constant force $\vec{F}$ is given by

$$W=\vec{F}·Δ\vec{R},\tag{1}$$

where $Δ\vec{R}$ is the displacement of the object. As a reminder, the dot product between any two vectors $\vec{A}$ and $\vec{B}$ is defined as

$$\vec{A}·\vec{B}≡ABcosθ,$$

where $θ$ is the angle between those two vectors. From this definition, you can see that if the two vectors $\vec{A}$ and $\vec{B}$ are perpendicular to each other, then $cos(±90°)=0$ and, thus, $\vec{A}·\vec{B}=0$.

This basically means that if an object moves from $\vec{R}(t_0)$ to $\vec{R}(t)$ with a displacement of $Δ\vec{R}$, if a constant force $\vec{F}$ was acting on that object during its displacement and if $\vec{F}$ was perpendicular to $Δ\vec{R}$, then that force would have done no work on the object. For example, the Moon revolves around the Earth in a roughly circular path and the force of gravity $\vec{F}_g$ exerted on the Moon by the Earth is always towards the center of Earth. Since this force is always acting on the Moon in a direction perpendicular to its displacement, it follows that the force $\vec{F}_g$ doesn't do any work on the Moon.

We can rewrite Equation (1) in the following way:

$$W=\vec{F}·Δ\vec{R}=FΔRcosθ=(Fcosθ)ΔR=F_RΔR,\tag{2}$$

where $F_R$ is the component of the force $\vec{F}$ which is acting in the same direction as $Δ\vec{R}$. Equation (2) tells us that it is only the component of force that is parallel to $Δ\vec{R}$ which does work on the object. Therefore, if I apply the force $\vec{F}$ to a box moving to the right as in Figure 1, the y-component of force $F_y$ will contribute zero work to the object. It is only the component of force $F_x=Fcosθ$ which is doing any work on the box. By changing the angle $θ$ at which $\vec{F}$ (while keeping the magnitude $F$ the same) is applied in a way that increases $cosθ$ (in this example, this can be accomplished by rotating $\vec{F}$ clockwise), the same force $\vec{F}$ will do more work on the object.