# 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.