# If work is zero, when is force nonzero?

Feb 18, 2017

Normally, when we talk about force and displacement, we refer to the definition of work in a one-dimensional representation:

$w = - \vec{F} \Delta \vec{x}$

where work is negative with respect to the worker who is say, pushing a box.

One can specify that this force is parallel to the displacement, i.e. $\vec{F} = {\vec{F}}_{\text{||}} = \vec{F} \cos \theta$. When $\theta = {0}^{\circ}$, the work is pointing exactly in the direction of the displacement.

On the other hand, if we imagine a force entirely perpendicular to the desired displacement, there is no component of the force that causes the displacement.

${\vec{F}}_{\bot} = \vec{F} \sin \left({0}^{\circ}\right) = \vec{F} \cos \left({90}^{\circ}\right)$

(since $\sin \left(x + \frac{\pi}{2}\right) = \cos \left(x\right)$.)

When $\theta = {90}^{\circ}$, one pushes perpendicular to the desired displacement, but $\sin {0}^{\circ} = \cos {90}^{\circ} = 0$.

So, I don't think such a force is nonzero.

Feb 18, 2017

$\implies \theta = {90}^{o}$ so $W = 0$