# Work done by vertical force

We know that work is defined by the dot product of force and displacement,which mathematically means $$W=Fs\cos \theta$$ in scaler. But let's suppose an object is moving in $$x$$ axis. It's vertical velocity $$v_y=0$$. If a force is applied in $$y$$ direction which is $$F$$,wouldn't $$v_y$$ increase as a result of acceleration $$a_y$$? If $$v_y$$ increases,that means the object will also start moving in the $$y$$ direction. So we have a motion in $$y$$ axis which is along the direction of force. So even though $$0=W=Fs\cos 90$$ makes sense mathematically,how can $$s$$ even be $$0$$ in light of the above explanation? I would be very grateful if someone kindly clears the concept.

• If you have a force in y direction and a movement in y direction, you have work not 0 , your angle $\phi$ is no longer 0. Nov 5, 2022 at 11:58
• You explained why the velocity in the y direction is not zero. Why do you think it is 0?
– Matt
Nov 5, 2022 at 11:58

You have the initial conditions

$$$$\textbf{v} = \begin{pmatrix} v_{x} \\ 0 \\ 0 \end{pmatrix}$$$$

You then apply a force,

$$$$\textbf{F} = \begin{pmatrix} 0 \\ F_{y} \\ 0 \end{pmatrix}$$$$

for some time $$T$$. During that time, the body will accelerate in the $$y$$ direction according to $$a_{y}(t) = F_{y}(t)/m$$. While the force is applied it will do work

$$$$\begin{split} W &= \int \textbf{F}\cdot\text{d}\textbf{s} \\ &= \int_{0}^{s_{y}} F_{y} \text{d}s^{\prime}_{y} \end{split}$$$$

where $$s_{y}$$ is the total distance the object moves in the $$y$$ direction while being acted on by the force $$F_{y}$$. To perform this integration, we would have to perform a change in variables from $$a \rightarrow t$$ if we only know the temporal information regarding the force. As you can see, because of the dot product in the definition of work done the motion in the $$x$$ direction will have no bearing on the final result.

• where $s_y$ is the total distance the object moves while being acted on by the force $F_y$, I think you should clear that you mean "... is the total distance in y direction" in the post Nov 5, 2022 at 13:39
• I've now clarified this point Nov 6, 2022 at 18:10

Think of a train cart moving in its track.

• Push along with the track and force is parallel to displacement. Angle is zero and we get the largest possible work: $$W=\mathbf F\cdot \mathbf d=Fd\cdot\cos(0)=Fd.$$

• Push perpendicular to the track and force is perpendicular to displacement. But the cart cannot move this way, so a normal force from the train rails will appear to balance out the force. We thus still have displacement only along the track as before. Angle is $$90^\circ$$ and we see that this force does no work: $$W=\mathbf F\cdot \mathbf d=Fd\cdot\cos(90^\circ)=Fd\cdot 0=0.$$ The fact that the cart is moving and having displacement is not due to this force $$\mathbf F$$.

Pushing upwards is the same scenario: Also here, force and displacement are perpendicular. This time the counteracting force is gravity.

Maybe you push hard enough to overcome gravity so the cart is moved upwards and lifted out of its tracks. Then you are turning its displacement vector $$\mathbf d$$, so that it now no longer is pernpendicular to the force, and thus you will get a work value done by this force.

But if the force is not strong enough to overcome gravity and the cart thus does not move upwards, then the cart will just keep its usual displacement and the force is doing no work to help. This displacement is then not due to this force.

F is along y. So, along y, F.ds=F.dy is not zero. So, F is doing work in that direction of displacement. But, along x, F.ds=F.dx=0. So, F is not doing work in that direction of displacement.

In the situation you describe, the angle $$\theta$$ is originally 90 degrees since the object is only moving along the x-axis. However, by applying a force in the y-direction, you instantaneously change the motion of the object to be some superposition of x- and y- displacements. Thus, your $$\theta$$ will no longer be 90 degrees after applying the force.

If we recreate your situation but instead our force in the y-direction is not great enough to cause a net y-acceleration => y-velocity => y-displacement (instantaneously) then indeed no work is done by this force.

Note: in introductory mechanics we assume forces act instantaneously, but of course we know this is not possible in more modern physical theories (i.e. relativity)!