# Why the discrepancy between total work and the sum of individual works?

The work done on a body is equal to the some of works done by all the forces acting on the body.

Suppose forces $F_1$ and $F_2$ are acting on a body. The component of displacement of the body in $F_1$'s direction is $d_1$, and in $F_2$'s direction is $d_2$. So, the work done is $$W = F_1 d_1 + F_2 d_2$$

But the same work can be calculated by the dot product of resultant force with resultant displacement. So, $$W = (F_1 + F_2) \cdot (d_1 + d_2) = F_1 d_1 + F_2 d_2 + F_1 \cdot d_2 + F_2 \cdot d_1$$ These are two different expressions of the same work and are equal only if $F_1 \cdot d_2 + F_2 \cdot d_1 = 0$ which is true only when $F_1$ and $F_2$ are perpendicular.

It is clearly stated in my book that the work done on a body is equal to the sum of the works done by the individual forces acting on the body. And this is true even when the forces act simultaneously. I've used this fact in solving problems in which gravity and friction simultaneously act on the body. In those problems, I don't calculate the dot product of the resultant of gravity and friction with the resultant displacement. Instead, I just add the works done by friction and gravity to calculate the change in kinetic energy. Why the discrepancy?

• Could you please write your formulas using $\LaTeX$? – Spirine Dec 23 '16 at 8:40
• @Sprine: I've no idea what LATEX is. And, I'm also looking for some way to change the paragraph. – Dove Dec 23 '16 at 8:59
• We write formulas in this site with MathJax not $\LaTeX$. They are not exactly the same thing. – Frobenius Dec 23 '16 at 9:11
• @Frobenius: What is MathJax and how do I use it? – Dove Dec 23 '16 at 9:21
• MathJax is a restricted version of $\LaTeX$ for use in sites as Physics SE. So, you must learn $\LaTeX$ first. The software is free to download. There are many help textbooks free to download. You could proceed step by step and you need time, you can't learn all this stuff in one day or week. Note that if you see in this site a formula then right-clicking and selecting -->"Show Math As" -->"TeX commands" you can also see the TeX code for this formula. I suggest to be a user in TeX-LaTeX Stack Exchange where you can post questions about LaTeX (but not MathJax). – Frobenius Dec 23 '16 at 10:03

Obviously work is force * distance and assuming a case in which $F_1$ is not perpendicular to $F_2$ (so $d_1$ is not perpendicular to $d_2$)

then

$F_1$ is not just moving through $d_1$ there is a component of $F_1$ that is moving through $d_2$. Also a component of $F_2$ that is moving through $d_1$.

So now in what I realise is a horrifying reversal no politician would be permitted (sorry again), I am suggesting that your first equation is not complete, and the second equation has it.

• I think there was a mistake in the question. I thought over it for sometime and I think the first expression for work is the correct one. I'm writing my thoughts in my answer. Please tell me if I'm correct – Dove Dec 24 '16 at 5:34

This answer misunderstood the problem - see my other answer.

There isn't a discrepancy. For dot product think cosine, or projection of the force into the direction of motion prior to multiplication by it. The second expansion you made is wrong because

$$(F_1+F_2)\cdot(d_1+d_2)=|F_1+F_2||d_1+d_2|cos\theta$$

This must always have a magnitude less than or equal to $|F_1+F_2||d_1+d_2|$

Not sure what went wrong - I think you may have tried to apply the standard quadratic expansion that works for multiplication, but does not hold for the dot product operator because that operator has a property of "no cancellation", which is explained here. https://en.wikipedia.org/wiki/Dot_product#Properties

I think the distributive law gives

$$W = (F_1 + F_2) \cdot (d_1 + d_2) = F_1 \cdot d_1 + F_2 \cdot d_2 + F_1 \cdot d_2 + F_2 \cdot d_1$$

• I just confirmed that dot product is distributive over vector addition by googling it. There was a pdf file for the proof. So, I don't think there's anything wrong with my second expansion. – Dove Dec 23 '16 at 11:32
• There is definitely something wrong with the expansion, the only question trying to find out what it is. – JMLCarter Dec 23 '16 at 11:42
• edit my answer. – JMLCarter Dec 23 '16 at 11:58
• But that's the same thing. I've assumed d1 and d2 to be the components of displacement d along F1 and F2. So, F1•d1=F1d1 and F2•d2=F2d2 – Dove Dec 23 '16 at 12:56
• Right, I have to apologise for missing that key part of the problem. So working back through it I'll append a different answer. – JMLCarter Dec 23 '16 at 13:04

I think that there is a mistake in the question. In the question, I've assumed that $d_1$ and $d_2$ are the components of the resultant displacement $d$ along $F_1$ and $F_2$. So, these are just the projections of $d$ along the directions of the two forces. I've derived the second expression for the work done by the dot product of resultant force with resultant displacement. I've used $d_1+d_2$ as the resultant displacement. But the truth is that vector sum of $d_1$ and $d_2$ will be equal to the resultant displacement only when $d_1$ and $d_2$ are perpendicular. Suppose $d_1$ makes an angle a with d and $d_2$ makes an angle b with d. Then the angle between $d_1$ and $d_2$ will be a+b. So, the square of the magnitude of resultant of $d_1$ and $d_2$ is : $d^2cos^2a+d^2cos^2b+2d^2cosacosbcos(a+b)$. I think this will be equal to the square of the resultant displacement , i.e. $d^2$ only when a+b=90 degrees. So, we can't assume that the resultant displacement is the vector sum of its projections along $F_1$ and $F_2$. Another way of seeing this is: If we assume one component of d along $F_1$ to be $d_1$ then if we have to calculate a second component of d such that if this second component is vectorily added to $d_1$, we'll get d, then that second component would be $d-d_1$ which won't necessarily be in the direction of $F_2$. And even if it is in the direction of $F_2$, it's magnitude won't necessarily be equal to dcosb where b is the angle between d and $F_2$. So, these projections won't necessarily add up to give the resultant displacement.