How to Calculate Work Done by a Force Along a Path?

[squelch]

Homework Statement

Find the work done by the force $\vec{F}=6\hat{i}+8\hat{j}$ if the object starts from the origin, moves along the x-axis to the point (1,0), then moves in the y direction to the point (1,2). Find the work if instead the object moves diagonally from the origin to the point (1,2).

Homework Equations

$W=\int{\vec{F}\cdot d\vec{s}}$

The Attempt at a Solution

I'm supposing that the answer for the "first part" (the object is traveling horizontally and vertically only) is given by:

$W=[\int^{1}_{0}{(6\cdot 1)\hat{i}+(8\cdot 0)\hat{j}}]+[\int^{2}_{0}{(6\cdot 0)\hat{i}+(8\cdot 2)\hat{j}}]$

..but am not sure if this is the correct procedure, and am afraid to proceed to the second part of the question if I'm misunderstanding what appears to be the much simpler part of the question. That doesn't even look like an integrand is supposed to to me.

edit:
Am I simply overthinking this, and need to find the angle formed by the right triangle and the magnitude of F and work it out as
$W=\int{|\vec{F}|cos(\theta)\cdot d\vec{s}}$

 

[BiGyElLoWhAt]

$W=[\int^{1}_{0}{(6\cdot 1)\hat{i}+(8\cdot 0)\hat{j}}]+[\int^{2}_{0}{(6\cdot 0)\hat{i}+(8\cdot 2)\hat{j}}]$

..but am not sure if this is the correct procedure, and am afraid to proceed to the second part of the question if I'm misunderstanding what appears to be the much simpler part of the question. That doesn't even look like an integrand is supposed to to me.

edit:
Am I simply overthinking this, [STRIKE]and need to find the angle formed by the right triangle and the magnitude of F and work it out as [/STRIKE]
$[STRIKE]W=\int{|\vec{F}|cos(\theta)\cdot d\vec{s}}[/STRIKE]$

The first ones almost right. When you take the dot product of 2 vectors, what kind of quantity do you get? What kind of quantity do you have there?

Remember:

$<a,b,c> \cdot <d,e,f> = ad + be + cf$

 

[squelch]

Well, you would get a ... forgive me if I'm using the wrong vocabulary here ... scalar vector, right?

I'm honestly not sure what I'm missing. I see the $\vec{F}_x = 6\hat{i}$, $\vec{F}_y = 8\hat{j}$, $d\vec{s}_x = 1\hat{i}$, and $d\vec{s}_y = 2\hat{j}$

and then did $\vec{F}_x \cdot d\vec{s}_x$ and $\vec{F}_y \cdot d\vec{s}_y$, but I'm not sure that I'm expressing it in the integrand correctly.

 

[BvU]

Yes, you want scalars, and you are almost there. When you write out e.g. the first term in the first dot product, you forget a $\hat i$:
$$6\hat i \cdot 1\hat{i} = 6 * 1\, (\hat i \cdot \hat i) = 6 * 1 * 1$$and now you have a scalar, so you can add it to other scalars.

If you want, you could write out the whole thing to show you understand it:
$$(6\hat \imath + 0 \hat \jmath)\cdot ( 1\hat{\imath} + 0 \hat \jmath) = \, ...$$where there also appears a term $(\hat \imath \cdot \hat \jmath)$

 

[squelch]

Yes, you want scalars, and you are almost there. When you write out e.g. the first term in the first dot product, you forget a $\hat i$:
$$6\hat i \cdot 1\hat{i} = 6 * 1\, (\hat i \cdot \hat i) = 6 * 1 * 1$$ and now you have a scalar, so you can add it to other scalars.

If you want, you could write out the whole thing to show you understand it:
$$(6\hat \imath + 0 \hat \jmath)\cdot ( 1\hat{\imath} + 0 \hat \jmath) = \, ...$$where there also appears a term $(\hat \imath \cdot \hat \jmath)$

You lost me at $(\hat i \cdot \hat i)$, but I'm probabaly just misunderstanding notation. Does this yield $\hat{i}^2$? I'm not sure what to do with that.

I'm also totally open to viewing a recommended video on this. I have trouble naming these concepts, which makes it difficult for me to look for reference material.

 

[BiGyElLoWhAt]

It's not a scalar vector, to the best of my knowledge that's not even a thing. It's just a scalar. i hat is a vector. If you look at the definition of a dot product a (dot) b, you see that it is magnitude(a)*magnitude(b)*cos(angle inbetween)

With i (dot) i, what do you get if you use this definition?

$\vec{a} \cdot \vec{b} = |a|*|b|*cos(\theta)$
$\hat{i} \cdot \hat{i} = ?$ and just to see what happens (it's good to do at least once)
$\hat{i} \cdot \hat{j} = ?$

What it seems like to me is that you're treating $a \cdot b$ the same as you would treat $a*b$
These are not the same thing.

 

[BvU]

You lost me at $\hat\imath\cdot\hat\imath$

Well, that is pretty quick. I don't know how to put it more explicitly. $\hat\imath$ is a unit vector in a certain direction. Length 1 and more or less by definition $\hat\imath\cdot\hat\imath\equiv 1$. Not a square, but a dot product (and yes, more experienced folks write $\hat\imath^2=1$ for good reasons, comparable to $\vec r \cdot \vec r = |\vec r | |\vec r| \cos 0 = |\vec r |^2$, in shorthand $r^2$ ).

You were evaluating $W=\int{\vec{F}\cdot d\vec{s}} = (6\hat \imath + 0 \hat \jmath)\cdot ( 1\hat{\imath} + 0 \hat \jmath) \,+ \, ... = \, ...$. Write it out as if you were still in primary school and you're fine.

 

[squelch]

I sought out a bit of outside help and reviewed your replies. Thanks for being so patient with me so far, I know that this is pretty basic stuff.

Please tell me if this procedure is sensible:

$$\begin{array}{l} W = \int_0^1 {{{\vec F}_x} \cdot d{{\vec s}_x} + } \int_0^2 {{{\vec F}_y} \cdot d{{\vec s}_y}} \\ W = 6x\hat i|_0^1 + 8x\hat j|_0^2\\ W = 6\hat i + 16\hat j \end{array}$$

 

[BvU]

No good. W is not a vector. There can be no single $\hat\imath$ or $\hat\jmath$ in terms in the answer.

 

[BvU]

Notice I made a mistake too:
$(6\hat \imath + 0 \hat \jmath)\cdot ( 1\hat{\imath} + 0 \hat \jmath) = \, ...$ should be
$(6\hat \imath + 8 \hat \jmath)\cdot ( 1\hat{\imath} + 0 \hat \jmath) = \, ...$ and now applying the primary school rules still gives the same answer for the first term in W, namely 6. No $\hat\imath$!

one of the rules being $(\vec a + \vec b)\cdot(\vec c + \vec d) = \vec a \cdot \vec c + \vec b \cdot \vec c + \vec a \cdot \vec d + \vec b \cdot \vec d$

 

[squelch]

So what I'm missing is that I should have dotted two ihats and two jhats together, and wound up with

W=6+16=22?

 

[BiGyElLoWhAt]

Yup Yup