How to find the work done by a tilted force on a block as it's pulled up an incline?

Jun 2017
218
6
Lima, Peru
The problem is as follows:

The figure from below shows a block which is pulled from point $A$ to point $B$ by a force $F=50\,N$ and a constant direction. Find the work in Joules that is made between points $A$ and $B$. (Hint: You may use the triangle $7-24-25$ for $16^{\circ}-74^{\circ}-90^{\circ}$)




The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&-310\,J\\
2.&-250\,J\\
3.&+310\,J\\
4.&+250\,J\\
5.&+280\,J\\
\end{array}$

Since the angle they use is $8^{\circ}$.

I could use the identity for half angle to obtain the relationships in the given triangle.

$\sin 8^{\circ}=\sqrt{\frac{1-\cos 16^{\circ}}{2}}$

$\sin 8^{\circ}=\sqrt{\frac{1-\frac{24}{25}}{2}}$

$\sin 8^{\circ}=\sqrt{\frac{\frac{1}{25}}{2}}$

$\sin 8^{\circ}=\frac{\sqrt{2}}{10}$

$\cos 8^{\circ}=\sqrt{\frac{\frac{49}{25}}{2}}$

$\sin 8^{\circ}=\frac{7\sqrt{2}}{10}$

But other than that I'm still stuck.

I can tell the distance between $A$ and $B$ as:

$AB=7\sec 8^{\circ}=\frac{7}{\cos 8^{\circ}}$

$AB=\frac{7}{\frac{7\sqrt{2}}{10}}=\frac{10}{\sqrt{2}}=5\sqrt{2}$

However that's how far I went with this problem.

I'm stuck as I don't know how to use the information provided of the force with the angle given.

My intuition tells me that I could naively say okay:

$W= F\times d = 50\cos 37^{\circ} \times 5 \sqrt{2}$

But I'm certain that this will not be the answer and neither appears in the alternatives. Can somebody help me here please?. :eek: :help:
 
Last edited:

romsek

Math Team
Sep 2015
2,679
1,480
USA
you don't really seem to know what you're doing.

$F \cdot dl = 50 \cos(37^\circ - 8^\circ) = 50 \cos(29^\circ)$

$W = \displaystyle \int F \cdot dl = 50 \cos(29^\circ) \sqrt{7^2 + 1^2} = \\

50 \cos(29^\circ) 5\sqrt{2} J$

Note we end up with more potential energy than we started and identical
velocities (0 m/s) thus we must have had a positive amount of work done.

Your intuition was almost correct but you used the symbol for cross product,
not dot product, and that disqualified you. :p
 
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Jun 2019
439
236
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Your intuition was almost correct but you used the symbol for cross product,
not dot product, and that disqualified you. :p
-_- *coughs* Don't point out the missing vector and unit vector symbols. Don't point out the missing vector and unit vector symbols. Don't point out the missing vector and unit vector symbols. *coughs*
 

romsek

Math Team
Sep 2015
2,679
1,480
USA
-_- *coughs* Don't point out the missing vector and unit vector symbols. Don't point out the missing vector and unit vector symbols. Don't point out the missing vector and unit vector symbols. *coughs*
eh whatever. I don't strive to be lookagain over at FMHF.

as long as stuff is understandable and there is no true abuse of notation going on I'm happy.

But $W = F \times d$ is just wrong
 
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Jun 2019
439
236
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But $W = F \times d$ is just wrong
Well, I mean, it's arguably less wrong than $F \cdot dl = 50~N \cos(29^\circ)$

$W = F \times d$ is at least defensible if you interpret it as scalar multiplication (not a cross product) and you add the caveat that $F$ and $d$ be parallel.
 
Jun 2017
218
6
Lima, Peru
Sorry for my conceptual errors...

you don't really seem to know what you're doing.

$F \cdot dl = 50 \cos(37^\circ - 8^\circ) = 50 \cos(29^\circ)$

$W = \displaystyle \int F \cdot dl = 50 \cos(29^\circ) \sqrt{7^2 + 1^2} = \\

50 \cos(29^\circ) 5\sqrt{2} J$
I'm sorry for my conceptual errors, I often have to go back to my physics book for theoretical reference (which is Rex&Wolfson's Essential College Physics btw). Anyways...

I took the last part as an added excercise to test if such answer could be among the alternatives given:

$W= 50 \cos(29^\circ) 5\sqrt{2}$

Since $\cos 29^{\circ}$ isn't exactly given I thought that I could use the angle difference identity to obtain an approximation for this result. Provided the hint mentioned in the problem.

$\cos 29^{\circ}=\cos \left(37-8\right)^{\circ}$


$\cos 29^{\circ}=\cos 37^{\circ}\cos 8^{\circ}+\sin 37^{\circ} \sin 8^{\circ}$


$\cos 29^{\circ}=\frac{4}{5}\frac{7\sqrt{2}}{10}+\frac{3}{5}\frac{\sqrt{2}}{10}$


$\cos 29^{\circ}=\frac{\sqrt{2}}{50}\left(28+3\right)$


$\cos 29^{\circ}=\frac{\sqrt{2}}{50}\left(31\right)$


So by pluggin in this in the earlier expression would yield:

$50\cdot 5 \cdot \sqrt{2} \left(\frac{31\sqrt{2}}{50}\right)=\left( 155\cdot 2\right )=+310\,J$

Which goes to the third alternative and that must be the answer.

Note we end up with more potential energy than we started and identical
velocities (0 m/s) thus we must have had a positive amount of work done.

Your intuition was almost correct but you used the symbol for cross product,
not dot product, and that disqualified you. :p
I get that we have more potential energy since it climbed the incline to a given height, but what do you mean by identical velocities? Does it mean that it did not have any sort of acceleration?. Can you explain to me this part please?

I'm glad that my intuition was right. Regarding the symbol. Excuse my poor handling of notations. I typically use this "cross" to all sorts of multiplications to specify when a pair of numbers or elements be multiplied. However I had forgotten that in maths and physics it has a special status and it can lead to confusions as the "cross" in vectors for multiplication is one thing and the "dot" is another. Btw I'm well aware about the difference of both, one implies having the result of another vector while the other refers to the product of two vectors produces a scalar. It is just that as I mentioned. I tend to prefer to use the "cross" over the "dots" since when I do this by hand I typically confuse it with the dot that separates decimals in real numbers. I hope you may understand. :)

-_- *coughs* Don't point out the missing vector and unit vector symbols. Don't point out the missing vector and unit vector symbols. Don't point out the missing vector and unit vector symbols. *coughs*
Sorry, I never thought using vector notations in this problem.

But I believe you were referring that I should have used:

$\vec{F}=\left\langle F\sin 29^{\circ}, F \cos 29 ^{\circ}\right\rangle$

or

$\vec{F}=\left\|F\right\|\sin 29^{\circ}\vec{i}+\left\|F\right\|\cos29^{\circ} \vec{j} $

I don't know if is a matter of preferences but if I were to use vectors I tend to prefer brackets over using the unit vectors. :rolleyes:

Well, I mean, it's arguably less wrong than $F \cdot dl = 50~N \cos(29^\circ)$
Maybe I arrived a bit late for the time when this comment was posted but this was not what I wrote. But I could understand that the intended meaning was a force of 50 newtons and the N there was just stressing that?.

$W = F \times d$ is at least defensible if you interpret it as scalar multiplication (not a cross product) and you add the caveat that $F$ and $d$ be parallel.
Thanks for that. From my point of view as someone who's studying these things after a long time, I find it hard to get back and pick up concepts which I forgotten so if I make mistakes like these I'm glad that you guys point it out so I can realize what sort of thing I just did. :)
 
Jun 2019
439
236
USA
Sorry, I never thought using vector notations in this problem.
My comment wasn't aimed at you. It was a gentle jab at romsek for criticising your potentially confusing notation while simultaneously making a potentially confusing notation error.

He wrote, in essence, $F\cdot dl = F \cos\theta$, which does not make sense dimensionally. It only makes sense if he meant $\vec{F} \cdot \hat{dl} = F \cos\theta$ (the projection of the force in the direction of travel).
$\vec{F} \cdot \vec{dl}$ would be infinitesimally small and have units of work.


As for the identical velocities, he was assuming that friction or something else counteracts force F so that the velocity is constant as it climbs the incline. This does not have to be the case; the work done by force F is $W = \int \vec{F} \cdot \vec{dx}$, regardless of 1st Law considerations. He was using the 1st Law to demonstrate why the answer made physical sense, though.