Work problem - Rope, pulley and brick (applied integration)

[themli]

If a brick is pulled across the floor by a rope thruogh a pulley, 1 meter above the ground - and work = W, where W = 10N, (in Newton).Show that the horizontal component of W, which is pulling the brick has the size
\frac{10x}{\sqrt{1+x^2}} (*)
Use this to calculate the amount of work needed to move the brick from x = 10 to x = 2.

This is what I have so far:
In terms of the expression (*), I'm thinking the 10x has to do with W being equal to 10N, and the W-side (call it Z) being the hypotenuse: \sqrt{1+x^2}

Let the angle by the brick be θ:
\cos θ=\frac{x}{\sqrt{1+x^2}}

So is the formula found by taking cosθ×10?

In terms of calculating work, I'm not sure. I first thought taking the definite integral of (*) from 2 to 10?

 

[BvU]

You're doing fine. Yes, the formula is good and since the force varies with x you need to integrate. Carry on !

 

[themli]

Can I integrate it as 10*cosine? Or should I use the original expression using integration by parts (and other methods)?

 

[BvU]

Write down the integral you want to calculate ($\ \int F \, dx\$). The integrand can be written as $\ 10 \cos\theta\$ but then you still need to do something ...

 

[themli]

Write down the integral you want to calculate ($\ \int F \, dx\$). The integrand can be written as $\ 10 \cos\theta\$ but then you still need to do something ...

Clearly, since 10\int_2^{10}cosθ gives a negative number... I have no clue about physics, so I might be missing something about the work?

In the text, the brick is moving from 10 to 2, so maybe i should do:

10\int_{10}^{2}cosθ \approx 14.53

How's that?

Alternatively, I know that W = F \times s or W = F(s), so I need to find some value for displacement

 

[BvU]

Once you have the right expression for the integral, there isn't much physics going on anymore: it's pure math.

What do you make of $dx$ ?

 

[themli]

Once you have the right expression for the integral, there isn't much physics going on anymore: it's pure math.

What do you make of $dx$ ?

Ah right - above I just used dθ , but It depends on the integrand, so dx means I need the integrand to be in terms of x. So I should use the expression \frac{x}{\sqrt{1+x^2}}, thus 10\int_{10}^2 \frac{x}{\sqrt{1+x^2}} dx

 

[BvU]

What I meant is that if you want to use theta as integration variable, you should not forget to express $dx$ in terms of $\theta$. Perhaps keeping $x$ as integration variable is easier...?

 

[themli]

Yes, that was just a typo on my part - I did integrate using dθ with cos (I did get a number above, which I assume is wrong?) and dx using the x-expression, which I haven't calculated yet, looks like a tricky integral, trig substitution maybe?

Btw about the limits of integration, from the illustration I made, it looks like we are moving in the negative direction, from 10 to 2 (this is the way it was stated in the problem as well). I know under the fundamental theorem you change the sign when flipping the limits of integration.

 

[BvU]

Turns ot both ways are pretty straightforward (goes to show that I'm also not that comfortable with integrals). Good exercise to actually do it both ways ! That also gives you an extra checking opportunity.

And you know you have the wrong sign if the work comes out negative

 

[themli]

10\int_{2}^10 \frac{x}{\sqrt{1+x^2}} dx Let u=1+x^2, du=2x dx so we have

5\int_{x=2}^{x=10} \frac{1}{\sqrt{u}} du

10\sqrt{u} = 10\sqrt{1+x^2}

From x = 2 to 10

\approx 78.14

 

[BvU]

Looks good. Now perhaps $\ \displaystyle \int 10\cos\theta\, dx \$ with $\ \theta = \cot x\$ ?

[edit] Sorry, wrong way around : should of course be $\ \cot\theta = x \$

 

[themli]

Let θ= cotx , dθ=-csc^2xdx so we have

-10\int_{2}^{10} cos(cotx)csc^2x dx

Let u = cotx, du=-csc^2x dx

-10\int_{x=2}^{x=10} cos(u) dx = -10(sin(cotx)) from 2 to 10

Is that negative? No idea how I would calculate \sin(\cot x), it's a composite function but..

 

[BvU]

Does that yield 78.14 ?

 

[themli]

No, so something's wrong. The answer I got was 1.441\times (-10) = -14.41

 

[BvU]

Step from $\ d\theta\$ to $\ dx\$ :$$
d\theta = -csc^2 x \, dx \Rightarrow \ dx = ... d\theta $$

 

[themli]

Step from $\ d\theta\$ to $\ dx\$ :$$
d\theta = -csc^2 x \, dx \Rightarrow \ dx = ... d\theta $$

 

[themli]

Step from $\ d\theta\$ to $\ dx\$ :$$
d\theta = -csc^2 x \, dx \Rightarrow \ dx = ... d\theta $$

dθ = -csc^2x dx
dx = \frac{dθ}{-csc^2x}

 

[themli]

I miscalculated something in the post above, the integral of \sin u du Is equal to
14.4 not negative. I missed a negative sign from the csc^2. But it's still not 78.14.

 

[BvU]

You need to express $\ \csc^2 x \$ in terms of $\theta$

 

[themli]

You need to express $\ \csc^2 x \$ in terms of $\theta$

Not sure how to do that. At the point θ to dθ, I take dθ/dx and get -\csc^2x dx. Isn't it equivalent to when I use u-substitution, and I express the term I differentiate still using x, with the point of cancelling the already existing x-terms? So same here, I replaced θ with x-terms.

And also, is 78.14 the wrong answer?

 

[BvU]

Not sure how to do that. At the point θ to dθ, I take dθ/dx and get -\csc^2x dx. Isn't it equivalent to when I use u-substitution, and I express the term I differentiate still using x, with the point of cancelling the already existing x-terms? So same here, I replaced θ with x-terms.

You have $\displaystyle \int \cos\theta\, dx\$ with $$
x = {\cos \theta\over\sin \theta}\ \Rightarrow \ dx = -{1\over \sin^2 \theta} \;d\theta $$ and you integrate from $\operatorname{arccot}10$ to $\operatorname{arccot}2$

And also, is 78.14 the wrong answer?

Well compare the two answers ...

(the value seems quite reasonable: about 10 N times 8 m would be a first guess when you completely ignore the angle)

 

[themli]

You have $\displaystyle \int \cos\theta\, dx\$ with $$
x = {\cos \theta\over\sin \theta}\ \Rightarrow \ dx = -{1\over \sin^2 \theta} \;d\theta $$ and you integrate from $\operatorname{arccot}10$ to $\operatorname{arccot}2$

Well compare the two answers ...

(the value seems quite reasonable: about 10 N times 8 m would be a first guess when you completely ignore the angle)

Got it! Got the same answer, 78.138. That was a strange integral lol. I can't thank you enough!

However, when the brick is dragged up, the angle could be calculated, because we now the length of the sides (1, 8, 10.1) by pythagorean theorem. Should I take that into account? Since 10N \times 10.1 = 101. That is the z side is 10.1 long.

Edit: Nvm, I was only trying to find the horizontal work.