This is a question I found in Mechanics for Engineers by Beer & Johnston.

A 600-lb crate is supported by the rope and pulley arrangement as shown below. Rope and pulley system

Write a computer program which can be used to determine, for a given value of $\beta$, the magnitude and direction of the force $F$ which should be exerted on the free end of the rope. Use this program to calculate $F$ and $\alpha$ for values of $\beta$ from 0 degrees to 30 degrees at 5 degree intervals.

For writing the program, I need to obtain a relation between $F$, $\beta$ and $\alpha$. The only thing which I was able to think of was that the component $Fcos\alpha=600 lb$ I tried to relate the angles $\alpha$ and $\beta$ by using high school geometry but I did not obtain any result which I could use.

Can anyone help me with the physics part of the question?

  • 1
    $\begingroup$ $F \cos \alpha$ projects to horizontal (adjacent over hypotenouse). I think you want $F \sin \alpha$ to find the vertical component. $\endgroup$ – ja72 Feb 28 '12 at 18:45
  • $\begingroup$ I don't see how to calculate $\alpha$. For each given $\beta$ and given $\alpha$ you can calculate $F$ and the tension $T$. There are two equations here, so only two quantities can be calculated. Consider the case where $\alpha=0$ and then it is obvious how to solve the problem. $\endgroup$ – ja72 Feb 28 '12 at 19:07

Another good trick to solve this would be to use Lami's theorem

With this you get $$\frac{F}{\sin(\pi-\beta)=\sin\beta}=\frac{600}{\sin(\frac\pi2+\beta-\alpha)}$$

While this isn't enough to solve it, Lami's theorem is a useful trick for your toolbox.


enter image description here

The first step in any mechanics problem is to draw a free body diagram. Here is a free body diagram of the problem showing all of the forces on the 600 lbs weight, which we idealize as a point. It shows the force, F, the angles α and β, the tension, T, on the rope, and the weight, w = 600lbs. We then break F and T into vertical and horizontal components. Since the problem says the weight "is supported by the rope and pulley" we assume everything is in equilibrium, which tells us that the sum of all forces is zero. We can break this down to say that the sum of the vertical components of the forces is zero, and the sum of the horizontal components is zero. Therfore, from the vertical components we get:

T*cos(β) + F*sin(α) = 600 lbs

From the horizontal components we get:

T*sin(α) = F*cos(β)

BOTH of these equations must be true at equilibrium. I think this should be enough to get you started on your program. HTH.

By the way, someone (ja72) mentioned above that there are two equations and three unknowns. Unless I'm missing something, I agree. You cannot calculate F and α for various values of β without some other assumption, either about the relationship between α and β (for example we could assume they are symmetrical so β = (π/2) - α ), or about the value of T. Note that if we do assume the are symmetrical, then T = F.

What I would suggest is one of two things: (a) write the program so that the user specifies α and then the program calculates F for that α and the seven β values that the problem asks for, or (b) for each of the seven values of β calculate the value of F for several values of α, for example for α = 0, α = 90 (i.e. π/2), α = 45 (i.e. π/4) and α = (π/2) - β.


I can't really understand how the T works as a pulley, but I can give you a general idea.

You don't need a relation between $F,\alpha,\beta$, you need a relation between $\beta$ and one other variable. Using a relation between three variables will complicate things. Whenever you have to get a closed form for multiple variables, daisy-chain them. Calculate the first variable, then the second using the value of the first, and so on. Sometimes (not here) you need to solve simultaneous equations, then teach your program to use Cramer's rule.

Using vectors will simplify it a lot.

Hint:Split F into two components, $F_x,F_y\,|\, \vec{F}=F_x\hat{i}+F_y\hat{j}$. For a vector $\vec{F}$, $\tan\alpha=\frac{F_y}{F_x}$.

  • $\begingroup$ I've modified the image. It now looks closer to the picture in the text book. $\endgroup$ – Green Noob Feb 28 '12 at 13:27
  • $\begingroup$ I got $Fcos\alpha=600$ - a relation between $F$ and $\alpha$ but how do I get a second relation involving $\beta$? Any suggestions? And thanks for answering :) $\endgroup$ – Green Noob Feb 28 '12 at 13:31
  • $\begingroup$ Use the vector method. Equate forces in the horizontal direction to get $F_x$. Same in the vertical direction. $F=\sqrt{F_x^2+F_y^2}$, normal vector rules. I already gave the formula for alpha. $\endgroup$ – Manishearth Feb 28 '12 at 13:35
  • $\begingroup$ Like I said, try to daisy-chain the equations. Calculate F without using $\alpha$. Then calculate $\alpha$. Relating F and $\alpha$ won't help till you have a relation between F and $\beta$ only or $\alpha$ and $\beta$ only. The issue here is that F will always pop up along with $\alpha$ in any equation you use in this situation. So, you rewrite it in terms of two new variables, $F_x,F_y$, which don't appear in the same equation. $\endgroup$ – Manishearth Feb 28 '12 at 13:39
  • $\begingroup$ Whenever you have a force and an angle, writing it as a vector takes two variables ($F,\alpha$), and spits out two better variables $F_x,F_y$. In this manner, you can easily get rid of annoying angles and trig functions. $\endgroup$ – Manishearth Feb 28 '12 at 13:40

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