# one simple and one tricky statics question

• Sep 19th 2009, 09:17 PM
hairy
one simple and one tricky statics question

I suppose we will begin with the simple one:
http://dl.getdropbox.com/u/594924/33.jpg
Solving for T is simple enough, but for the life of me I can't come up with a useful expression to help me get any forces on the pulleys, at least as far as combining the resultant with trig functions of the angles going on. The back of the book says the force on pulleys A and D is 14.9 newtons and the Force on B and C is 40.8 newtons. This doesn't help me reverse-engineer the expression they used, however. I think the slightest hint on this problem would get me pointed in the right direction. it looks simple but i've never done a problem like this before (and there are no complete solutions in my book).

The second problem is a little trickier.
http://dl.getdropbox.com/u/594924/61.jpg

I'm sure I'll be setting up the equilibrium equations and solving them as a system, with the obvious AB = 2AD = 2AC substitution that they hand you, but expressing the equations as a function of d? Should that even be part of the strategy? Let me show you what I have for equilibrium:
http://dl.getdropbox.com/u/594924/eqns.jpg

The z angles are the angles between AC and AD vectors and the xy plane. These are also the angles used in the first trig factors in my x and y sums. The 71.57 and 108.435 are the angles AC and AD and the positive y axis, and the 198.435 and 161.565 are between the vectors and positive x axis.

But I have no strategy from here. I'm going to spend more time tomorrow on it but any thoughts of yours in the meantime would be very helpful. I think in the meantime I'm going to try to balance the equations without worrying about d, and if i can do that while satisfying AB = 2AD = 2AC, maybe it'll just work out.

thanks again
• Sep 20th 2009, 01:06 AM
Hello hairy

Welcome to Math Help Forum!
Quote:

Originally Posted by hairy

I suppose we will begin with the simple one:
http://dl.getdropbox.com/u/594924/33.jpg
Solving for T is simple enough, but for the life of me I can't come up with a useful expression to help me get any forces on the pulleys, at least as far as combining the resultant with trig functions of the angles going on. The back of the book says the force on pulleys A and D is 14.9 newtons and the Force on B and C is 40.8 newtons. This doesn't help me reverse-engineer the expression they used, however. I think the slightest hint on this problem would get me pointed in the right direction. it looks simple but i've never done a problem like this before (and there are no complete solutions in my book).

I don't have time to look at #2 now, but you're right, #1 is simple. The principles you can use are:

• Assuming the pulleys are frictionless, the tension in the string is the same all the way round.

• Each pulley is in equilibrium under the action of 3 forces: the tensions in the two parts of the string, and a reaction force exerted by the table. This reaction force, then, is equal and opposite to the resultant force exerted by the string.

So, let the reaction force at each pulley have two components at right angles to each other in convenient directions (e.g. for pulley A along BA and along DA) and then resolve in each of these two directions.

PS Indeed that works OK.

I presume that you have found the tension in the string, $\displaystyle T$, to be $\displaystyle \frac{25}{\cos 30}$

Then let the components of the reaction force at A be $\displaystyle X_A$ and $\displaystyle Y_A$ along DA and AB, respectively. Then resolve in these two directions, and get:

$\displaystyle X_A = T\sin30$ and $\displaystyle Y_A + T\cos30=T$

Solve for $\displaystyle X_A$ and $\displaystyle Y_A$. Then find the magnitude of the reaction (and hence the resultant force exerted by the string on the pulley at A) by calculating $\displaystyle \sqrt{{X_A}^2+{Y_A}^2} = 14.94$ N

Similarly at B, but it's even easier because the components of the reaction force are equal to each of the tensions along BA and BC. The resultant force therefore has magnitude $\displaystyle \sqrt{T^2+T^2}= 40.82$ N.
• Sep 20th 2009, 08:54 AM
hairy
ah, i see. i'm sure this is going to come up again, and i'll remember it. thanks.
• Sep 20th 2009, 09:17 AM
Hello hairy
Quote:

Originally Posted by hairy
...
The second problem is a little trickier.
http://dl.getdropbox.com/u/594924/61.jpg

I'm sure I'll be setting up the equilibrium equations and solving them as a system, with the obvious AB = 2AD = 2AC substitution that they hand you, but expressing the equations as a function of d? Should that even be part of the strategy? Let me show you what I have for equilibrium:
http://dl.getdropbox.com/u/594924/eqns.jpg

The z angles are the angles between AC and AD vectors and the xy plane. These are also the angles used in the first trig factors in my x and y sums. The 71.57 and 108.435 are the angles AC and AD and the positive y axis, and the 198.435 and 161.565 are between the vectors and positive x axis.

But I have no strategy from here. I'm going to spend more time tomorrow on it but any thoughts of yours in the meantime would be very helpful. I think in the meantime I'm going to try to balance the equations without worrying about d, and if i can do that while satisfying AB = 2AD = 2AC, maybe it'll just work out.

thanks again

There are one or two issues with your trigonometry that I'd like to clear up. First, they concern the angles around the point A.

So, suppose E is the point vertically below D in the horizontal plane through A; and F is the point in this plane vertically below the mid-point of CD. Then in $\displaystyle \triangle AFE$:

$\displaystyle \angle F = 90^o, AF = 6, FE = 2$

$\displaystyle \Rightarrow AE = \sqrt{36+4}=\sqrt{40}$

If $\displaystyle \angle FAE = \phi$, then $\displaystyle \tan\phi = \frac{2}{6}$

$\displaystyle \Rightarrow \phi = 18.43^o$

I think you have already worked this out - but look at what happens next:

In $\displaystyle \triangle ADE$:

$\displaystyle AD=\sqrt{40+9}=7$

and if $\displaystyle \angle DAE = \theta, \sin\theta= \frac{3}{7}$

$\displaystyle \Rightarrow \theta = 25.38^o$, not as you have (I think) calculated: $\displaystyle \theta =\tan^{-1}(\tfrac12)=26.57^o$.

Next, I think you can find the distance $\displaystyle d$ in a single equation, by looking at the components of the tensions along the $\displaystyle x$-line.

Suppose that AB makes an angle $\displaystyle \psi$ with the horizontal. Then, since $\displaystyle AB = \sqrt{d^2+36},\, \cos\psi=\frac{6}{\sqrt{d^2+36}}$

Suppose also that the tension in AD and AC is $\displaystyle T$; the tension in AB is therefore $\displaystyle 2T$. Then, using your notation:

$\displaystyle \sum F_x=2T\cos\psi - T\cos\theta\cos\phi-T\cos\theta\cos\phi=0$

$\displaystyle \Rightarrow \frac{6}{\sqrt{d^2+36}}=\cos\theta\cos\phi$

Knowing $\displaystyle \theta$ and $\displaystyle \phi$, you can now solve for $\displaystyle d$.

Finally a revised version of the vertical forces equation will enable you to find the tensions in the strings:

$\displaystyle \sum F_z= 2T\sin\psi + 2T\sin\theta-490.5 = 0$

Knowing $\displaystyle d$ and the fact that $\displaystyle \sin\psi = \frac{d}{\sqrt{d^2+36}}$, can you complete it now?

• Sep 20th 2009, 10:10 AM
hairy
Quote:

Originally Posted by Grandad
Hello hairyThere are one or two issues with your trigonometry that I'd like to clear up. First, they concern the angles around the point A.

So, suppose E is the point vertically below D in the horizontal plane through A; and F is the point in this plane vertically below the mid-point of CD. Then in $\displaystyle \triangle AFE$:

you've already lost me. I don't see how AFE form a triangle if they are all points on the x axis.
• Sep 20th 2009, 12:59 PM
Hello hairy
Quote:

Originally Posted by hairy
you've already lost me. I don't see how AFE form a triangle if they are all points on the x axis.

What makes you think that they all lie on the $\displaystyle x$-axis? E is vertically below D, whereas the $\displaystyle x$-axis passes through F, the point vertically below the mid-point of CD.

If you still can't see it, look again at the diagram. The $\displaystyle x-z$ plane is a plane of symmetry. It contains the line AB, and passes through the mid-point of CD, and therefore has C and D lying equidistant from it, one on either side. Does the vertical line through C pass through the $\displaystyle x$-axis? No. Well, neither does the vertical line through D.

• Sep 20th 2009, 05:44 PM
hairy
Quote:

Originally Posted by Grandad
Hello hairyWhat makes you think that they all lie on the $\displaystyle x$-axis? E is vertically below D, whereas the $\displaystyle x$-axis passes through F, the point vertically below the mid-point of CD.

If you still can't see it, look again at the diagram. The $\displaystyle x-z$ plane is a plane of symmetry. It contains the line AB, and passes through the mid-point of CD, and therefore has C and D lying equidistant from it, one on either side. Does the vertical line through C pass through the $\displaystyle x$-axis? No. Well, neither does the vertical line through D.