# Basic Statics Question

## Homework Statement

http://home.exetel.com.au/peleus/EIDEprob.jpg

2. The attempt at a solution

Basically did the standard sum of moments around point A to try and get the vertical component of point D.

-6.2*997 + 9.2*Dy - 12.4*1472

Dy = 2655.89

From there working out Dx

Dx = 2655.89 / tan 50

Dx = 2228.56

Then doing some Pythagoras finding CD to be 3467N.

I think I'm going wrong for a few possible reasons,

1 - It's possible that the distances I'm using aren't perpendicular to the forces, I tried working out the perpendicular forces however and got a very similar answer.

2 - The weight of the bar and weight on the end act straight down, so I might need to somehow figure out a component of their force, but I have no idea how.

If anyone could show me the working for the correct answer for this, it would be much appreciated.

Cheers.

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Doc Al
Mentor
Basically did the standard sum of moments around point A to try and get the vertical component of point D.

-6.2*997 + 9.2*Dy - 12.4*1472
You didn't take the angle of the beam into consideration when finding the torques due to the weights.

Also, if the tension in the cable is T, then the component perpendicular to the beam is Tsin50.

Hint: Use Torque = rFsinθ to find the magnitude of the torques.

I haven't learnt particular form yet.

Any chance you can show me how to do it so I can learn by your working?

I completely understand the idea behind hinting along so someone can work it out for themselves but I think it will just be easier if I can see someone elses working so I'll know what to do with all the future examples.

Cheers.

Doc Al
Mentor
I haven't learnt particular form yet.
You can always just find the component of the weight perpendicular to the beam. Find the angle that the beam makes with the vertical. If that angle is θ, then the component of weight perpendicular to the beam is Wsinθ. (Thus the torque due to the weight = rWsinθ.)

Edit - Wrong, redoing working.

Ok, if someone can confirm this for me I'll be forever in your debt.

Using the sine rule we find angle ACD to be 57 deg.

This leaves angle CAD to be 73 deg. This also means the beam is 17 deg above the horizon.

The result of this is that the weight's have 73 (72.9652 to be precise) degrees of angle between them and the weights.

--- How the text book go their answer ---

Sum of moments around point A, counter-clockwise is positive.

9.2 * Dy - 1471.5 * 12.4 * sin72.9652
Dy = 1896.31
Dx = 1591.19

D resultant is 2575.45N (probably lost 0.05N in a rounding error somewhere to their 2576N answer).

--- The text book didn't use the weight figure ---

Sum of moments around point A, counter-clockwise is positive.

-6.2 * 977.5 * sin72.9652 + 9.2 * Dy - 12.4 * 1471.5 * sin72.9652
Dy = 2526.16
Dx = 2119.70

D resultant is 3297.67N

If anyone could confirm my working is sound, you will avert me having a nervous breakdown.

Thank you.

Last edited:
Doc Al
Mentor
Using the sine rule we find angle ACD to be 57 deg.

This leaves angle CAD to be 73 deg. This also means the beam is 17 deg above the horizon.

The result of this is that the weight's have 73 (72.9652 to be precise) degrees of angle between them and the weights.
Looks good.

--- How the text book go their answer ---

Sum of moments around point A, counter-clockwise is positive.

9.2 * Dy - 1471.5 * 12.4 * sin72.9652
Dy = 1896.31
Dx = 1591.19

D resultant is 2575.45N (probably lost 0.05N in a rounding error somewhere to their 2576N answer).

--- The text book didn't use the weight figure ---
Looks like the book left out the weight of the beam. Oops!

Sum of moments around point A, counter-clockwise is positive.

-6.2 * 977.5 * sin72.9652 + 9.2 * Dy - 12.4 * 1471.5 * sin72.9652
Dy = 2526.16
Dx = 2119.70

D resultant is 3297.67N

If anyone could confirm my working is sound, you will avert me having a nervous breakdown.