Moments and weights of a table lamp

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  • #1
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Homework Statement


The weight of rod AB is 4[N] and that of piece D is 10[N]. the moment at joint C is 1.8[Nxm]. what is the weight of rod AC.

Homework Equations


Moments: ##M=F\cdot L##

The Attempt at a Solution


Rod AB doesn't contribute to the moment at C:
$$10\cdot 15+180=W_AC\cdot 5\rightarrow W_AC=66[N]$$
It's 6[N] in the book.
 

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  • #2
SteamKing
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Homework Statement


The weight of rod AB is 4[N] and that of piece D is 10[N]. the moment at joint C is 1.8[Nxm]. what is the weight of rod AC.

Homework Equations


Moments: ##M=F\cdot L##

The Attempt at a Solution


Rod AB doesn't contribute to the moment at C:
Are you sure about this assumption?

Draw a free body diagram of rod AC. How would you simulate the loads applied to rod AC by the rod AB and the lamp D?

$$10\cdot 15+180=W_AC\cdot 5\rightarrow W_AC=66[N]$$
It's 6[N] in the book.
 
  • #3
BvU
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[edit] sorry, no good.
 
  • #4
collinsmark
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Homework Statement


The weight of rod AB is 4[N] and that of piece D is 10[N]. the moment at joint C is 1.8[Nxm]. what is the weight of rod AC.

Homework Equations


Moments: ##M=F\cdot L##

The Attempt at a Solution


Rod AB doesn't contribute to the moment at C:
$$10\cdot 15+180=W_AC\cdot 5\rightarrow W_AC=66[N]$$
It's 6[N] in the book.
You are correct that rod AB does not contribute to the moment at C. :smile:

But your equation isn't set up right.

You've added the total moment at point C to the individual contribution of piece D. That's not right.

Instead, you need sum all the individual contributions, and that sum will equal the total moment at point C (180 [N⋅cm]). Hint: Keep in mind the direction of each contribution; some will be positive and others negative. All contributions sum to 1.8 [N⋅m], but 1.8 [N⋅m] is not part of that sum.

---

By the way, on a separate note, if you wanted to keep both 'A' and 'C' in subscript in [itex] W_{AC} [/itex] when using [itex] \LaTeX [/itex], use curly brackets:

Code:
 W_{AC}
 
  • #5
Suraj M
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You've added the total moment at point C to the individual contribution of piece D. That's not right.
did you mean to say piece AC instead of piece D.
 
  • #6
collinsmark
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did you mean to say piece AC instead of piece D.
No, I actually meant to say piece D. My intention was to point out the particular place where things were going awry with the OP's equation.

The OP set up the equation as:
$$10\cdot 15+180=W_AC\cdot 5$$
Breaking that down, that's

[tex] \underbrace{10 \cdot 15}_{\mathrm{contribution \ from \ piece \ D}} + \underbrace{180}_{\mathrm{total \ moment \ at\ C}} = \underbrace{W_{AC} \cdot 5}_{\mathrm{contribution \ of \ piece \ AC}} [/tex]
but that's not correct.

Instead it should be:

[tex] \Sigma \mathrm{(each \ individual \ contribution)} = 180 [/tex]

[And again, some individual contributions may be positive and others negative depending if they are directed clockwise or counterclockwise.]
 
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  • #7
BvU
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[edit] sorry, no good.
Seeing SK's reply, I (too?) quickly erased my post, which was:

How did you decide which side of the equals sign you would place the 1.8 [N m] ?

But the more I look at it, the less I like this exercise. The book answer would give me a torque on C of 0.15 x (-10) + (-0.05) x (-6) = -1.2 [N m]
(x to the right, y up, positive torque towards the viewer)

And personally I wouldn't like to answer -6 [N] for this exercise on a test.

Oops, did I spoil something now ?
 
Last edited:
  • #8
collinsmark
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And personally I wouldn't like to answer -6 [N] for this exercise on a test.

Oops, did I spoil something now ?
Oh, I see what you are saying now. Yes, something is wrong with the book; unless piece AC has some sort of upwards "buoyant" force attributed to it, giving it a negative weight. :eek:
 

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