Solve "Moments of Force" with Uniform Meter Rule & Pulley

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Homework Help Overview

The problem involves a uniform metre rule supported at the 40 cm mark and a string at the 100 cm mark, with a mass of 20 g hanging from the string. The goal is to determine where a 50 g mass should be suspended on the rule to achieve balance.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of moments and forces, questioning the calculations and assumptions made regarding the reaction force and the placement of the masses. There is confusion about the inclusion of various forces and moments in the calculations.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem setup and the calculations involved. Some guidance has been offered regarding the use of moments about different points, but no consensus has been reached on the correct approach.

Contextual Notes

Participants are navigating through the implications of using different points for taking moments and the necessity of including all relevant forces in their calculations. There is a mention of homework constraints regarding the use of weights versus masses.

look416
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Homework Statement



A uniform metre rule of mass 100 g is supported by a knife-edge at the 40 cm mark and a string at the 100 cm mark. The string passes round a frictionless pulley and carries a mass of 20 g as shown in the diagram.

At which mark on the rule must a 50 g mass be suspended so that the rule balances?

Homework Equations





The Attempt at a Solution


using the \sum F=0
therefore, R at the knife-edge + T at the 100 cm = W
R = 0.8
then take the moment at the c.g, which is at 0.5
therefore R(z)=T(50)
in the end,what i got is z=0.3125
where T=0.5N
wrong again...
 
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Hi look416! :smile:

(have a sigma: ∑ :wink:)

I don't understand what you've done …

there should be moments of three forces …

and where does T = 0.5N come from? :confused:

(btw, no need to convert to N … they're all weights, so just use the masses :wink:)
 
lolz
maybe I am not good in explaining XD
anyway
if i take the head of the ruler as the moment
which means
RX + 50(100) = 50(100)
which results in RX = 0
=.=
T = 0.5N is because the question demand for the X cm when the load is 50g
for 50g x 10 x 10^-3 = 0.5N
 
look416 said:
RX + 50(100) = 50(100)

uhh? :redface: where's the 40cm? where's the 20g? and what's R? :confused:
 
=.=
so, i have to find the value of R by taking moments as the head of the ruler
if so,
R(40)+20(100)=50(100)
R=75g
then??
 
look416 said:
=.=
so, i have to find the value of R by taking moments as the head of the ruler

Is R the reaction force at the knife-edge?

If so, you don't find it by taking moments, you find it by adding the vertical components of force to zero …

which I thought you'd already done, and found it was 80g (weight) ??
R(40)+20(100)=50(100)

You haven't included the 50 g mass in this moments equation. :confused:

(btw, if you take moments about the knife-edge you won't need to find R anyway)
 
but there said at which mark on the rule must a 50 g mass be suspended so that the rule balances?
so shouldn't i have to included the knife edge in my calculation?
not taking it as the moments ...
 
look416 said:
but there said at which mark on the rule must a 50 g mass be suspended so that the rule balances?
so shouldn't i have to included the knife edge in my calculation?
not taking it as the moments ...

(i wish you wouldn't say "taking it as the moments" … you take moments of forces about a point :wink:)

Yes, the 50g mass, and its unknown position, must be included (unless you take moments about that position).

You can take moments about any point …

it can be either end of the ruler, or the knife-edge, or the point where the 50g is, or indeed anywhere else, either on or off the ruler.

But in this case it's easier to use the knife-edge, since that avoids working out the reaction force, R. :smile:
 

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