Moments question

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abdul 580
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#1
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#1
A uniform plank of length 1.5 m and mass 9.0 kg is placed horizontally on two narrow
vertical supports as shown. A block, X, of mass 3.0 kg is placed at the end of the
plank immediately above the centre of the right-hand support.
Calculate the magnitude of the downward force on
the right-hand support,

How would you do this question
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Callicious
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Posting the figure would help.

Generally speaking, you can use "Sum of moments = 0" and "Sum of forces = 0" for the system (since, I am assuming at least, the system isn't evolving.)
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abdul 580
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#3
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#3
It's question 5 b) i)
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Callicious
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(Original post by abdul 580)
It's question 5 b) i)
Name:  234`24`45.PNG
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For the first part you can use sum/moments about the mass X to get an expression for the weight on the other pillar (then just use sum of forces equal to zero.) Remember that

\sum_i \tau_i = 0 \newline
where the tau are the individual moments of the forces involved. This will apply for any choice of pivot in the system assuming it's static.


For the second part, you have two pivots that you can use for moment equations- solve the sum/moments=0 about each pivot and you can get an expression for the reaction force on the opposing pivot.

I recommend posting the image directly instead of a PDF/etc attachment (unless the attachment just has the singular question rather than a whole sleuth of them.)
Last edited by Callicious; 4 weeks ago
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_sabz_
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#5
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#5
use the principle of moments and take moments about the left hand to work out your right handed support magnitude
all anticlocklwise being
- 1.5 x (magnitude of right support)
then clocklwise being
- 0.75 x 9g + 3g x 1.5

and by rearranging that you should be able to do the first part!
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abdul 580
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#6
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#6
(Original post by _sabz_)
use the principle of moments and take moments about the left hand to work out your right handed support magnitude
all anticlocklwise being
- 1.5 x (magnitude of right support)
then clocklwise being
- 0.75 x 9g + 3g x 1.5

and by rearranging that you should be able to do the first part!
Thank you so much

(Original post by Callicious)
Name:  234`24`45.PNG
Views: 7
Size:  79.6 KB
For the first part you can use sum/moments about the mass X to get an expression for the weight on the other pillar (then just use sum of forces equal to zero.) Remember that

\sum_i \tau_i = 0 \newline
where the tau are the individual moments of the forces involved. This will apply for any choice of pivot in the system assuming it's static.


For the second part, you have two pivots that you can use for moment equations- solve the sum/moments=0 about each pivot and you can get an expression for the reaction force on the opposing pivot.

I recommend posting the image directly instead of a PDF/etc attachment (unless the attachment just has the singular question rather than a whole sleuth of them.)
Thank you so much
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