As level Non-Uniform Plank question.

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The discussion revolves around solving a physics problem involving a non-uniform plank supported at both ends with a woman standing on it. For part (a), the equilibrium condition leads to the conclusion that the reaction force at B is 75 kg. In part (b), the center of mass of the plank is calculated by considering the moments about point A, ultimately yielding a distance of approximately 4.75 m from point A. Participants emphasize the importance of correctly identifying forces, including the weight of the woman and the plank's weight. Overall, the conversation focuses on applying equilibrium principles and moment calculations to find the required values.
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Homework Statement



A non-uniform plank of wood AB has length 6m and mass 90kg. The plank is smoothly supported at its two ends A and B, with A and B at the same horizontal level. A woman of mass 60kg stands on the plank at the point C, where AC=2m. The plank is in equilibrium and the magnitudes of the reactions at A and B are equal. the plank is modeled as a non-uniform rod and the woman as a particle.

Find

(a) the magnitude of the reaction on the blank at B,

(b) the distance of the centre of mass of the plank from A.

Homework Equations



Moment = Fx

The Attempt at a Solution



For part A i was thinking total down = total up, if that's the case you have 90g+60g = 150g if that's the case would b = 75g as its equilibrium?

also I would take moments about point A for question (b) however I do no know what i would do from there.

Help would be appreciated
 
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Welcome to PF!

Hi templarbaza! Welcome to PF! :smile:
templarbaza said:
… the magnitudes of the reactions at A and B are equal.

For part A i was thinking total down = total up, if that's the case you have 90g+60g = 150g if that's the case would b = 75g as its equilibrium?

Correct! (well, except that kg would be better! :biggrin:)
also I would take moments about point A for question (b) however I do no know what i would do from there.

Call the distance from A to the centre of mass "x", and it should work out. :wink:

What do you get? :smile:
 
well, I did 90x2 + 75x6 = 530

then I am not quite sure, do you divide it by 150?

if so i got x = 3.53m
 
templarbaza said:
well, I did 90x2 + 75x6 = 530

then I am not quite sure, do you divide it by 150?

if so i got x = 3.53m

(if you're using x for distance, use * for times)

Have you drawn the forces?

What are the three forces, and where are they?
 
hmm I am not sure,

ac = 2m

c = 60kg

60 * 2 = 120 (woman x distance from point a)

75 * 6 = 450 (a x force at point B) < --- not sure about that and from this point onward.

570 / 120 = x

x = 4.75m

Im not sure I know you have a force created by the woman, one by the plank what is the third force?

If I have the right method but am using the wrong information i would like some more hints =]
 
Last edited:
templarbaza said:
Im not sure I know you have a force created by the woman,

Don't talk like that, it will confuse you …

it isn't "a force created by the woman", it's the weight of the woman.
one by the plank what is the third force?

The other two forces are the reaction force at B, and the weight of the plank at its centre of mass.
 
Thanks for the help. so by accounting for the weight of the woman and the reaction force at B and dividing by the weight of the plank would give the Center of Mass (x)
 
(just got up :zzz: …)
templarbaza said:
Thanks for the help. so by accounting for the weight of the woman and the reaction force at B and dividing by the weight of the plank would give the Center of Mass (x)

Yes. :smile:
 
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