# Mechanics help!!!

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#1
A plank AE, of length 6m and weight 100N, rests in a horizontal position on supports at B and D, where AB = 1m and DE =1.5m. A child of weight 145N stands at C, the mid-point of AE. The child is modelled as a particle and the plank as a uniform rod. The child and the plank are in equilibrium. Calculate the magnitude of the force exerted by the support on the plank at B and D.

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7 years ago
#2
(Original post by Victoria Brown)
A plank AE, of length 6m and weight 100N, rests in a horizontal position on supports at B and D, where AB = 1m and DE =1.5m. A child of weight 145N stands at C, the mid-point of AE. The child is modelled as a particle and the plank as a uniform rod. The child and the plank are in equilibrium. Calculate the magnitude of the force exerted by the support on the plank at B and D.

Using Moments
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#3
(Original post by TenOfThem)
Using Moments
I tried taking moments from points B and D and it doesn't seem to be working
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7 years ago
#4
(Original post by Victoria Brown)
I tried taking moments from points B and D and it doesn't seem to be working
In addition, remember that Forces Up (the reactions) = Forces Down (the mg s)
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7 years ago
#5
(Original post by Victoria Brown)
I tried taking moments from points B and D and it doesn't seem to be working
If it's still not coming out, post some working so we can see what's going on.
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7 years ago
#6
(Original post by Victoria Brown)
I tried taking moments from points B and D and it doesn't seem to be working
Take moments from A. That's how I did the question. M1 right? Also resolve vertically.

The child stands at the mid-point so 245N is acting there, and resolving vertically gives RB + RD = 245N.

See how you go from there. I actually really like this topic
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7 years ago
#7
First of all a tip :
1) Draw a diagram as it always helps .

And as the rod is uniform you will find it easier to solve this question!

See the attached image and take moments at either A or B . Any point you want . Find the answer .
Then using upward forces = downward forces find the answer for the magnitude of another reaction .

Hope you understood .! And this question was asked earlier.
http://www.thestudentroom.co.uk/show....php?t=1155126

Please see attach image, it will clear all your doubts. If you have any questions feel free to ask me any questions .

Actually providing full solutions is Bad!
Sum of anti clockwise moment = Sum of clockwise moment
If rod is uniform the weight of rod acts at the center .

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7 years ago
#8
also, if there are two or unknowns , take moments around known forces, makes things easier
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7 years ago
#9
(Original post by TenOfThem)
In addition, remember that Forces Up (the reactions) = Forces Down (the mg s)
That depends where you take the moments from, surely? Forces up on the left of your pivot + forces down on the right of your pivot = forces up on the right of your pivot + forces down on the left of your pivot. Is there any more mathematical way to express this? And don't quote me sum of anticlockwise moments = sum of clockwise moments, I know that, but still ... perhaps a method that factors in the possibility of some moments being at angles?
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7 years ago
#10
That depends where you take the moments from, surely? Forces up on the left of your pivot + forces down on the right of your pivot = forces up on the right of your pivot + forces down on the left of your pivot. Is there any more mathematical way to express this? And don't quote me sum of anticlockwise moments = sum of clockwise moments, I know that, but still ... perhaps a method that factors in the possibility of some moments being at angles?
No

Because I am not taking moments when I point this fact out

I am resolving

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3 years ago
#11
how did you get the distance of CB as 2 and CD as 1.5
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3 years ago
#12
(Original post by yussifmumin)
how did you get the distance of CB as 2 and CD as 1.5
C is the midpoint of AE,so AC = 3.

You're told AB=1, so BC = 3-1 =2.

Similar argument the other end.
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1 month ago
#13
will there be a reaction force for the child (at C), and why?
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1 month ago
#14
(Original post by Hamza Chowdhury)
will there be a reaction force for the child (at C), and why?
Yes. To my understanding, the child is standing on the plank, and their weight is acting on them, not the plank, However, since they are motionless and in contact with the plank, we know that there is a contact force, or reaction force. There is a reaction force that acts on the child which is equal to their weight (such that they are stationary), and also that child must apply a reaction force on the plank which is equal to their weight, but (and again, only to my understanding) we call that force the weight because it is equal in magnitude and direction, just acting on different objects.

Just to make things clearer, the reaction force of the child acting on the plank is what people might call weight, but it is not actually the weight force
Last edited by monsieurberlin; 1 month ago
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1 month ago
#15
(Original post by monsieurberlin)
Yes. To my understanding, the child is standing on the plank, and their weight is acting on them, not the plank, However, since they are motionless and in contact with the plank, we know that there is a contact force, or reaction force. There is a reaction force that acts on the child which is equal to their weight (such that they are stationary), and also that child must apply a reaction force on the plank which is equal to their weight, but (and again, only to my understanding) we call that force the weight because it is equal in magnitude and direction, just acting on different objects.

Just to make things clearer, the reaction force of the child acting on the plank is what people might call weight, but it is not actually the weight force
so I don't need to consider an upwards reaction force for the child in my equations?
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1 month ago
#16
Shouldn't do - you only need to consider forces which act on the plank
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1 month ago
#17
(Original post by monsieurberlin)
Shouldn't do - you only need to consider forces which act on the plank
thanks
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