# Centre of gravity / torque / moments question?Watch

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Thread starter 7 years ago
#1
1. The problem statement, all variables and given/known data

Imagine a patient lying down on a bed , you want to weigh her without moving her, but the scale can only support the pair of legs of the bed at one end - or the pair at the other end , but not simultaneously. So he puts the scale at point X (the pairs of bed legs either side of the her feet ) at it was 700N , at point Y (the pair of legs either side of her head) the scale read 800N , The Dr knows from previous determination the bed weighs 90kg.

QUESTIONS

i)the total force exerted by the floor on the bed? A) 1500 - this i understand

ii) the mass of the patient is ? A) 60kg - again i understand this

III) Suppose the length of the bed , the horizontal distance between X and Y is L.

At what horizontal distance from Y is the center of gravity of the patient , bed and bedding considered a single body located?

I don't understand the answer? How is it derived? Using moments I guess?

I understand the moment force is the force x perpendicular distance from the centre?

As the patient is stationary all forces should cancel out? I was trying to come up with an equation that would solve for the answer but can't for the life of me.
0
7 years ago
#2
Take moments about X
Let the length of the bed be L
The force of 700N acts at the end
Let the distance of the CofG from X be x
The weight of the bed plus bits is 1500N and acts at a distance x fromX
One moment is clockwise the other is anticlockwise.

The answer to the question of 7/15L gives you a big clue when you consider that the weight is 1500N and the force at the end opposite X was 700N
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Thread starter 7 years ago
#3
Ok so I think I solved it. But don't really understand how......

I said the distance from the head (800N side) of the bed to the centre of mass was (X) and therefore the distance from the foot (700N side) of the bed to the centre of mass was (L-X).

I then assumed that as the bed is stationary, the forces exerted at each end must be equal, so I said:

the moment forces acting at each end must be equal, so:

800(X) = 700(L-X)
800(X) = -700(X) + 700L
1500(X) = 700L
(X) = (7/15)L

However when I've seen this sorts of questions explained in the past they've always talked about clockwise and anti-clockwise rotation, and using negatives for clockwise. If I'd have used negatives for clockwise here, I'd have had to have plugged -700 in the first line of my equation, which would have come out wrong. Would it not?
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Thread starter 7 years ago
#4
(Original post by Stonebridge)
Take moments about X
Let the length of the bed be L
The force of 700N acts at the end
Let the distance of the CofG from X be x
The weight of the bed plus bits is 1500N and acts at a distance x fromX
One moment is clockwise the other is anticlockwise.

The answer to the question of 7/15L gives you a big clue when you consider that the weight is 1500N and the force at the end opposite X was 700N

Just wanted to quote you so you notice I attempted my own reasoning just below your post. And its not the same as yours

Is my reasoning also correct though as it came to the right answer?
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Thread starter 7 years ago
#5
(Original post by byebyebadman)
However when I've seen this sorts of questions explained in the past they've always talked about clockwise and anti-clockwise rotation, and using negatives for clockwise. If I'd have used negatives for clockwise here, I'd have had to have plugged -700 in the first line of my equation, which would have come out wrong. Would it not?
Actually I think I only have to use negatives if I'm saying the sums of the forces = zero. Whereas I've said they equal each other. So no need for zeros.

Anyway, I'm still confused as to why my working out is different than yours in the post above?
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7 years ago
#6
Sorry, just read the question again. I got the patient's head and feet (X and Y) the wrong way round!
You can take moments about X or Y
The anticlockwise moment equals the clockwise moment when the bed doesn't move.

If you take them about Y it's easier.
You then get
700 L = 1500 x

If you take them about X you get

800 L = 1500 (L-x)

Apologies if I confused you initially.
1
Thread starter 7 years ago
#7
(Original post by Stonebridge)
Sorry, just read the question again. I got the patient's head and feet (X and Y) the wrong way round!
You can take moments about X or Y
The anticlockwise moment equals the clockwise moment when the bed doesn't move.

If you take them about Y it's easier.
You then get
700 L = 1500 x

If you take them about X you get

800 L = 1500 (L-x)

Apologies if I confused you initially.
So my equations way of working it was correct. I just added a few unneeded steps?
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7 years ago
#8
(Original post by byebyebadman)
So my equations way of working it was correct. I just added a few unneeded steps?
You took moments about the c of g
It also gives the right answer but is not the usual way of doing this.
Only the mark scheme will tell you if that method is ok. It should be fine.

As I say, it's easier if you take them about Y.
0
1 month ago
#9
Can you explain Q1?
0
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