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# Weight Questions watch

1. I'm currently doing the Forces chapter on - The Principle of Moments now. I'm doing the summary questions but don't understand anything, I read through the pages 3 times and still don't know how to apply the text to the questions.

1. A child of weight 200N sits on a seesaw at a distance of 1.2m from the pivot at the centre. The seesaw is balanced by a second child sitting on it at a distance of 0.8m from the centre. Calculate the weight of the second child.

I have no clue on how to work out the weight, can someone please explain it to me, thanks.
2. neither do I, But with modern Britain I'm sure just putting the answer "obese" will be correct
3. The principle of moments states that the weight of the 200N child times its distance from the pivot is equal to the weight of the other child times its distance from the pivot. Now you have, there, three known and one unknown quantity, and the equation to put them in.
4. M=Wx

Moment = Weight * Distance from pivot.

Because the seesaw is in balance the two forces must be equal either way around the pivot. So W1X1=W2X2. (unsure how to subscript).
5. Thanks I did this and...this is the result:

200*1.2 = 240 (W1D1)
W1D1 = W2D2
240 = 0.8 x X
X = 240/0.8 = 300

That is the correct answer, but is it the correct working out?
6. Yes, that's the correct working.
7. ^ thank you very much both of u
and lol @ the obese comment, only if I'd get marks for that!
8. (Original post by Flav)
neither do I, But with modern Britain I'm sure just putting the answer "obese" will be correct
HAHA
9. Ok, here's another question:

A metre rule, pivoted at its centre of mass, supports a 3N weight at its 5.0cm mark, a 2.0N weight at its 25cm mark and a weight W at its 80cm mark

1. Calculate the weight W

This is what I have done:
W1D1 = W2D2 + W3D3
5*3 = 2*25 + 80*Y
15 = 100 + 80y
now what do i do ?
10. On the one side of the centre pivot you have the weight W
On the other you have the 2N and the 3 N
This is balanced.
You have to use the distance from the centre of the rule.
So the 3N weight at the 5cm mark is actually 45cm from the pivot. (The pivot is at 50cm)
Same for the others.
Then 3*d1 + 2 *d2 = W*d3 (with d1 d2 and d3 measured from the centre)
11. 2 more questions

1. a unifrm metre rule of weight 1.2N rests horizontally on two knife-edges at the 100mm mark and the 800mm mark. Sketch the arrangement and calculate the suppoer on the rule due to each knife edge.

2. A uniform beam of weight 230N & a length of 10m rests horizontally on the tops of 2 brick walls, 8.5m apart such that a length of 1.0m projects beyond one wall and 0.5m projects beyond the other wall.

a) Calculate the support force of each wall on the beam
b) the force of the beam on each wall
12. One at a time! (Question 2 is exactly the same method as Q1)
The diagram is very useful. Without it, it's a bit tricky to explain.
You need to understand the Principle of Moments to do these (and the previous) questions.
If the ruler is a rest we know that there is no overall, resultant force on it. This includes forces, called moments, that tend to cause rotation.
You need to know that a moment is equal to the force times the perpendicular (right angle) distance from the pivot.
OK so far?
Let's assume the ruler has its zero mark on the left and 1000mm on the right of the diagram.
For the ruler on the knife edges, there is a downward force of 1.2N, its weight, which acts at the mid point. (500mm)
To find the force of the right hand knife at the 800mm mark, take the pivot to be the knife at the 100mm mark.
Now write down the equation that says the clockwise turning forces (moments) are equal to the anticlockwise moments.
The weight is clockwise, and the knife pushing up at the 800mm mark is the anticlockwise. Remember, it's force times distance from pivot.

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