mg2004
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I thought I worked it out correctly but apparently not.
The question is as following:

The lorry is placed so that the two front wheels are on one weighing platform and the two back wheels on another; the masses recorded are 1350, k, g,1350kg and 1450, k, g,1450kg respectively. The axles are 3.00m apart and at the same height.

Find the centre of mass from the front axle.

What additional mass m, would have to be placed 50cm in front of the front axle to make the weights borne by the axles equal?
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mqb2766
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(Original post by mg2004)
I thought I worked it out correctly but apparently not.
The question is as following:

The lorry is placed so that the two front wheels are on one weighing platform and the two back wheels on another; the masses recorded are 1350, k, g,1350kg and 1450, k, g,1450kg respectively. The axles are 3.00m apart and at the same height.

Find the centre of mass from the front axle.

What additional mass m, would have to be placed 50cm in front of the front axle to make the weights borne by the axles equal?
Upload what you tried?
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mg2004
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(Original post by mqb2766)
Upload what you tried?
I chose a point 'x' (which is a certain distance from the midpoint of 1.5m) to be the centre of mass and went off the basis that the sum of the clockwise and anticlockwise moments was 0, because the object is in equilibrium. So, I set the clockwise and anticlockwise moments equal to each other and described the distances in terms of x, then I solved for x and added this to 1.5m to arrive at what I thought was the answer. But, obviously I've done something wrong and I'd appreciate any help.
(g is of course the gravitational field strength assumed as 9.81m/s/s in my answer)
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mqb2766
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How do you get to the top of the second image?
The weights on each axis are fairly similar, so x must be fairly small.

Also its not wrong to write (3-(1.5-x)), but 1.5+x is simpler? A clear diagram helps. Similarly, just cancel "g" in the fraction rather than multiplying it out and making the numbers larger.

Edit - a final one, you could simply note that taking moments about the com would mean that the axis distances from the COM were in the same ratio as their masses. Its easy to solve then transform to the actual distances.
Last edited by mqb2766; 3 weeks ago
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mg2004
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(Original post by mqb2766)
How do you get to the top of the second image?
The weights on each axis are fairly similar, so x must be fairly small.

Also its not wrong to write (3-(1.5-x)), but 1.5+x is simpler? A clear diagram helps. Similarly, just cancel "g" in the fraction rather than multiplying it out and making the numbers larger.

Edit - a final one, you could simply note that taking moments about the com would mean that the axis distances from the COM were in the same ratio as their masses. Its easy to solve then transform to the actual distances.
Ahh thank you, I've done that now and I reached the correct conclusion. I didn't consider using the ratio of the masses (but now I realise that that is much simpler) so your advice was very helpful to me.
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