# Centre of gravity

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for question 3b in the 2012 A2 Paper, why have they used what they did to work out the load at the handle?

And for 3d, the mark scheme says A lower centre of gravity is best to stop the case falling over but then they have given masses which are based higher with 14kg at B and 4kg at C and CD is horizontal

Thanks

for question 3b in the 2012 A2 Paper, why have they used what they did to work out the load at the handle?

And for 3d, the mark scheme says A lower centre of gravity is best to stop the case falling over but then they have given masses which are based higher with 14kg at B and 4kg at C and CD is horizontal

Thanks

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#2

(Original post by

http://www.physics.ox.ac.uk/olympiad...rs.html#BPhOP1

for question 3b in the 2012 A2 Paper, why have they used what they did to work out the load at the handle?

And for 3d, the mark scheme says A lower centre of gravity is best to stop the case falling over but then they have given masses which are based higher with 14kg at B and 4kg at C and CD is horizontal

Thanks

**runny4**)http://www.physics.ox.ac.uk/olympiad...rs.html#BPhOP1

for question 3b in the 2012 A2 Paper, why have they used what they did to work out the load at the handle?

And for 3d, the mark scheme says A lower centre of gravity is best to stop the case falling over but then they have given masses which are based higher with 14kg at B and 4kg at C and CD is horizontal

Thanks

(1 x 14) + (5 x 1/2) + (0 x 4) = 16.5

i.e. the mass at the handle end multiplied by 1

plus the mass of the case multiplied by 1/2 (because it's half way between the handle and the wheel)

plus the mass at the wheel end multiplied by 0 (because it's zero distance from the wheel)

and they've shortened it by removing anything that will always result in +0

---

I think the answer it gives for 4d is just an error, you want the greatest mass in corner C.

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the full calculation for example 1 goes

(1 x 14) + (5 x 1/2) + (0 x 4) = 16.5

i.e. the mass at the handle end multiplied by 1

plus the mass of the case multiplied by 1/2 (because it's half way between the handle and the wheel)

plus the mass at the wheel end multiplied by 0 (because it's zero distance from the wheel)

and they've shortened it by removing anything that will always result in +0

---

I think the answer it gives for 4d is just an error, you want the greatest mass in corner C.

**Joinedup**)the full calculation for example 1 goes

(1 x 14) + (5 x 1/2) + (0 x 4) = 16.5

i.e. the mass at the handle end multiplied by 1

plus the mass of the case multiplied by 1/2 (because it's half way between the handle and the wheel)

plus the mass at the wheel end multiplied by 0 (because it's zero distance from the wheel)

and they've shortened it by removing anything that will always result in +0

---

I think the answer it gives for 4d is just an error, you want the greatest mass in corner C.

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#4

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why don't u multiply it by the distance from the handle end instead?

**runny4**)why don't u multiply it by the distance from the handle end instead?

force at wheel + force at handle = total weight force

you could do it this way and then subtract from the total weight - I think that'd also be perfectly valid.

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That'd give you the force at the wheel.

force at wheel + force at handle = total weight force

you could do it this way and then subtract from the total weight - I think that'd also be perfectly valid.

**Joinedup**)That'd give you the force at the wheel.

force at wheel + force at handle = total weight force

you could do it this way and then subtract from the total weight - I think that'd also be perfectly valid.

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#6

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but why would u take the mass of d instead of b for eg 3 and 4

**runny4**)but why would u take the mass of d instead of b for eg 3 and 4

if you want to calculate the force at the wheel and subtract from the total it's the other way around - you'd multiply any mass at A and D by zero because they're both directly underneath the handle.

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(Original post by

if you're directly calculating the force at the handle you multiply any mass at B or C by zero - because B and C are directly above the wheel.

if you want to calculate the force at the wheel and subtract from the total it's the other way around - you'd multiply any mass at A and D by zero because they're both directly underneath the handle.

**Joinedup**)if you're directly calculating the force at the handle you multiply any mass at B or C by zero - because B and C are directly above the wheel.

if you want to calculate the force at the wheel and subtract from the total it's the other way around - you'd multiply any mass at A and D by zero because they're both directly underneath the handle.

B and C are directly above the wheel?- they still must add load to the suitcase?

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#8

(Original post by

but why does it matter if

B and C are directly above the wheel?- they still must add load to the suitcase?

**runny4**)but why does it matter if

B and C are directly above the wheel?- they still must add load to the suitcase?

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(Original post by

the case pivots around the wheel and when the masses are above the wheel the force due to gravity acts through the wheel. so the moment due to the weight of those masses is zero.

**Joinedup**)the case pivots around the wheel and when the masses are above the wheel the force due to gravity acts through the wheel. so the moment due to the weight of those masses is zero.

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