Question

http://papers.xtremepapers.com/CIE/Cambridge%20International%20A%20and%20AS%20Level/Mathematics%20%289709%29/9709_s13_qp_51.pdf
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Answer 1

All you need to consider are the two masses of the rods given, AB and BC. No need to worry about the hypotenuse in this case.

The mass is proportional to the length of the rod, and is also proportional to the force exerted on it by gravity (i.e. its weight). If it takes 4N to support BC, meaning the weight of BC is 4N, then the weight of AB is 0.3/0.2*4 = 6N. The masses, therefore, are 4/9.8 =0.408 kg and 6/9.8 = 0.612 kg for BC and AB respectively.

For i, we're just interested in the horizontal distance of the centre of mass from AB, so we only need to consider the x-coordinate of each mass. To make things easy, we'll use AB as the reference point, so it's x-coordinate will be 0 m. Then, since BC is uniform, we can consider the entire mass of BC to be focused at a point halfway along its length. In other words, the x-coordinate of the mass of BC is -0.1 m (relative to the reference point). Now we can plug these numbers into the following equation for the x-coordinate of the centre of mass:

\[x_c={m_{AB}x_{AB}+m_{BC}x_{BC} \over m_{AB}+x_{BC}}\]
\[x_c={0.612 \times 0 + 0.408 \times (-0.1) \over 0.612+0.408}\]
\[x_c=-0.04\]
The centre of mass is located 0.04 m to the left of AB, which is exactly what you'd expect considering so much of the mass is concentrated so close to AB.

I agree with your answer!

Answer 2

@matt101 I dont get this part of your answer, "The masses, therefore, are 4/9.8 =0.408 kg and 6/9.8 = 0.612 kg for BC and AB respectively". appreciate your help and thanks :)

Answer 3

We want the masses for the centre of mass equation. We have the forces in newtons, and because these forces are actually the weights of each bar, I just divided by acceleration due to gravity to leave the mass.

Truthfully I could have plugged the forces rather than the masses into the centre of mass equation (though technically then we'd be finding the centre of gravity, although in this case it's located at the same point as the centre of mass). You'd get the same answer because the 9.8's of the numerator and denominator would reduce each other out, effectively leaving you with the masses anyways.

Answer 4

first, you shouldn't be bringing gravity into this.

imagine this shape was floating through space, you know, the thing that crashes into George Clooney's shuttle in the movie Gravity. it would still have mass, wouldn't it? and it would still have a centre of mass -- otherwise known on earth as centre of gravity. but here is no gravity so what does that mean for your methodology?

imagine now that the structure is 200 miles high - its tip about the orbit of the international space station. your calculation then would need to factor in the fact that gravity has changed over the length of the structure. it would also be wrong as the question asks for the centre of mass. mass is a property of matter.

to calculate the centre of mass, just worry about mass, and calculate it as a moment of mass. m1 • d1 + m2 • d2 = (m1+m2) • d. simples. you could probably do it in your head that way.

also, in terms of the mass itself, you should safely assume in this question that 1 m length of structure has mass 1kg. the structure is described as uniform. by assuming a unit mass (ie that 1 m has mass 1kg) you can forget totally about that as well as it is the same across the calculation. it could be 3 kg/m or 1,000 kg/m, it really doesn't matter, as it is the same for the entire structure.

the centre of mass for a *uniform* structure depends upon geometry, not density or gravity.

i hope that makes sense.