[goggy]

I'm going through past papers and this question is really confusing me.

A rectangular frame consists of four uniform metal rods. AB and CD are vertical and each is 40cm long and has mass 0.2kg. AD and BC are horizontal and each is 60cm long. AD has mass 0.7kg and BC has mass 0.5kg. The frame is freely hinged at E and F where E is 10cm above A, and F is 10cm below B.

(i) Sketch a diagram showing the directions of the horizontal components of the forces acting on the frame at E and F.

(see attachment)

(ii) Calculate the magnitude of the horizontal component of the force acting on the frame at E.

It would be pointless if I found the moment at E, so I assume I have to take it at F.

This is where I'm confused. :s I have the markscheme and it says it is "0.5x30g + 0.7x30g + 0.2x60g = 20X". Does this mean one of the rod's masses don't matter? Why is it 0.5x30g? Isn't BC 10cm away from F? Any help would be appreciated.

[Roy064]

(Original post by goggy)
This is where I'm confused. :s I have the markscheme and it says it is "0.5x30g + 0.7x30g + 0.2x60g = 20X". Does this mean one of the rod's masses don't matter?

Yes, the mass of AB doesn't matter as the mark scheme is taking moments around either E or F, therefore the moment of the weight of AB around E or F is 0 as the line of action goes through the points E and F.

(Original post by goggy)
Why is it 0.5x30g? Isn't BC 10cm away from F? Any help would be appreciated.

It's the line of action again, uniform rod means the weight acts in the centre of the rod, and the moment around either E or F is 30cm away from the line of action of the weight of BC and AD.

I could draw up some extra diagrams if you need extra help?

[goggy]

(Original post by Roy064)
Yes, the mass of AB doesn't matter as the mark scheme is taking moments around either E or F, therefore the moment of the weight of AB around E or F is 0 as the line of action goes through the points E and F.



It's the line of action again, uniform rod means the weight acts in the centre of the rod, and the moment around either E or F is 30cm away from the line of action of the weight of BC and AD.

I could draw up some extra diagrams if you need extra help?

That makes sense. And yes, if you would be so kind! I would appreciate it greatly.

[Roy064]

(Original post by goggy)
That makes sense. And yes, if you would be so kind! I would appreciate it greatly.

Yeah sure, here are the diagrams of why each moment is included in the equation the way it is (quite a big picture) http://i.imgur.com/V65i1.jpg

From what I can see the confusion lies in why the mass of AB doesn't count, so a simple way of thinking of it is:

moments are being taken around F, so the turning point is F.

So if you apply the other forces (weights of the other rods) to a straight rod hinged at F (imaginary, not included on the diagram but you can draw your own to visualise it a little better), they will turn that said rod clockwise, and the horizontal force at E will turn it anti-clockwise. The weight of AB is pushing on the actual point, so will have no effect, or to apply the theory a little, the perpendicular distance from the weight of AB to F is 0, so has no 'turning effect'.