M2 Moments help

I am revising M2 for Jan, currently doing MEI M2 June 2006 paper.

There is a question on moments acting on two rods jointed together.

http://www.mei.org.uk/files/papers/m206ju_rw67.pdf Question 2i

There is one point which if I take CW moments I get the answer to be 35N, Mark Scheme says 15N, I will attempt to craw it out in ASCII (although the paper itself may make things easier)

A______P____B____R_____Q______|C

A + B are hinges
P + Q are centres of mass, (90 + 75N respectively)
R is a support
C is a verticle cable.

I have worked out B to have 60N upwards.

Taking moments CW about R I get:
0.5*60 + 1*75 -3C = 0

Solving to get C = 35N

The mark scheme says:
75*1 - 3C - 0.5*60 = 0
Solving to get C = 15N.

The only reasons for the difference I can think of is that B is to the left of R so the distance is negative but surely the moment still being CW means it should be positive.

Please help!

tazzik

OP

Shameless self bump.

It's not that the distantce 0.5 is considered negative (it doesn't matter which direction you measure the lever arm in, but whether the force would cause rotation in a cw or anti-cw direction that determines whether a moment is + or -), but you have the force acting on BC at B in the wrong direction. Consider the equilibrium of the hinge itself at B: if you have one force acting on it upwards (the force applied to the hinge by AB) of 60N, then for the hinge to be in equilibrium, what can you say about the force applied to the hinge B by BC? And how does this affect your calculation when you now take moments about R?

Reply 3

tazzik

OP

2

TheNack

It's not that the distantce 0.5 is considered negative (it doesn't matter which direction you measure the lever arm in, but whether the force would cause rotation in a cw or anti-cw direction that determines whether a moment is + or -), but you have the force acting on BC at B in the wrong direction. Consider the equilibrium of the hinge itself at B: if you have one force acting on it upwards (the force applied to the hinge by AB) of 60N, then for the hinge to be in equilibrium, what can you say about the force applied to the hinge B by BC? And how does this affect your calculation when you now take moments about R?

I'm not sure I understand, the first part tells me to calculate the force of hinge (B) on the rod AB (60N Upwards) so the force at B is 60N upwards, a cw moment about R no?

Reply 4

TheNack

tazzik

I'm not sure I understand, the first part tells me to calculate the force of hinge (B) on the rod AB (60N Upwards) so the force at B is 60N upwards, a cw moment about R no?

It's a little hard for me to explain without drawing you a diagram, but if you are considering the sections AB and BC separately, then you are effectively cutting through the hinge at B to create 2 separate systems. Its important to think about what the hinge is doing in the problem: it transfers horizontal (and vertical) forces between AB and BC. When you remove the hinge at B to consider AB and BC separately, you have to introduce 2 equal and opposite vertical forces to replace the hinge so that each of AB and BC is still in equilibrium. These 2 forces are "internal" to the system, that is they are the forces that B applies to AB and BC, and they cancel out at the hinge so that it is in equilibrium.

If B applies a force of 60N upwards on AB, what force must it apply on BC for there to be equilibrium? I hope this makes sense to you, ill try to draw a diagram to make it clearer.

Reply 5

tazzik

OP

2

OOOHHHH I see

Hence why the mark scheme calls for an understanding of N3L,

Excellent, thank you very much!

Reply 6

TheNack

tazzik

OOOHHHH I see

Hence why the mark scheme calls for an understanding of N3L,

Excellent, thank you very much!

What do you mean N3L? And here's my diagrammatical explanation, made with the help of paint:

Consider the whole system ABC:

Spoiler

Make a cut at B and consider AB and BC separately. The hinge at B transfers a horizontal force between AB and BC (it would also transfer a vertical force if any were applied to the system):

Spoiler

B_up is calculated by considering the equilibrium of AB about A (take moments about A and solve for B_up). In order for these internal forces to disappear when you put the hinge back (they are effectively reactions to the applied forces P and Q), there must be an equal and opposite force applied to BC:

Spoiler

Reply 7