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# M2 Moments on a bar with two supports watch

1. I'd like some help on this question. It's from MEI M2 Ex3A, Q3.

A uniform horizontal bar of mass 5kg has length 30cm and rests on two vertical supports, 10cm and 22cm from its left hand end. Find the magnitude of the reaction force at each of the supports.

I don't know how to start. Help appreciated, and will be repped!

Many thanks
2. You know the weight force will act 15cm from A. Now take moments about A to get an equatin involving the forces at C and D, and the weight. Then resolve vertically to get another equation involving the forces at C and D and the weight. Now solve your two equations simultaneously.
3. Ok, so there is a moment at A of
15*5g + 10C + 22D = 0, (1)
and vertically, C + D + 5g = 0 (2)

Since g = -9.8,
(1) -147 + 10C + 22D = 0
(2) C + D - 49 = 0

substituting (2) into (1),
C = 49 - D,
-147 + 10(49 - D) + 22D = 0
343 + 12D = 0, D = 28.6N

That's the right answer! I didn't realise it was that simple! I LOVE YOU TTOBY!!!
4. (Original post by Blackjamin)
Ok, so there is a moment at A of
15*5g + 10C + 22D = 0, (1)
and vertically, C + D + 5g = 0 (2)

Since g = -9.8,
(1) -147 + 10C + 22D = 0
(2) C + D - 49 = 0

substituting (2) into (1),
C = 49 - D,
-147 + 10(49 - D) + 22D = 0
343 + 12D = 0, D = 28.6N

That's the right answer! I didn't realise it was that simple! I LOVE YOU TTOBY!!!

Forgive me for reviving an old thread.. I must be missing something incredibly apparent here.. but..

How can 15*5g = -147
When g=-9.8
15*5*-9.8=-735..

And if the distance ought to be in metres;
0.15*5*-9.8=-7.35
if (1) is 15*5g +10C + 22D = 0
and 10C = 10(49-D) = 490 -10D
How does eq. (1) = -147 + 10(49 - D) + 22D =(-147+490)+12D= 343+12D

Since g=-9.8
15cm*5*g = -735
Which gives (1) as -735+490+12D= -245+12D=0
Thus D=245/12=20.416... which is unlikely.

Should not the distance be in Metres?
0.15m*5g=-7.35
(1) 482.65+12D=0
D=-40.22083, evidently wrong, can't have a negative force.

15*5*-9.8=-735
I do not understand

HOW am I getting this wrong?
Forgive me for reviving an old thread.. I must be missing something incredibly apparent here.. but..

How can 15*5g = -147
When g=-9.8
15*5*-9.8=-735..

And if the distance ought to be in metres;
0.15*5*-9.8=-7.35
if (1) is 15*5g +10C + 22D = 0
and 10C = 10(49-D) = 490 -10D
How does eq. (1) = -147 + 10(49 - D) + 22D =(-147+490)+12D= 343+12D

Since g=-9.8
15cm*5*g = -735
Which gives (1) as -735+490+12D= -245+12D=0
Thus D=245/12=20.416... which is unlikely.

Should not the distance be in Metres?
0.15m*5g=-7.35
(1) 482.65+12D=0
D=-40.22083, evidently wrong, can't have a negative force.

15*5*-9.8=-735
I do not understand

HOW am I getting this wrong?
Just had a look through the workings more closely. Yes, the 15*5g = -147 is a mistake - they must have forgotton to times by the 5. So maybe they were just lucky that their answer was correct.

However, it is not necessary in this question to convert all the distances to metres - any conversions that you do perform will get cancelled out at the end.

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