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# Edexcel M1 Moments Ex 5D Q4 (b) watch

1. Question referenced in title if you have the book but if not here it is:

A non-uniform rod AB of length 5m and mass 15kg rests horizontally suspended from the ceiling by two vertical strings attached to C and D, where AC = 1m and AD = 3.5m.

(a) Given that the centre of mass is at E where AE = 3m, find the magnitudes of the tensions in the strings.

(This part was fine for me, answers are Tc=3g and Td=12g)

(b) When a particle of mass 10kg is attached to the rod at F the rod is just about to rotate about D.

Find the distance AF.

Here is how my diagram looks:

Everything same as you'd have in part a, with a point F with the 10g (N) pointing down, drawn at a point between C and E. I called AF the distance x (m). Also, Tc=0 now, due to the rod being on the verge of rotating about D.

I tried taking moments about C, with respect to clockwise turning effects being positive, and ended up with: (10g)(x-1)+(15g)(2)-(12g)(2.5)=0. This gave x=0 which obviously is wrong.

I then tried taking moments about D, taking clockwise as positive and ended up with: -(15g)(0.5)-(10g)(3.5-x)=0. This gave x= 4.25m, correct answer... But I don't understand what was wrong with taking moments about C =/

Can anyone tell me what's wrong with taking moments about C?
2. (Original post by jamb97)
Question referenced in title if you have the book but if not here it is:

A non-uniform rod AB of length 5m and mass 15kg rests horizontally suspended from the ceiling by two vertical strings attached to C and D, where AC = 1m and AD = 3.5m.

(a) Given that the centre of mass is at E where AE = 3m, find the magnitudes of the tensions in the strings.

(This part was fine for me, answers are Tc=3g and Td=12g)

(b) When a particle of mass 10kg is attached to the rod at F the rod is just about to rotate about D.

Find the distance AF.

Here is how my diagram looks:

Everything same as you'd have in part a, with a point F with the 10g (N) pointing down, drawn at a point between C and E. I called AF the distance x (m). Also, Tc=0 now, due to the rod being on the verge of rotating about D.

I tried taking moments about C, with respect to clockwise turning effects being positive, and ended up with: (10g)(x-1)+(15g)(2)-(12g)(2.5)=0. This gave x=0 which obviously is wrong.

I then tried taking moments about D, taking clockwise as positive and ended up with: -(15g)(0.5)-(10g)(3.5-x)=0. This gave x= 4.25m, correct answer... But I don't understand what was wrong with taking moments about C =/

Can anyone tell me what's wrong with taking moments about C?
You've assumed that Td remains the same in part (a) and (b). But attaching a mass is going to change the tensions. And you can see that in the question since Tc has gone from 3g to 0.

So in (b), Td is now an unknown and Tc is 0 so the best thing to do is take moments about D so the unknown tension isn't part of your working.

(You can find Td by resolving vertically if you like but there's no point).
3. (Original post by jamb97)

I tried taking moments about C, with respect to clockwise turning effects being positive, and ended up with: (10g)(x-1)+(15g)(2)-(12g)(2.5)=0. This gave x=0 which obviously is wrong.
?
The tension in String D is going to change once the force F is added, you'll need extra tension in D to keep the thing in equilibrium. You'd need to resolve vertically to find the new tension, or more easily, resolve about D so that you can ignore the tension in D.

Edit: Nevermind... ninjaed. I swear this was unanswered for an hour and it gets two replies within 5 minutes.

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