M1 - Edexcel, 21st May - ULTIMATE THREAD

Love your car :biggrin:

Ask me in a PM to takeover the helper role when you get bored or tired :smile:

Reply 82

That's amazing EvenStevens. Yeah, you've helped so much! Thanks for taking your time to help. More +rep when I can.

Reply 83

I was just doing some practice from my M1 book on moments, and I got stuck on this question. (Heinemann exc 6B q9 p143)

"A uniform rod AB has length 6m and mass 4 kg. It is resting in equilibrium in a horizontal position on supports at points X and Y where AX = 2m and AY = 4.5m. A particle of mass M kg is placed at point C where AC = 5m. Given that the rod is on the point of tilting about Y, calculate the value of M."

I guess i'm stumped because no forces have been given except the downwards 4g.

Any help would be much appreciated as other than moments, i feel i am pretty solid on M1!

Reply 84

Okay. So rod is at point of tilting about Y therefore reaction is at X=0 so you can ignore that and you do have another downward force, Mg.

Reply 85

So, (4g x 3) + (5mg) = something

I really have no idea.

Reply 86

Okay you don't really need to know the reactions so it's best if you take moments at Y. Remember X=0 so ignore that. So:
Moments about Y is (0.5*Mg) = (1.5*4g)
Equal because clockwise moment = anticlockwise moment.
0.5Mg = 6g
"g" can cancel out because it is in each term.
0.5M=6
M = 12
I hope this helps.

Reply 87

jdfan

8

A boy kicks a football vertically upwards from a height of 0.6m above the ground with a speed of 10.5m/s.

a) Find the greatest height above the ground reached by the ball

For this I got 6.225m

b) Calculate the length of time for which the ball is more than 2m above the ground

I did this part like so:

v^2 = 10.5^2 - (2 \times 9.8 \times 1.4)

to find the velocity of the particle at 2m (9.1m/s) and then found the time from this point to the top to be 0.93 which I multiplied by two to get an answer of 1.86s. Is there a quicker way than this?

Thanks

Reply 88

EvenStevens

OP

17

If it's on the point of tilting but actually not tilting, it's still in equilibrium but the normal on the X pivot is 0.

So if you take moments around pivot Y clockwise, you should get...

-(1.5 \times 4g) + (0.5 \times mg) = 0

0.5mg = 6g

, cancel the 'g's,

0.5m = 6

Therefore m = 12.

Is that right?

Reply 89

Inquilaab

9

Gimothy

I have a hard copy. I don't really want to type all the questions out, but if you tell me a topic ill find the relevant question for you.

Never mind , I just found it on TSR.
Here's the link

Reply 90

Any more questions??

You can find papers and respective MS's on www.freewebs.com/sohanshah

Reply 91

Thank you Junker and Even Stevens, the two end of the rod A and B and all of the distances were confusing me.

Thanks!

Reply 92

Adjective

Ooh, I'm going to join the help squad.

Reply 93

Adje

Ooh, I'm going to join the help squad.

Me you and EvenStevens it is. We'll keep a rota :biggrin:

.

Its my shift currently :smile:

Having said that, I will also do it until 3 am approx :biggrin:

Reply 94

I've forgotten moments! My question is on moments where there are 2 supports with reactions or vertical cables, how do you tell the clockwise moments and anticlockwise moments apart. I can't believe I forgot this but i'm just blank...

Reply 95

Adjective

I'm giving sohanshah a chance to answer before I pounce :p:

Reply 96

Adje

I'm giving sohanshah a chance to answer before I pounce

Please pounce, I really need to know. Is it just the way the arrows are?

Reply 97

Adjective

You should always take moments from one of the supports. Any forces acting in opposite directions on opposite sides of the support from which you're taking the moment, act to tilt the rod in the same way. You treat the other support as you would treat any other force, and it acts in the 'obvious' direction - i.e. the direction that would hold the rod up. Normally it's this force you're expected to find initially.

Reply 98

Adje

You should always take moments from one of the supports. Any forces acting in opposite directions on opposite sides of the support from which you're taking the moment, act to tilt the rod in the same way. You treat the other support as you would treat any other force, and it acts in the 'obvious' direction - i.e. the direction that would hold the rod up. Normally it's this force you're expected to find initially.

Good job you did pounce. I've received saddening news and am trying to conquer it: freeexampapers is trying not to be shut down :frown:

Reply 99