[EvenStevens]

Part c)

When the speed of the vehicles is the towbar breaks (OH NOEZ!). Assuming that the resistance to the motion of the car remains of constant magnitude 300N...

find the distance moved by the car from the moment the towbar breaks to the moment when the car comes to rest...

Alright, again, let's draw our diagram to help us visualise this...



Firstly... we need to know the deceleration of the car. This is going to have to be another F=ma job. Let's resolve again, because that's the direction the car is going.

So



Okay, so the car is accelerating at (I.e. it's decelerating at )

Notice I bolded the word "CONSTANT" in the question, this is a clue you can use SUVAT equations as SUVATs only work when the forces are constant.

Right, so now we know the (de)acceleration of the car, we can find out the distance moved before it comes to a halt..

So what do we know overall?
Looking back at the question, it was travelling at 6m/s, it rests at 0m/s and we've found the acceleration.

Initial velocity = u = 6
Final velocity = v = 0
Acceleration = a = -1/3
Displacement = s = ?

We can use

Substitute your values in...







Part d) to follow if requested.

If not, I'll leave it at that.

Hope this solves your connected particle trouble, Junker.

[Kyalimers]

Love your car

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

[Junker]

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

[12ian34]

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!

[Junker]

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.

[12ian34]

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

I really have no idea.

[Junker]

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.

[jdfan]

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: 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

[EvenStevens]

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...



, cancel the 'g's,

Therefore m = 12.

Is that right?

[Inquilaab]

(Original post by 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

[Kyalimers]

Any more questions??

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

[12ian34]

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

Thanks!

[Adjective]

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

[Kyalimers]

(Original post by Adje)
Ooh, I'm going to join the help squad.

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

Its my shift currently

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

[Junker]

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...

[Adjective]

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

[Junker]

(Original post by 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?

[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.

[Kyalimers]

(Original post by 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

[12ian34]

Here, the question asks to find out what X and Y are.

As it is in Limiting equilibrium (question says), Sum of moments = 0.

3g + 5g - X - Y = 0

(this is resolving vertically)

So, 8g = X + Y

Then, around X, (3g * 1) - (5g * 3) + (2 * Y) = 0

So, 2Y = 12g
So, Y = 6g

Then, sub into 8g = X + Y

So, 8g = X + 6g

X = 2g.

Ans: X = 19.6133N Y = 58.8399N
Now, which bit are you confused about, because some of those minuses can be tricky.

I think i have explained that correctly, I hope this helps.