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# Mechanics Help! watch

1. (Original post by 2776)
Ive found the value of t when p returns to O. It is 0 or 2.

So it must be 2.

Put that into equation, I get, v= -8.

Does that make sense? as -8 means it is reversing, so it must be true.
yeah that's right.
2. (Original post by king of swords)
yeah that's right.
Also Ive found out that t= 6/4 is the time for p to reach A.

But how do u find the maximum speed whilst P is travelling from O to A?
3. (Original post by 2776)
Also Ive found out that t= 6/4 is the time for p to reach A.

But how do u find the maximum speed whilst P is travelling from O to A?
for v to be a maximum, dv/dt = 0 and d2v/dt^2 = -ve....for a maximum.
4. (Original post by king of swords)
for v to be a maximum, dv/dt = 0 and d2v/dt^2 = -ve....for a maximum.
Imagine a rectangle. At one edge is a pulley. A is on teh rectangle atached to B which is dangled over the pulley. Each is 0.6kg. Now I tried finding the tension in teh string/acceleration as the B falls. But all I found is that 1.2a = 0?

i.e.

0.6a= T-0.6g

0.6a = 0.6g -T

so when u add them together u get 1.2a = 0.
5. (Original post by 2776)
Imagine a rectangle. At one edge is a pulley. A is on teh rectangle atached to B which is dangled over the pulley. Each is 0.6kg. Now I tried finding the tension in teh string/acceleration as the B falls. But all I found is that 1.2a = 0?

i.e.

0.6a= T-0.6g

0.6a = 0.6g -T

so when u add them together u get 1.2a = 0.
Is the rectangle rough if so what is the value for μ?
6. (Original post by king of swords)
Is the rectangle rough if so what is the value for μ?
rectangle is smooth, btw ive given u rep, for being sooooo helpful.
7. (Original post by 2776)
rectangle is smooth, btw ive given u rep, for being sooooo helpful.
well:
0.6a = T - 0.6g --->the vertical bit
T = 0.6a --->the horizontal bit (ouch: 0.6a = 0.6a - 0.6g )
1.2a = 2T - 0.6g
8. (Original post by king of swords)
well:
0.6a = T - 0.6g --->the vertical bit
T = 0.6a --->the horizontal bit (ouch: 0.6a = 0.6a - 0.6g )
1.2a = 2T - 0.6g
forget that:
0.6a = 0.6g - T....better
T = 0.6a
so T + 0.6a = 0.6g = 1.2a
so a = 0.5g so T = 0.3g
9. (Original post by king of swords)
well:
0.6a = T - 0.6g --->the vertical bit
T = 0.6a --->the horizontal bit (ouch: 0.6a = 0.6a - 0.6g )
1.2a = 2T - 0.6g
Nope, no more,
10. (Original post by king of swords)
forget that:
0.6a = 0.6g - T....better
T = 0.6a
so T + 0.6a = 0.6g = 1.2a
so a = 0.5g so T = 0.3g
How much rep did i give u? check
11. (Original post by 2776)
Nope, no more,
yeah i have look at my reply to correct that post a little further up.
12. (Original post by 2776)
How much rep did i give u? check
7.
13. (Original post by king of swords)
7.
Last ques: Which school do u go to?
14. (Original post by 2776)
Last ques: Which school do u go to?
Bradford Grammar School....i'm guessing you are at Leeds.
15. (Original post by king of swords)
Bradford Grammar School....i'm guessing you are at Leeds.
Ur right!. U know adshur is at BGS as well
16. Has this whole thread been based on that first post? If so, surely that must be the longest thread to answer 1 question?
17. (Original post by 2776)
Ur right!. U know adshur is at BGS as well
Yeah she's my best friend lol .
18. (Original post by king of swords)
Bradford Grammar School....i'm guessing you are at Leeds.
Right, last ques: when A has moved a distance of 2 metres it becomes detached from the string. From this instant B takes a further 0.2 seconds to reach the floor. Find h. (i.e. the distance between B and the floor from rest)
19. (Original post by king of swords)
Yeah she's my best friend lol .
lol

I have found that after moving 2 m, the velocity is root 2g.
20. (Original post by 2776)
Right, last ques: when A has moved a distance of 2 metres it becomes detached from the string. From this instant B takes a further 0.2 seconds to reach the floor. Find h. (i.e. the distance between B and the floor from rest)
When A moves 2m B also moves 2m. From then on B accelerates only due to gravity since there is no tension left in the string.
So:
s = ?, u = ?, a = g, t = 0.2
so s = tu + ½at^2.....then h = s + 2.

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Updated: December 7, 2003
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