#1
This may seem like a really stupid question (especially as the exam is just over a week away) but...
When trying to use F=MA I found myself asking what the F actually means with regards to 'where' it is in the problem.

Could do a) but not the rest..

0
9 years ago
#2
(Original post by Closer*)
This may seem like a really stupid question (especially as the exam is just over a week away) but...
When trying to use F=MA I found myself asking what the F actually means with regards to 'where' it is in the problem.

Could do a) but not the rest..

wouldnt f = (the downwards force of B) - (the frictional force at A)

Thats what i would try!
0
9 years ago
#3
(Original post by Closer*)
This may seem like a really stupid question (especially as the exam is just over a week away) but...
When trying to use F=MA I found myself asking what the F actually means with regards to 'where' it is in the problem.

Could do a) but not the rest..

Try drawing free-body force diagrams for particles A & B. You can then use F=ma three times:
-Vertically on A
-Horizontally on A
-Vertically on B

Coupled with you should now be able to solve for tension and acceleration.

By the way, this should really be in the Maths Academic Help forum, rather than the university course forum. I've let a mod know and they should move it for you.
0
#4
Oo yeh dint even realise that id posted it there!

Thanks for the help i will rep you both!

but I dont understand wether to minus one weight and add another orrr
0
9 years ago
#5
F = Force!
0
9 years ago
#6
F = Resultant force
0
#7
(Original post by wow09)
F = Resultant force
OOoooo i see! il have another go thankyou!
0
9 years ago
#8
(Original post by wow09)
F = Resultant force
You owned me sir!
REP!
0
9 years ago
#9
Ok. Consider particle B. It is moving down, so we'll let downwards be our positive direction. On your diagram, you should have 2 forces acting on B. Weight (or mbg) downwards and tension (T) upwards. It is moving downwards with acceleration a so F=ma gives us:

(Forces acting down) - (forces acting up)=mass * (acceleration acting down)
mbg-T=mba

---

Now on A: A is moving also with acceleration a to the right, so we'll let right be our positive direction. On your diagram, you should have 4 forces acting on A. Tension to the right, friction (f) to the left, weight (mag) acting down and reaction force (R) acting up. So F=ma horizontally gives us:

(forces acting right) - (forces acting left)=mass * (acceleration acting right)
T-maa=maa

A is not moving in the vertical direction so its acceleration here is 0. Can you write down this equation on your own? Can you finish it off from here?
0
#10
Thankyou Agrippa, it really has made things clearer

I get 1.4-T= 0.5a and 0.2g-T=o.2a but i dnt seem to get the right answer.. (my teachers have it as 0.8)
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#11
can anyone tell me where i've gone wrong?
0
9 years ago
#12
It should be T - 1.4 = 0.5a
0
#13
(Original post by .x.alexa.x.)
It should be T - 1.4 = 0.5a
Thankyou so much! lol that was annoying me for ages!!
0
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