All Mechanics DOUBTS HERE!!!

[hesho21]

Guys plz question no.3 in the attached paper here . It made me madd!! Thanx in advance

[nid404]

Here you go....

Since the system is in equilibrium, the sum of horizontal components should be equal to zero.
That is

W₂sin60^o- W₁sin40= 0   Call this eq 1

The sum of vertical components should also be zero         
W₁cos40+ W₂cos60 - 5N= 0                            Call this eq 2

Now you rearrange eqn 1 to get either W1 in terms of W2 or the other way round. I'll go with the first

Eq 1 W₂sin60^o= W₁sin40
W₁=W₂sin60^o/ sin40

Substitute W1 in eqn 2

W₂sin60^o/ sin40 (cos 40)  + W₂cos60= 5
 1.53 W₂= 5
  W₂=3.26N

Now substitute the value of W₂ in eq 1

W₁sin40= 3.26 sin 60
W₁=4.4N

[immortal]

Ur method is confusing,what happens if dey r not in equilibrium
By the way is it ok 2 solve this using da data 2 form a triangle.
V wud get   5/sin80=W₁ /sin60=W₂/sin40.

[nid404]

You can use the triangle method too...

[hesho21]

Thanks sooo much that really helped  and ure explanation was great, good luck

[solo_G]

okay guys i have a doubt but its for cambridge M1..its from nov 09 varient one.....Qs 6 part (iii)........i cant get why was U substituted as -2 in the equation s=Ut+1/2 at*2?! can someone explain the entire question cuz i keep getting the required time in negative....i would appreciate any help.thanx in advance

[astarmathsandphysics]

Something wrong with that entire paper. My copy too. Just picures with no questions

[solo_G]

Something wrong with that entire paper. My copy too. Just picures with no questions

sorry about that.....imma upload another copy

[solo_G]

sorry people but i have another doubt...its from CIE nov 08...the first question part (i) and part (ii).....i know its just a two marks question but its driving me crazy.....thanks in advance.

[astarmathsandphysics]

Will try to answer it 1st thing in morning. To bed now

[hesho21]

So for the 8N force
parallel to the 10 N force u have 8cos(theta)
perpendicular to the 10N force u have 8sin(theta)

Now u want to find the resultant
(a) parallel to the 10 N force ==> 10-8cos(theta)
(b)perpendicular is just 8sin(theta)

Hope it helped

[solo_G]

So for the 8N force
parallel to the 10 N force u have 8cos(theta)
perpendicular to the 10N force u have 8sin(theta)

Now u want to find the resultant
(a) parallel to the 10 N force ==> 10-8cos(theta)
(b)perpendicular is just 8sin(theta)

Hope it helped

aha .....that was helpful hesho21.....i got it now....makes sense thanks man!

[falafail]

sorry people but i have another doubt...its from CIE nov 08...the first question part (i) and part (ii).....i know its just a two marks question but its driving me crazy.....thanks in advance.

(i)(a) you have to resolve the forces horizontally, so its 10 - 8cos?
(i)(b) you have to resolve the forces vertically, so it's 8sin?

(ii) the x-component squared + the y-component squared will = the resultant squared.
(10 - 8cos?)²  +  (8sin?)²  =  8²
then just simplify and you'll get cos? = 5/8

[solo_G]

(i)(a) you have to resolve the forces horizontally, so its 10 - 8cos?
(i)(b) you have to resolve the forces vertically, so it's 8sin?

(ii) the x-component squared + the y-component squared will = the resultant squared.
(10 - 8cos?)²  +  (8sin?)²  =  8²
then just simplify and you'll get cos? = 5/8

yeah falafail.....i got it and i did the last question just like you said....thanks for replying ...ehm ehm i wana ask smthn else....in nov09 varient 1 question 6 part(iii).......can u explain?!

[falafail]

yeah falafail.....i got it and i did the last question just like you said....thanks for replying ...ehm ehm i wana ask smthn else....in nov09 varient 1 question 6 part(iii).......can u explain?!

right okay,
so in part (ii) you found the height above the ground of P and Q, and their speed, when the string breaks.
that's s_P= 3 m, s_Q= 7 m, and their speed is 2 ms^-1.
you use the equation s = ut + 1/2 at², except here they're accelerating due to gravity only so a=10
so for P:
3 = 2t + 1/2 (10)t²
solving for t you get t = 0.6 or t = -1 (obviously the second is invalid)
for Q:
7 = -2t + 1/2 (10)t²
here it's -2 because it's in the opposite direction to P. remember this is velocity, you have to take into consideration the speed AND the direction.
so yeah, solve for this one you get t=1.4 or t=-1

1.4 - 0.6 = 0.8