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#1 2011-03-27 04:23:32

minimath
Guest

M1 Mechanics Resolving... Help!

Hi, I am really struggling with this particular example in this topic!

m1stuck.png

I am trying to understand how this example works.

I know they are resolving upwards, so the Qsin70° is because a right-angled triangle is formed at the top-right, so the 'height' is Qsin70°.

For the bottom, this is the part I don't understand. If the downward force is already 5 N, why are we finding 8cos75°??
They have said "To solve problems on the equilibrium of a particle, use the fact that the resolved forces in any two perpendicular directions amount to 0".

Can someone explain where the -8cos75° comes from?

We are taking "UP" to be positive, so we have Qsin70° - 5 = 0, since resolved forces in any two perpendicular directions amount to 0. The 5 is subtracted because it's in the opposite direction.

So why are we finding 8cos75°?

I would really appreciate it if someone could go through this problem step by step in detail, and specifically why they are doing each thing, because I really don't get this. I understand everything up to this point.

This is really frustrating! My M1 exam is in 52 days, and I only have 5 topics left in M1 to cover (there are 17 in total)...but this topic is really aggravating me. I really want to understand this so I can finish up M1 with complete understanding of it, so I can start doing past papers.

Thanks for your help

#2 2011-03-27 05:16:49

samuelbm
Member
Registered: 2011-03-26
Posts: 6

Re: M1 Mechanics Resolving... Help!

Hello minimath
I'll try my best to help you but english is my second language so I'm sorry for the mistakes

First to solve this problem we have to understand that because all the four vectors are equilebrate their total will be equal to zero. To give you an image, think of two people pulling on a rope to make the other felt, in fact as long as no one move we know that both use the same strengh on the rope but in opposite way so the total strengh is zero. (like in your problem)

The diffenrence in my problem and yours, mine just include vector parallel to the ground (or ''X's'' axis) but you have to mind vertical vector's too.

second of all, you have to understand vector can be split into two constituent, x and y.

So, let's try it!

horizontally:
Pcos0 + Qcos70 + 8cos195 + 5cos270=0
(cos270=0)-> Pcos0 + Qcos70 + 8cos195=0
(cos0=1)-> P+ Qcos70 + 8cos195=0

vertically:
Psin0 + Qsin70 + 8sin195 + 5sin270 = 0
(sin270=-1)-> Psin0 + Qsin70 + 8sin195 -5= 0
(sin0=0)->  Qsin70 + 8sin195 -5= 0

we last with those two equation:
P+ Qcos70 + 8cos195=0
Qsin70 + 8sin195 -5= 0

we can see that there is two unknown variable in the horizontally equation, the P and de Q
In fact it is impossible to solve an equation with two variables.
The second equation has only one unknown variable, the Q
so if we isolate it we got:

1) Qsin70 + 8sin195 -5= 0
2) Qsin70 + 8sin195 =5
3) Qsin70 = 5-8sin195
4) Q=(5-8sin195)/ sin70 ≈7,52

We just found the Q, now the P
we take the other equation and we introduce the knew value

1) P+ Qcos70 + 8cos195=0
2) P= -Qcos70 - 8cos195
3) Q=7,52
4) P=-7,52cos70 -8cos195≈5,15

so,
Q=7,52
P=5,15

Hope I've been useful, samuelbm

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