Mechanics!

[Zenzenzen]

I was wondering if someone could help me with this M1 problem:

http://www.xtremepapers.me/CIE/International A And AS Level/9709 - Mathematics/9709_s10_qp_43.pdf

I have read the mark scheme, but I still do not understand parts 2 and 3. All help appreciated!

 

[arlery]

(i) Firstly total mass is 200+ 250 = 450g
Resultant = 4500.
Friction = 3150
so 3150 = 4500*muu
so muu = 3150/4500 = 0.7

F= ma
C.O.F. is 0.2 so Friction will be 4500*0.2 = 900N

F= ma
900 = 450 a
a = 2 m/s^2

(iii) total C.O.F. is 0.2 +0.7 =0.9
So friction = 0.9*4500 =4050N
therefore, P max = 4050 N

 

[Zenzenzen]

F= ma
C.O.F. is 0.2 so Friction will be 4500*0.2 = 900N

Shouldn't we take the R from only the top box in this equation? Wouldnt it be 2000*0.2?
Why are we taking R from both of the boxes?

 

[arlery]

No, because this is the C.O.F. of between the boxes, hence the R will total mass*g.

 

[Zenzenzen]

So no matter where the COF is we always take the total R? Like when its on bottom and middle, R stays the same.
Thanks, and by the way did you understand that other pulley problem?

 

[arlery]

Yes we take total R because if it was just one box on the floor, then we took R by multiplying C.O.F. by mg, here though, since there are two boxes, the C.O.F. will be multiplied by the total R. So yeah, the R stays same.

Yeah I did, but I don't understand this question from Nov 2010,p 42, Q. 3

 

[Zenzenzen]

Thats the one I was talking about! I really hope something like this does not come on the test tomorrow >_>

 

[arlery]

True same here. Can you please solve it for me?

 

[Zenzenzen]

Sigh... looks like im failing tmws test.

 

[PDallas]

Question 3 from P41 is quite easy if you move the referential. Then you just solve for x and for y

x: W = 7.3 x sin (x) (=) x: W = 4.8N
y: 5.5 = 7.3 x cos (x) (=) y: x = 41.1

Hope you got it
Don't woory the exam is not that hard

 

[arlery]

I still don't get it.

 

[Zenzenzen]

Question 3 from P41 is quite easy if you move the referential. Then you just solve for x and for y

x: W = 7.3 x sin (x) (=) x: W = 4.8N
y: 5.5 = 7.3 x cos (x) (=) y: x = 41.1

Hope you got it
Don't woory the exam is not that hard

Thanks! I finally understood it! This is more trigonometry than MECHANICS. Haha

 

[arlery]

There's an easier way.

7.3^2 = 5.5^2 + W^2
7.3^2-5.5^2 = W^2
W = square root answer
W = 4.8 N

 

[abrraza]

(i) Firstly total mass is 200+ 250 = 450g
Resultant = 4500.
Friction = 3150
so 3150 = 4500*muu
so muu = 3150/4500 = 0.7

F= ma
C.O.F. is 0.2 so Friction will be 4500*0.2 = 900N

F= ma
900 = 450 a
a = 2 m/s^2

(iii) total C.O.F. is 0.2 +0.7 =0.9
So friction = 0.9*4500 =4050N
therefore, P max = 4050 N

that isnt Resultant that is Reaction or Normal..

 

[arlery]

What do you mean?

 

[abrraza]

i mean that 4500 is reaction force not resultant .. i hope u get it

 

[arlery]

Oh yeah, sorry my mistake, I wrote resultant by mistake, I meant the normal force.