mechanics NOV 2007 Q7

[omerkc1]

mechanics NOV 2007 Q7......does anyone know how to solve it?

 

[khuhroproezaX]

Friction is 25N and is acting towards A... Which means the x component of 75N force is greater than the x component of TN force... so first of all find TN..
as specified it is about to move, so 25+T=75cos30
and solving it we find T, which is 80N
so.. now you need to find the force exerted by the slab/block on the ground to find the reaction force......
so now find the y component of 80N(T) and 75N force.. you would get something like 106.28N
Its weight is 20x10=200N
now to find the force exerted 200-106.28=93.22N
now the friction force is 25N and the formula is F=uR (where F is fricitonal force, u is coefficient and R is reaction force, which in normal case is massXgravitational force)
so 25/93.22 = 0.268 .. which i guess is the answer too ....

 

[khuhroproezaX]

my bad.. .its May/June not NOV sorry i apologize :S
wait i will do that in a moment

 

[khuhroproezaX]

Okay im leaving the first part its easy and i cant do it here.. i would recommend marking scheme if you find it difficult (well you shouldnt)
so for other parts here it is how it should be done..
The question says the object Q is moving downwards, which should mean P is moving upwards.... so this also means.. the Tension in the string is greater than the Frictional force Plus the downward force applied by the object P... T-F> the downward force...
downward force is 0.13gsin14.25 =0.32 N (sinQ=14/65)
now for the 3rd part.. first find the contact force by cos rule... 0.13gcos14.25 = 1.26 N
and F=uR so Frictional force is 0.6x1.26= .75N
so weight+Fricitonal force = 1.076N (0.13gsin14.25 =0.32 + |0.13gcos14.25 = 1.26x0.6=0.75| )
so now
arrange it this way:-
T-1.076=0.31a
1.1-T=0.11a (0.11g-T=0.11a)

(in above eq T-1.076 is because downward force is less than the upward force which is T, and in the 2nd eq... weight is more than T)
You get A=0.1ms^-2
I hope you get it.. if still there remains some room for confusion do let me know i will help you further....

 

[ammarelahi]

can anyone tell whats the direction of tension in a string when a weight is applied at its centre as in N09 P41 Q4 ??? Please !!

 

[ammarelahi]

anyone ?

 

[khuhroproezaX]

In such case tension would be a counteracting force, So it would apply in the direction of the string... Do you want me to solve the question for You.... I havnt done much i'm concentrating more on Chem :S

 

[ammarelahi]

yes please especially the last part by resolving components at P

 

[khuhroproezaX]

Okay first of all... resolve the vectors by finding out the angles first...
i hope you didnt face any problem with that.. but still i will do them
apply Sin (it is a triangle with 90 at one side and you can apply cos and tan too as you have everyside)
SinQ=P/H gives, sin^-1(40/50) =53.13
and then you draw a vertical line through P separating the components and the forces.
Now you find out the angle for both sides, they are 36.87 for 30cm and 53.13 for 40cm
so now they are in equilibrium that means...
Y component of both should equal to 5
write the equation
Xcos36.87+ycos53.13=5 (i have used alphabet X for 30cm force, and Y for 40cm force)
And then you can also conclude that
the X component should equal too!
Xsin36.87=ysin53.13
Find X and Y
they are X=4 and Y=3
For second part:
You can see that component Y(40cm wali force) is holding on ring S, that means x component of Y it is exerting on P is also S's Frictional Force! (please dont confuse between force X and component x)
So now find the x component of Y, which should be like 3sin53.13 = 2.4N
so now moving on to the 3rd part..
You know F=uR where F is frictional force, u coefficient, and R reaction force)
You found out the friction force in part two, we will use it here!
2.4=0.6R
Find R which is 3.2N
Normally in cases when there are no external forces, R is weight BUT over here there do exist external forces.
The y component of Y force is the force... Understand that weight and Y component both are acting downwards on the RING...
so R is actually Weight + y component..
find out y component which is 1.8 so form an equation as 1.8+W=3.2
now you get W= something as 1.4 wait its exact 1.4 so yeah there you go!
need any further assistant let me know.. IF you find an error tell me... i'm so really tired and exhausted so..

 

[yubakkk]

can any one explain me q.5 i of may 2006
and
may 2004 q.4 why 2.5 is taken as hypotenus
please help me who knows////

 

[omerkc1]

man in Q4 may 20004 its mentioned in the question that the thing slides 2.5m to the bottom....
as for Q5 2006....the force exerted by the pully is 4(2)^1/2 .....so it means that it is the normar reaction force the pully has when the tension acts on it ......so you should get the answers like.........
2(Tcos45)=4(2)^1/2......note that we used "2"tcos45 becaue there are two components of tension acting in the same direction..........

 

[ammarelahi]

thanks but isnt the y-component of of tension in PS acting upwards on the ring S ??

 

[ammarelahi]

can anyone help me with P43 J10 Q7 ??

 

[ammarelahi]

anybody? here is the link http://www.xtremepapers.me/CIE/Internat ... _qp_43.pdf

 

[arlery]

Q7 (i) Total R = m1g + m2g
=200g + 250g
= 2000+ 2500 = 4500 N
Friction = 3150
so 3150 = COF*4500
so COF = 4500/3150 = 0.7

(ii) Friction = 2000* 0/2
friction = ma
400 = ma
400/200 = a
a = 2 m/s^2
(ii) P - Friction = total mass * acceleration
Pmax = 3150 + 450(2)
= 4050