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Mathematics: Post your doubts here!

A

awzm guy

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assalamoalaikum !!!!

im currently doing my AS Levels.

i have a dbt in Question no.4 paper 6(statistics 1) 2004 may/june Qp.
 
Haya Ahmed

Haya Ahmed

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aftsre.jpg

People how to sovle the (iii) part .. thanks :)
 
MafaldaC

MafaldaC

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Can someone help me please?
4

Particles A and B, ofmasses 0.2 kg and 0.3 kg respectively, are connected by a light inextensible string.

The string passes over a smooth pulley at the edge of a rough horizontal table. Particle A hangs freely

and particle B is in contact with the table (see diagram).

(i) The system is in limiting equilibrium with the string taut and A about to move downwards. Find

the coefficient of friction between B and the table. [4]

A force now acts on particle B. This force has a vertical component of 1.8N upwards and a horizontal

component of X N directed away from the pulley.

(ii) The system is now in limiting equilibrium with the string taut and A about to move upwards.

Find X. [3]
Cant answer question ii and whats the difference of A going down or up?
 
Thought blocker

Thought blocker

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MafaldaC said:
Can someone help me please?
4

Particles A and B, ofmasses 0.2 kg and 0.3 kg respectively, are connected by a light inextensible string.

The string passes over a smooth pulley at the edge of a rough horizontal table. Particle A hangs freely

and particle B is in contact with the table (see diagram).

(i) The system is in limiting equilibrium with the string taut and A about to move downwards. Find

the coefficient of friction between B and the table. [4]

A force now acts on particle B. This force has a vertical component of 1.8N upwards and a horizontal

component of X N directed away from the pulley.

(ii) The system is now in limiting equilibrium with the string taut and A about to move upwards.

Find X. [3]
Cant answer question ii and whats the difference of A going down or up?
Click to expand...
abcde credits. (Part ii)
i)
T = F and F = μR
T = μR
From the figure you can see T = 2N and R = 3N and μ = T/R = 2/3
tension-png.3793

ii)
As the system is in limiting equilibrium, the tension would equal A's weight i.e. 2N.
As A is about to move upwards, B is about to move rightwards. So the frictional force now will be in the same direction as the tension.
First, analyse the vertical forces.
In equilibrium, R + 1.8 = 3. => R = 1.2 N.
Considering forces horizontally, X = 2 + 2/3 (1.2) = 2.8 N.
 
Thought blocker

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Haya Ahmed said:
aftsre.jpg

People how to sovle the (iii) part .. thanks :)
Click to expand...
I just know one thing here :
f (x) = sin (x) / x

dy/dx = [ d/dx (sin (x) ) * x - (sin (x) ) * d/dx (x) ] / x²

dy/dx = [ cos (x) * x - (sin (x) ) ] / x²

dy/dx = [ xcos (x) - sin (x) ] / x²

dy/dx = ( x cos (x) / x² ) - ( sin (x) / x² )

dy/dx = ( cos (x) / x ) - ( sin (x) / x² )
 
Haya Ahmed

Haya Ahmed

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Thought blocker said:
I just know one thing here :
f (x) = sin (x) / x

dy/dx = [ d/dx (sin (x) ) * x - (sin (x) ) * d/dx (x) ] / x²

dy/dx = [ cos (x) * x - (sin (x) ) ] / x²

dy/dx = [ xcos (x) - sin (x) ] / x²

dy/dx = ( x cos (x) / x² ) - ( sin (x) / x² )

dy/dx = ( cos (x) / x ) - ( sin (x) / x² )
Click to expand...
LOL ... you didn't take numerical solutions ? :D
 
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