# M3 questions

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#1
A and B are two fixed points on a smooth horizontal table. A particle P is in equilibrium, attached to A by means of a light elastic string of natural length l and modulus of elasticity lamda and to B by means of a light elastic string of natural length 3l and modulus of elasticity lamda. Given that the extension in the string AP is 0.5l, find the distance AB.

A particle P of mass 2kg is attached to the mid-point of a light elastic string of natural length 2m and modulus of elasticity 20N. One end of the string is attached to point A of a ceiling and the other end is attached to point B on the floor 3m vertically below A. Calculate the tension in AP and the tension in PB.

A ball of mass 0.5kg is attached to one end of a cord of unstretched length 0.6m whose other end is fixed. When a horizontal force of magnitude QN is applied to the ball holding the ball inn equilibrium, the cord increases its length by 10% and is inclined at an angle of arcsin 3/5 to the downward vertical. Calculate the value of Q and the modulus of the cord.

I found 1-11 okay but 12-15 got me.

I really appreciate this.
0
15 years ago
#2
i)AP a=l L=L x=(1/2)l

BP a=3l L=L Let d be the distance AB=>x=d-(9/2)l

Equilibrium=>equal tensions

T=((L(1/2)l)/l)=(1/2)L (i)
T=((L(d-(9/2)l))/3l) (ii)

(i)=(ii)=>(1/2)L=((L(d-(9/2)l))/3l)=>(3/2)l=d-(9/2)l=>d=6l

Newton.
0
#3
(Original post by Newton)
i)AP a=l L=L x=(1/2)l

BP a=3l L=L Let d be the distance AB=>x=d-(9/2)l

Equilibrium=>equal tensions

T=((L(1/2)l)/l)=(1/2)L (i)
T=((L(d-(9/2)l))/3l) (ii)

(i)=(ii)=>(1/2)L=((L(d-(9/2)l))/3l)=>(3/2)l=d-(9/2)l=>d=6l

Newton.
Thankyou, could you explain the part in bold?
0
15 years ago
#4
(Original post by Widowmaker)
Thankyou, could you explain the part in bold?

A_________________P_____________ _______________B
|<-------->|<----->|<--------->|<---------------->|
_____l______ (1/2)l______x_____________3l_______ __

d=l+(1/2)l+x+3l=>x=d-(9/2)l

Newton.
0
#5
(Original post by Newton)
A_________________P_____________ _______________B
|<-------->|<----->|<--------->|<---------------->|
_____l______ (1/2)l______x_____________3l_______ __

d=l+(1/2)l+x+3l=>x=d-(9/2)l

Newton.
Ah I see.
There are two other questions if you have time.
0
15 years ago
#6
ii) Let the tension in AP=T1 and PB=T2

L=20N, x=1m, a=2m

=>T=(Lx/a)=10N

By looking at the situation you have an obvious systematic equation

T1=T2+2g => T1-T2=2g (i)

The net tension in each part of the string, T=10N

For the top half T1-2g=10 (ii) and for the bottom T2+2g=10 (iii)

(ii)+(iii)=>T1+T2=20 (iv)

(i)+(ii)=>2T1=39.8=>T1=19.8N

Substituting back into (iv):

T2=20-19.8=0.2N

Newton.
0
15 years ago
#7
[email protected]=(3/5)=>[email protected]=(4/5), [email protected]=(3/4)

The string is unstretched when l=0.6m=>a=0.6m

After the force has been applied:

Vertically:

[email protected]=mg=>(4/5)T=0.5g=>T=0.625gN=6.125

10% of 0.6m=0.06

Therefore:

(L0.06/0.6)=0.1L=6.125=>L=61.25N

Horizontally:

[email protected]=Q

=>(3/5)(6.125)=Q=3.675N

Newton.
0
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