You are Here: Home >< Maths

# M3 question watch

1. A and B are two fixed points on a smooth horizontal table with AB = 1.5m
A particle P is in equilibrium, attached to A by means of a light elastic strting of natural length 0.5m and modulus of elasticity lamda N and to B by means of a light elastic string of natural length 0.75m and modulus of elasticity 2 lamda N. Given that the tension in AP is 10N, find the value of lamda.
2. AP a=(1/2)m, L=LN, x=xm
BP a=(3/4)m, L=2LN x=(0.25-x)m

If the particle is in equilibrium both of the strings have equal tension

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

10=2Lx (i)
10=(8/3)L(0.25-x) (ii)

(i)=>5=Lx (iii)
(ii)=>(15/4)=0.25L-Lx=>Lx=0.25L-3.75 (iv)

(iii)/(iv)=>0.25L-3.75=5=>0.25L=8.75=>L=35N

Newton.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: February 6, 2005
Today on TSR

### University open days

Wed, 21 Nov '18
• Buckinghamshire New University
Wed, 21 Nov '18
• Heriot-Watt University
Wed, 21 Nov '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams