This discussion is closed.
Christophicus
Badges: 18
Rep:
?
#1
Report Thread starter 15 years ago
#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.
0
Christophicus
Badges: 18
Rep:
?
#2
Report Thread starter 15 years ago
#2
:confused:
0
Newton
Badges: 0
Rep:
?
#3
Report 15 years ago
#3
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.
0
X
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

What are you most likely to do if you don't get the grades you were expecting?

Go through Clearing (64)
41.03%
Take autumn exams (54)
34.62%
Look for a job (4)
2.56%
Consider an apprenticeship (4)
2.56%
Take a year out (21)
13.46%
Something else (let us know in the thread!) (9)
5.77%

Watched Threads

View All