IIT JEE , AIEEE Forum - Physics , Mechanics - Tension in the top of a string over a pulley

konichiwa2x

Hi,

Consider a fixed pulley of mass M. A light inextensible string pass over it. The ends of a string are connected to blocks of mass m1 and m2 respectively. Let T1 and T2 be the tensions in string on the left and right side respectively. What is the tension of the string passing over the rim of the pulley??

gautam_agni_sharma

sorry for the incomlete answer as the force due to m1 is t1 andm2 is t2 then the net force actin on the pulley would be t1+t2

sarthakvij

i think the answers should be T1+T2

select_100

i think dat it would depend on the masses...say if m1>m2nd the tension on the string having m1 is T1 then tension of the string on the rim will be T1-T2

ayush007

well its a good question with simple logic .
condition 1: if there is friction between the rim and the string ---
T1=t2e^u(pi).
u=coeff. of friction.
^ :represents to power.

now,
resultant force of t1 and t2 acts on the rim; T=( t2^2+t1^2+2t1t2 )^1/2=t2(1+e^u(pi)).
condition 2: no friction :
T=t2+t1.

konichiwa2x

but how did you get this: "T1=t2e^u(pi)."

Can you please explain?

ayush007

In the figure below you could see the division of the components of tensions which tend to be different due to the friction between the rim and string. here the tensions are different not only due to the mass of the pulley but also the friction between the rim and belt.

(Normal direction)

(Tangential direction)

konichiwa2x

thank you!

r_misra

hey, ayush , this was what i was searching out for ...well explained.
thank you..

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