A spring balance inside a elevator has a pulley connected to it.The pulley has 1.5kg on one and 3kg on the other side.The elevator is going up with an accelaration of g/10.The pulley and string are light and the pulley is smooth.What is the reading of the spring balance.

pls explain the solution. 

i hope u can take out the acc. of the system and tension in the pulleys by using g effective( i.e. when acc. upwards geff= g +a

for downwards geff= g-a) instead of g

now the spring balance reading will be double the tension using constraint

Q) A spring balance inside a elevator has a pulley connected to it.The pulley has 1.5kg on one and 3kg on the other side.The elevator is going up with an accelaration of g/10.The pulley and string are light and the pulley is smooth.What is the reading of the spring balance.



Solution) Let m₁ = 1.5kg

m₂ = 3kg be the masses attached to the string passing over the pulley.

Acceleration of the string w.r.t elevator = a

If tension in hte string = T

Then we have following equations



m₁(g + a) = T or 1.5 (g + a) = T

and m₂ (g - a) = T or 3(g - a) = T

Solving which we obtain a= 10/3 m.s^-2 and T = 20N

Thus force on the spring balance through which pulley is attached = 2T = 40N

40N of force is equivalen to mass M = 40/g = 4kg

Now when elevator is moving up with acceleration g/10, then effective force on the spring balance = M (g + g/10)

= 4 (10 + 1) = 44N

Thus reading on spring balance corresponds to mass of 44N that is = 44/g = 44/10 = 4.4 kg

T = tension in the string

A = accelaration of the blocks

a = accelaration of elevator

g = accelaration due to gravity

 now as the total system is in a moving frame a pseudo force starts acting on the particles in that system which is equal to  {mass of respective body x accelaration of system}.

Hence pseudo forces acting are '3a N' and '1.5a N'

3a+3g-T=3A ----- I

T-1.5g-1.5a=1.5A -----II

Eliminating A from both equations

taking g = 9.8 and a = .98

we get T = 21.56 N

hence on the spring balance

2T=mg

you get  m = 4.4 kg