IIT JEE , AIEEE Forum - Physics , Mechanics - H C Verma problem on force (Newton's laws)

priyesh

FOR QUESTION 31

let the tension in the sring connected to block of mass 2M be 2T

therefore my making FBD of pulley(connected with the block of mass M) we see that tension in the string connected to block of mass M is T

now let acceleration of block of mass 2M be a leftwards

therefore by constraint relation acceleration of block of mass M will be 2a downwards.

therefore equation for block of mass 2M is

2T = 2Ma --------------------------------------------------- 1

equation for block of mass M is

Mg - T = M(2a) => Mg - Ma = 2Ma (putting value of T from 1) => a = g/3

therefore acceleration of block of mass M is 2a = 2g/3 ans of a)

also by putting value of a in equation 1 we get T =Mg/3

therefore tension in string connected with block of mass M is T = Mg/3

& tension in string connected with block of mass 2M is 2T = 2Mg/3 = ans. of b)

now let the force exerted by clamp = F

now since net force on pulley has to be zero

making fbd of pulley A (see fig.)

sqrt(T^2 +T^2) -F = 0

=> F = root(2) * T (ans of c)

also tan(theta) = T/T = 1

therefore theta = 45 degrees

hence the force applied by the clamp is at angle of 45 degrees with horizontal