find the acceleration of the block of mass M in the situation shown in figure. All the surfaces are frictionless and the pulleys and the string are light.

Explanation

`M_a-2T = 0. T+Ma-Mg=0`
`rarr Ma=2T`
`rarr (Ma)/2+ma=Mg (because T= (Ma)/2)`
`gt T= (Ma)/2`
` rarr3Ma=2Mg`
` rarr a=(2g)/g`
)
a. Acceleration of mas, `
M= (2g)/3`
b. `Tension, T= Ma/2= M/2 xx (2g)/3=Mg/3`
c. Let, `R= resultant of tensions
= force exerted by teh clamp on the pulley
` :. R'= sqrt(T^2+T^2)= sqrt(2) t`
`:. R= sqrt(2) T= ((sqrt(2) Mg)/3`
`again, tan theta = T/T=1`
`rarr theta = 45^0`
So it is `(sqrt(2)Mg)/3` at an angle of `45^0` with horizontal.