Question

Consider the situation shown in figure (5−E14). Both the pulleys and the string are light and all the surfaces are frictionless. (a) Find the acceleration of the mass M; (b) find the tension in the string; (c) calculate the force exerted by the clamp on the pulley A in the figure.

Answer 1

Given :

Let the acceleration of mass  M be a.

So, the acceleration of mass 2Mwill be  a/2.

(a) 2M(a/2) − 2T = 0

⇒ Ma = 2T

T + Ma − Mg = 0

⇒Ma/2+Ma=Mg            

⇒3Ma=2Mg

⇒a=2g/3

(b) Tension:

T=Ma/2

=M/2×2g/3=Mg/3

(c) Let T‘ = resultant of tensions

∴T’=√T²+T²=√2 T

∴T’=√2T

=√2Mg/3

Again, tanθ=T/T=1

⇒θ=45°

So, the force exerted by the clamp on the pulley is  √2 Mg/3at an angle of 45° with the horizontal.

Answer 2

Answer:

(a)

The acceleration is given as,

T+Ma−Mg=0

Ma /2 +Ma−Mg=0

a= 2g/3

(b)

The tension is given as,

T= Ma/2

T=

$\frac{m \times \frac{2g}{3} }{2}$

T= Mg/3

(c)

Let the resultants of the tension is R.

It can be written as,

R' =

$\sqrt({ {t}^{2} } + {t}^{2} )$

R' = √2 Mg/3

The angle is given as,

tanθ= T/T

θ=45°

Thus, the force exerted by the clamp on the pulley is √2Mg/3 at an angle of 45 ∘with horizontal.