Question

The friction coefficient between the two blocks shown in figure is µ but the floor is smooth (a) what maximum horizontal force F can be applied without disturbing the equilibrium of the system? (b) suppose the horizontal force applied is double of that found in part(a) , find the acceleration of the two masses.

Suppose the entire system of this question is kept inside an elevator which is coming down with an acceleration a < g. Repeat parts (a) and (b).

Answer 1

First part.

(a). F = Fric. + T

Now, For T,

T = fric.

∴ F = fric. + fric.

⇒ F = 2 × friction.

⇒ F = 2 × μmg

∴ F = 2μmg

(b).  If the Horizontal force applied will becomes twice, then,

F = 4μmg

⇒ 4μmg - T - μmg = ma

⇒ 3μmg - T = ma

For T,

T - μmg = Ma

⇒ T = mμg + ma

∴ 3μmg - mμg + Ma = ma

⇒ 3μmg - mμg - Ma = ma

⇒ a(M + m) = gμ(3m- m)

∴ a = 2mgμ/(M + m)

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Now, If the System will be kept in the Elevator which is descending down then substitute the values of the g by (g - a).

(a). F = Fric. + T

Now, For T,

T = fric.

∴ F = fric. + fric.

⇒ F = 2 × friction.

⇒ F = 2 × μmg

∴ F = 2μmg

(b).  If the Horizontal force applied will becomes twice, then,

F = 4μm(g - a)

⇒ 4μm(g - a) - T - μm(g - a) = ma

⇒ 3μm(g - a) - T = ma

For T,

T - μm(g -a) = Ma

⇒ T = mμ(g - a) + ma

∴ 3μm(g - a) - mμ(g - a)+ Ma = ma

⇒ 3μm(g -a) - mμ(g - a) - Ma = ma

⇒ a(M + m) = (g - a)μ(3m- m)

∴ a = 2m(g -a)μ/(M + m)

Hope it helps.