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.
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(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)
Hope it helps.
Answered by
1
(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)
Hope it helps.
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