A large box of mass M is pulled across a horizontal frictionless surface by a horizontal rope with tension T. A small box of mass m sits on top of the large box. The coefficients of static and...
A large box of mass M is pulled across a horizontal frictionless surface by a horizontal rope with tension T.
A small box of mass m sits on top of the large box. The coefficients of static and kinetic friction between the two boxes are mu_s (static friction) and mu_k (kinetic friction) respectively. Find an expression for the maximum tension Tmax for which the small box rides on top of the large box without slipping.
wherem_u = mu_s ofcourse.
I am sorry but actually I thought the coefficient of kinetic friction was beteen the bottom surface and big box, but actually its frictionless, so actually you don't need to consider mu_k, as the small box is not sliding but siting steady over the big box, so there is no use ofm_kand so every where we putm_k =0 in the above result and we will get the right answer. So let me write it down completely:
Let us think that the "small" box is sitting steady over the "big" box , so the condition will be:
Acceleration of small box (a) = Acceleration of the big box -----(1)
Now let us first calculate the acceleration of the big box:
forces on the Big Box are:
(1) Tension T along +x direction say.
(2) (M+m) * a, along +x direction too. Because rope is pulling
big and small box together.
So we set the equation:
(M+m) * a = T
=> a = T /(M+m)
Now we go for the small box:
Forces on the Small Box are:
(i) Force exerted by the big box on small box and that is =
force of friction m*a along -x direction
(ii) m*a along +x direction
and these two forces are exactly equal that is why it is not moving.
mu_s * m * g = m * a
=> mu_s g = [T/(M+m)]
=> T = m_u * (M+m) * g
Which is the maximum tension in the rope.