# 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.

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### 2 Answers

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:

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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.

So

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.**