How do you evaluate the tension force if the coefficient of friction between the objects K and L is $0.6$?

Question

How do you evaluate the tension force if the coefficient of friction between the objects K and L is $0.6$?

So the system is accelerating, whence we have to consider that

$$\sum F_x = m_1a$$

$$F_k - T = 2a $$

$$\mu mg - T = 2a \implies 0.6 \times 2 \times 10 - T = 2a \implies 12-T = 2a$$

For the object L,

$$\sum F_x = m_2a$$

$$F - F_k = 6a $$

$$10 - 12 = 6a \implies a = -\dfrac{1}{3}$$

Plugging $a$ into the first equation

$$12-T = 2 \times -\dfrac{1}{3} \implies 12-T = -\dfrac{2}{3} $$

However, there won't be an integer solution from what I got above. Could you assist me?

Answer 1

It's a very easy to solve exercise but requiring to have very clear concepts in mind. Two hints.

1.- That the system is moving needs a proof (or a disproof). The block $M$ does not move in any case because, by hypothesis, it is attached to a wall by an inextendable string.

2.- The friction forces are reaction forces, so is, their magnitude depends on other forces in a, say, peculiar way.

Answer 2

To get the tension in the string you should consider the free body diagram of the top block. The tension pulls it to the left and the frictional force pulls it to the right.

Frictional force is a reactive force whose maximum value is $2\text{kg}\times 9.8\text{N/kg}\approx19.6\times 0.6\approx 11.8\gt 10$. Therefore, the block is sitting still and the frictional force between the two blocks is $10\text{N}$.

Since the tension balances the frictional force, the tension is $10\text{N}$.