Name: Friction on a Single Block
Uploaded: 2019-07-19
Duration: 3 min 19 s
Description: Why is it hard to move a heavy bookcase across a carpeted floor? What is the reason behind it? Let's find out in this interesting video. Have fun learning!

Question

The friction coefficient between the two blocks shown in the figure is $μ$ but the floor is smooth. Suppose the entire system is kept inside an elevator which is coming down with an acceleration of $a < g .$

(a) What maximum horizontal force $F$ can be applied without disturbing the equilibrium of the system?

(b) Suppose the horizontal force applied is double the value found in part (a). Find the accelerations of the two masses.

Accepted Answer

(a) Let the tension in the string be $T$ when the maximum external force is $F .$

Equation of the top block in the vertical direction is given by,

$m g - N = m a$

$N = m (g - a)$

Hence, maximum value of the frictional force will be, $f = μ N = μ m (g - a)$ .

Equations of the block in the horizontal direction will be,

$F = T + f . . . (1)$

and $T = f . . . (2)$

Solving these two equations and putting the values of all the forces, we get the required maximum external force for the equilibrium of the system, i.e., $F = 2 μ m (g - a)$ .

(b) The external force is doubled in magnitude. The two blocks being tied to the same string, will move with the same acceleration in opposite directions. Let the acceleration be $a' .$ Let the new tension in the strings be $T'$ . The value of the friction force remains unchanged as it is the kinetic friction in this case.

Writing force equations similarly for both the blocks, we get

$2 F - T' - f = m a' . . . (3)$

and $T' - f = M a' . . . (4)$

Adding these two equations and putting the respective values of all the forces, we get

$\Rightarrow 2 F - 2 f = (m + M) a'$
$\Rightarrow a' = \frac{2 μ m (g - a)}{(m + M)}$ .

Important Questions on Friction

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