Changes

Dragging motion for pile of blocks

, 00:42, 13 August 2011
no edit summary
If we denote by $N_{AB}$ the normal force between the blocks, then $N_{AB}$ acts upward on the upper block. The gravitational force $m_Bg$ acts downward. Since there is no net acceleration in the vertical direction, we get:
$\! N_{AB} = m_Bg \qquad (1.1)$ ===On the lower block=== If we denote by $N_A$ the normal force between the lower block and the floor, we obtain, by [[Newton's first law]], that: $N_A = N_{AB} + m_Ag \qquad (1.2)$ Plugging in (1) into (2) we get: $N_A = (m_A + m_B)g \qquad (1.3)$ Intuitively, the weight borne by the floor is the total of the weights of both blocks.
==Case of force on lower block==
{{fillin}}Suppose an external horizontal force $F$ is applied on the lower block. We determine the conditions under which both blocks start moving together, the two blocks start moving but not together, and neither block moves. ===Situation where neither block moves===
Denote by $f_{s1}$ the force of static friction between $A$ and the floor, and by $f_{s2}$ the force of static friction between $A$ and $B$.

We get the following equation in the horizontal direction:

$F = f_{s1} + f_{s2} \qquad (2.1)$

$f_{s2} \le \mu_{s2} N_{AB} \qquad (2.2)$

On the other hand, the horizontal component of the force diagram of $B$ gives:

$f_{s2} = 0 \qquad (2.3)$

Plugging in (1.3) and (2.2), we get:

$f_{s1} \le \mu_{s1} (m_A + m_B)g \qquad (2.4)$

Plug (2.3) and (2.4) in (2.1) and obtain:

$F \le \mu_{s1}(m_A + m_B)g \qquad (2.6)$

Thus, the ''maximum'' possible value of external force at which the system does not start accelerating is:

$F_{\max} = g[\mu_{s1}(m_A + m_B)] \qquad (2.7)$

{{quotation|'''ANSWER FORMAT CHECK''': For these kinds of situations, the critical values of external forces are always of the form $g$ times a homogeneous linear polynomial involving the masses, with coefficients drawn from various possible friction coefficient.

===Situation where blocks move together as a system===

Suppose both blocks have an acceleration of $a$ in the direction of the external applied force $F$. Denote by $f_{k1}$ the force of ''kinetic'' friction between $A$ and the floor, and by $f_{s2}$ the force of static friction between $A$ and $B$ (the friction is static because the blocks move together so there is no slipping between surfaces). We get:

$F = f_{k1} + f_{s2} + m_Aa \qquad (3.1)$

Also:

$\! f_{k1} = \mu_{k1}N_A \qquad (3.2)$

$\! f_{s2} \le \mu_{s2}N_{AB} \qquad (3.3)$

and finally, the horizontal force component on $B$ gives:

$\! f_{s2} = m_Ba \qquad (3.4)$

Plug in (3.2) and (3.4) into (3.1) and get:

$F = \mu_{k1}N_A + m_Ba + m_Aa \qquad (3.5)$

Plug in (1.3) in here to get:

$F = \mu_{k1}(m_A + m_B)g + (m_A + m_B)a \qquad (3.6)$

Rearranging, we obtain:

$\! a = \frac{F}{m_A + m_B} - \mu_{k1}g \qquad (3.7)$

{{quotation|'''INTUITIVE CHECK''': The value of acceleration is the ''same'' as what we would get if we assumed a single block of mass $m_A + m_B$ with coefficient of kinetic friction $\mu_{k1}$ with the floor.}}

Also, we note by plugging in (3.3) into (3.4) that:

$m_Ba \le \mu_{s2}N_{AB} \qquad (3.8)$

Plugging in (3.7) and (1.1) into (3.8) we get:

$F \frac{m_B}{m_A + m_B} - \mu_{k1}m_Bg \le \mu_{s2}m_Bg \qquad (3.9)$

which rearranges to:

$F \le (\mu_{s2} + \mu_{k1})(m_A + m_B)g \qquad (3.10)$

We already know from (2.7) a ''lower'' bound on $F$, so we get, combined, that:

$\mu_{s1}(m_A + m_B)g \le F \le (\mu_{s2} + \mu_{k1})(m_A + m_B)g \qquad (3.11)$

===Situation where blocks move at separate accelerations===

Suppose $a_A$ is the acceleration of the lower block and $a_B$ is the acceleration of the upper block. Denote by $f_{k1}$ the force of kinetic friction between $A$ and the floor and by $f_{k2}$ the force of kinetic friction between $B$ and the floor. We get:

$F = f_{k1} + f_{k2} + m_Aa_A \qquad (4.1)$

Also:

$f_{k1} = \mu_{k1}N_A \qquad (4.2)$

and

$f_{k2} = \mu_{k2} N_{AB} \qquad (4.3)$

The horizontal force component for $B$ gives:

$f_{k2} = m_Ba_B \qquad (4.4)$

Plugging in (4.2) and (4.3) into (4.1), we get:

$F = \mu_{k1}N_A + \mu_{k2}N_{AB} + m_Aa_A \qquad (4.5)$

Plugging in the values of $N_A$ and $N_B$ from (1.1) and (1.3), we get

$F = \mu_{k1}(m_A + m_B)g + \mu_{k2}m_Bg + m_Aa_A \qquad (4.6)$

This rearranges to

$a_A = \frac{1}{m_A}\left[F - g(\mu_{k1}(m_A + m_B) + \mu_{k2}m_B)\right] \qquad (4.7)$

Plugging in (4.3) and (1.1) into (4.4), we get:

$\mu_{k2}m_Bg = m_Ba_B \qquad (4.8)$

This rearranges to:

$\! a_B = \mu_{k2}g \qquad (4.9)$

Finally, the constraint on $F$ is that:

$F \ge (\mu_{s2} + \mu_{k1})(m_A + m_B)g \qquad (3.11)$
==Case of force on upper block==

{{fillin}}