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Dragging motion for pile of blocks

4,908 bytes added, 00:42, 13 August 2011
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If we denote by <math>N_{AB}</math> the normal force between the blocks, then <math>N_{AB}</math> acts upward on the upper block. The gravitational force <math>m_Bg</math> acts downward. Since there is no net acceleration in the vertical direction, we get:
<math>\! N_{AB} = m_Bg \qquad (1.1)</math> ===On the lower block=== If we denote by <math>N_A</math> the normal force between the lower block and the floor, we obtain, by [[Newton's first law]], that: <math>N_A = N_{AB} + m_Ag \qquad (1.2)</math> Plugging in (1) into (2) we get: <math>N_A = (m_A + m_B)g \qquad (1.3)</math> 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 <math>F</math> 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 <math>f_{s1}</math> the force of static friction between <math>A</math> and the floor, and by <math>f_{s2}</math> the force of static friction between <math>A</math> and <math>B</math>.
 
We get the following equation in the horizontal direction:
 
<math>F = f_{s1} + f_{s2} \qquad (2.1)</math>
 
<math>f_{s2} \le \mu_{s2} N_{AB} \qquad (2.2)</math>
 
On the other hand, the horizontal component of the force diagram of <math>B</math> gives:
 
<math>f_{s2} = 0 \qquad (2.3)</math>
 
Plugging in (1.3) and (2.2), we get:
 
<math>f_{s1} \le \mu_{s1} (m_A + m_B)g \qquad (2.4)</math>
 
Plug (2.3) and (2.4) in (2.1) and obtain:
 
<math>F \le \mu_{s1}(m_A + m_B)g \qquad (2.6)</math>
 
Thus, the ''maximum'' possible value of external force at which the system does not start accelerating is:
 
<math>F_{\max} = g[\mu_{s1}(m_A + m_B)] \qquad (2.7)</math>
 
{{quotation|'''ANSWER FORMAT CHECK''': For these kinds of situations, the critical values of external forces are always of the form <math>g</math> 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 <math>a</math> in the direction of the external applied force <math>F</math>. Denote by <math>f_{k1}</math> the force of ''kinetic'' friction between <math>A</math> and the floor, and by <math>f_{s2}</math> the force of static friction between <math>A</math> and <math>B</math> (the friction is static because the blocks move together so there is no slipping between surfaces). We get:
 
<math>F = f_{k1} + f_{s2} + m_Aa \qquad (3.1)</math>
 
Also:
 
<math>\! f_{k1} = \mu_{k1}N_A \qquad (3.2)</math>
 
<math>\! f_{s2} \le \mu_{s2}N_{AB} \qquad (3.3)</math>
 
and finally, the horizontal force component on <math>B</math> gives:
 
<math>\! f_{s2} = m_Ba \qquad (3.4)</math>
 
Plug in (3.2) and (3.4) into (3.1) and get:
 
<math>F = \mu_{k1}N_A + m_Ba + m_Aa \qquad (3.5)</math>
 
Plug in (1.3) in here to get:
 
<math>F = \mu_{k1}(m_A + m_B)g + (m_A + m_B)a \qquad (3.6)</math>
 
Rearranging, we obtain:
 
<math>\! a = \frac{F}{m_A + m_B} - \mu_{k1}g \qquad (3.7)</math>
 
{{quotation|'''INTUITIVE CHECK''': The value of acceleration is the ''same'' as what we would get if we assumed a single block of mass <math>m_A + m_B</math> with coefficient of kinetic friction <math>\mu_{k1}</math> with the floor.}}
 
Also, we note by plugging in (3.3) into (3.4) that:
 
<math>m_Ba \le \mu_{s2}N_{AB} \qquad (3.8)</math>
 
Plugging in (3.7) and (1.1) into (3.8) we get:
 
<math>F \frac{m_B}{m_A + m_B} - \mu_{k1}m_Bg \le \mu_{s2}m_Bg \qquad (3.9)</math>
 
which rearranges to:
 
<math>F \le (\mu_{s2} + \mu_{k1})(m_A + m_B)g \qquad (3.10)</math>
 
We already know from (2.7) a ''lower'' bound on <math>F</math>, so we get, combined, that:
 
<math>\mu_{s1}(m_A + m_B)g \le F \le (\mu_{s2} + \mu_{k1})(m_A + m_B)g \qquad (3.11)</math>
 
===Situation where blocks move at separate accelerations===
 
Suppose <math>a_A</math> is the acceleration of the lower block and <math>a_B</math> is the acceleration of the upper block. Denote by <math>f_{k1}</math> the force of kinetic friction between <math>A</math> and the floor and by <math>f_{k2}</math> the force of kinetic friction between <math>B</math> and the floor. We get:
 
<math>F = f_{k1} + f_{k2} + m_Aa_A \qquad (4.1)</math>
 
Also:
 
<math>f_{k1} = \mu_{k1}N_A \qquad (4.2)</math>
 
and
 
<math>f_{k2} = \mu_{k2} N_{AB} \qquad (4.3)</math>
 
The horizontal force component for <math>B</math> gives:
 
<math>f_{k2} = m_Ba_B \qquad (4.4)</math>
 
Plugging in (4.2) and (4.3) into (4.1), we get:
 
<math>F = \mu_{k1}N_A + \mu_{k2}N_{AB} + m_Aa_A \qquad (4.5)</math>
 
Plugging in the values of <math>N_A</math> and <math>N_B</math> from (1.1) and (1.3), we get
 
<math>F = \mu_{k1}(m_A + m_B)g + \mu_{k2}m_Bg + m_Aa_A \qquad (4.6)</math>
 
This rearranges to
 
<math>a_A = \frac{1}{m_A}\left[F - g(\mu_{k1}(m_A + m_B) + \mu_{k2}m_B)\right] \qquad (4.7)</math>
 
Plugging in (4.3) and (1.1) into (4.4), we get:
 
<math>\mu_{k2}m_Bg = m_Ba_B \qquad (4.8)</math>
 
This rearranges to:
 
<math>\! a_B = \mu_{k2}g \qquad (4.9)</math>
 
Finally, the constraint on <math>F</math> is that:
 
<math>F \ge (\mu_{s2} + \mu_{k1})(m_A + m_B)g \qquad (3.11)</math>
==Case of force on upper block==
 
{{fillin}}
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