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New to physics and confused:

Lets say we have two blocks one stacked atop the other, on a frictionless surface, with static friction between the two blocks. Lets say we apply an external pushing force to the bottom block.

The goal is to find the maximum value for this external force such that the top block doesn't slip off the bottom block.

What I don't understand is the motion equation for the bottom block (block 1):

$$ m_1a_1 = F_{ext} - F_f $$

My simple argument. If $ F_{ext} \leq F_{f_{max}} $ , we have $F_{ext}$ completely cancelled by the static friction which will match $F_{ext}$'s magnitude and we have no motion in the bottom block, but we have motion in the top block due to that friction force and, unintuitive, the top bock will fly forward off the bottom block.

Similarly, if $F_{ext} > F_{f_{max}}$ isn't that the definition of slipping?

By my logic there is no solution and some very unintuitive behavior, but I sillily can't seem to find the flaw in my argument.

Atticus Deutsch
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3 Answers3

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I will not solve the problem for you here, but I will give some hints.

Friction always opposes the actual or impending motion; therfore, you can determine the direction of the force of friction from the bottom block on the top block if the top block is forced say to the right. Use Newtons's third law to determine the force friction on the bottom block.

Look at the bottom block and determine ALL the forces acting on it.

For no slip, friction has a MAXIMUM value determined by the coefficient of static friction, but the force of friction can be less (for example if the force applied is very small).

For no slip, both blocks have the same acceleration.

Search for "block friction" on this site and you will find numerous questions an answers for this problem.

John Darby
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  • Thank you for your response. I'm still confused. We have for the top block $F_N\mu = m_2a_2$ implies $a_2 = g\mu = a_1$ then for the second block, $F_{ext} - F_{f_{max}} = m_1a_1 $ and substituting out $a_1$, $F_{ext} - F_{f_{max}} = m_1 g\mu $ which implies $F_{ext} > F_{f_{max}}$ which implies slipping by definition so we have a contradiction. – Atticus Deutsch Jan 10 '23 at 03:29
  • Fext applied to lower block minus the force of friction on lower block is net external force on LOWER block and is always positive to move the lower block; this says nothing about slipping of UPPER block. To keep upper block from slipping it must have same acceleration as lower block and acceleration of upper block is due only to force of friction on it. As Fext increases force of friction increases up to a max force of static friction beyond which friction becomes kinetic (less) and the upper block acceleration becomes less than the lower block acceleration due to slipping. Does this help? – John Darby Jan 10 '23 at 03:57
  • Fext acts on the lower block, not the upper block. Other related questions on this exchange compare case with Fext on upper or lower block. Use free body diagrams and clearly note which block the forces act on. I'll see if I can find one of these related questions and forward to you. – John Darby Jan 10 '23 at 04:03
  • See https://physics.stackexchange.com/questions/623025/block-on-block-problem-and-friction on this exchange. – John Darby Jan 10 '23 at 04:13
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Your error is in saying that $F_{ext}$ is completely cancelled by the force of friction.

Static friction always acts to prevent acceleration at its interface. It can sometimes be a lot more clear to start from that constraint on acceleration, and then determine what the force of friction would have to be, based on that.

At that point, you can determine if the interface can actually provide that much friction. It's just like analyzing a load on a structure.

Alex K
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I think you need to also look at Newton's Second Law for the top block. There, the horizontal component equation (in the case that the top block remains on the bottom block) is: $$ F_{fs} = m_B a $$ The magnitude of the frictional force can be anything up to its maximum value, $F_{fs}^{\rm max} = \mu_s F_N = \mu_s m_B g$ (using the vertical component of Newton II). Therefore there is a maximum acceleration of the the system: $$ a_{\rm max} = F_{fs}^{\rm max} / m_B = \mu_s g $$ Then you can apply the same "max" condition to your equation for the bottom block, and solve those two equations (eliminating $a_{\rm max}$) for $F_{\rm ext}$.

Ben H
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