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I was doing a practice problem in prepping for a midterm and then numbers that I got just seem implausible. I know that they are the 'right' answers, I'm just not sure if this situation could exist in real life.

The problem had to do with pulleys and things, but the bit I'm interested in is basically this: a block (A) is being pulled with a force and is accelerating at a rate of $0.5 m/s^2$. It is sliding on top of another block (B) and the co-efficient of friction between them is $0.3$. This block B is sliding on a table and the co-efficient of friction between the block and the table is $0.2$. Block A is 15 kg. Block B is 5 kg.

Block A: a = 0.5, mass = 15kg, $\mu_{A-B} = .3$

Block B: mass = 5kg, $\mu_{B-Table} = .2$

enter image description here

According to my calculations, the lower block should be accelerating at a rate of $0.981m/s^2$. This seems impossible, as the block pulling it isn't moving that fast. Can anyone explain why this is?

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closed as off-topic by sammy gerbil, ZeroTheHero, Yashas, Jon Custer, John Rennie Mar 27 '17 at 8:18

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  • $\begingroup$ Shouldn't these be called blocks B and C according to the diagram? $\endgroup$ – JMac Mar 21 '17 at 12:51
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It makes much more sense, if you remember that accelerated motion of the block A (or B in your diagram) does not imply that the blocks should be moving relative to each other. You should be able to figure out the answer yourself given the hint.

Edit:

Let's assume that the block A is sliding over the block B. From forces of friction you indeed get that friction with block A pulls the block B with force $F_{pull}=4.5g$, and friction with the table tries to stop the motion with force $F_{fr}=4g$, which results in block B being accelerated at $a=0.1g\approx 0.98m/s^2$.

Let us assume this situation is actually physically possible. What would happen as soon as A starts sliding over the block B? B will accelerate 2x faster, and soon will be moving faster than the block A. As this happens, the direction of friction force between A and B will be reversed. Since both friction forces are trying to stop the block B now, the system will presumably return to the initial state, when again B is stationary, and A is sliding over it. My analysis is, of course, extremely simplified, since friction forces in this jerky motion would change between dynamic and static all the time, and the force pulling the block A would also have to change rapidly to keep acceleration of the block A constant.

Or, you could simply call this situation physically impossible, and say that B slides together with the block A. The above analysis of the situation, if done properly, should also lead to the same conclusion. In that case static friction between blocks A and B will adjust accordingly to compensate friction force between B and the table and in addition it will ensure that B is also accelerating with $a=0.5m/s$. Your calculations have showed that static friction between blocks A and B is fully capable of that. Furthermore, the same is true for larger forces pulling the block A, as long as the acceleration of the block A remains $a \leq 0.1g$. If this criterion is not satisfied, block A will indeed start sliding over B, and acceleration of the block B will remain $a_B = 0.1g$.

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  • $\begingroup$ I'm not sure how that's relevant (though given the lack of OP's work shown it's anyone's guess). The accelerated motion of block A does imply that there is an attempt at relative motion between the two blocks so that friction force applies. $\endgroup$ – JMac Mar 21 '17 at 13:20
  • $\begingroup$ @JMac exactly, but it is not necessarily dynamic friction $\endgroup$ – Arturs C. Mar 21 '17 at 13:22
  • $\begingroup$ To be clear, I know the answer, I got it right. The answer key days I'm right, my math says I'm right. This was from a past midterm so I seriously doubt that the answer key is wrong. I didn't show much work because I'm certain I'm right. I'm asking why it is possible for this to happen or if it could in real life. $\endgroup$ – Sam Spade Mar 21 '17 at 23:17
  • $\begingroup$ @spadeANDarcher I have edited my answer to (hopefully) satisfy your interest $\endgroup$ – Arturs C. Mar 22 '17 at 12:06
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Acceleration will NOT be 0.98 units or 0.1g.

For B,

acceleration=(Frictional force between A and B - Frictional force between B and ground)/mass of B

\begin{align} RHS&=(M_A*\mu_{A-B} -(M_A +M_B)*\mu_{B-Table} )/M_B\\ &=(15\cdot0.3-(15+5)\cdot0.2)/5\\ &=-0.62 units <0 \end{align}

Which means friction with table will dominate and stop relative motion between itself and B hence stopping B from moving completely.

Note that frictional force is self adjusting so friction force with ground will just be equal to friction between A and B.

EDITED value of mass of A from 0.5 to 15 as corrected by Arturs C.(thanks for correction)

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  • $\begingroup$ I think you have an error here: mass of the block A is 15kg, not 0.5kg $\endgroup$ – Arturs C. Mar 21 '17 at 13:01
  • $\begingroup$ @monster You seem to have accidentally created two different accounts. Please see the help center for instructions on how to merge them. $\endgroup$ – ACuriousMind Mar 21 '17 at 19:14

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