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I was doing a question in Newton's laws of motion where I came across this question .

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The question said there was same frictional coefficient between all surfaces and there is no acceleration of the blocks and that they moved with uniform velocity. So we are all good in this case .

If the surfaces were frictionless then we would use constraints ( using constraints would not be necessary , it can be just observed in this question) and get that both accelerate with same acc. in opp. directions.

But I had the following doubts

Doubts:

[ Situation : If the blocks are acc. ]

[ These doubts are not only refering to this question but to all questions of this type]

1) If all the surfaces have friction in between them

Case 1) same frictional co-efficients

Case 2) different friction co-efficients

, then will we able to use constraint relations in them to find acceleration

2) If only some surfaces have friction between them and others are smooth.

I hope to gain better clarity with the help of you guys .

PS: I know some of you may find these questions silly but still getting valuable info. from you all will help greatly

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closed as off-topic by Bob D, Jon Custer, Kyle Kanos, Emilio Pisanty, GiorgioP Jul 28 at 19:40

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  • $\begingroup$ To me, at least, it is not clear what’s your question. You are talking vaguely about constraints $\endgroup$ – Bob D Jul 1 at 20:24
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Frankly, I think you have far too many variables in your situations and cases. To simplify matters I will assume no friction between the pulley and the cable. You should be able to generalize the situation to include it using the approach given below that assumes friction may or may not be between the blocks and supporting surface.

First, be aware that there are two coefficients of friction, the coefficient of static friction and the coefficient of kinetic or sliding friction. Static friction resists impending motion of an object (resists the start of motion). Kinetic friction resists the relative motion between surfaces when that motion is already under way. The coefficient of kinetic friction involving the same surfaces is generally less than the coefficient of static friction involving the same surfaces. Since the general situation you present is one in which the blocks are accelerating, motion has already started. Therefore we know that the external force F has caused the maximum static friction forces between surfaces to be exceeded, so that we need only concern ourselves with kinetic friction. The case you first introduced (from the book) is where after acceleration begins the applied force is reduced to exactly equal the kinetic friction force so that there will be no net forces on the blocks and they will move at constant velocity.

The figures below are free body diagrams of blocks A and B. Figure 1 shows the most general case where there is friction between blocks A and B and between block B and the supporting surface of B (friction between all surfaces except the pulley). Figures 2 and 3 modify the free body diagrams for the cases where there is only friction between one or the other contacting surfaces.

For all the figures, the following are the general constraints that should apply to all of your cases.

  1. If we can assume there is no stretching of the cable and no friction at the pulley then the tension $T$ is the same throughout the cable.

  2. The magnitude of the acceleration of block A will equal block B, or

$$a_{A}=a_{B}$$

  1. If there is friction between A and B the friction force B exerts on A will be equal and opposite to the friction force A exerts upon B, or

$$F_{fBA}=F_{fAB}$$

The other terms in the figures are as follows:

  1. $μ_{AB}$ is the coefficient of kinetic friction, if any, between blocks A and B.

  2. $μ_{BS}$ is the coefficient of kinetic friction, if any, between block B and its supporting surface.

  3. $M_A$ and $M_B$ are the masses of blocks A and B

  4. $g$ is the acceleration due to gravity.

Figure 1- (Friction all surfaces)-

Block A:

$$F-μ_{AB}M_{A}g-T=M_{A}a_A$$

Block B:

$$T-μ_{AB}M_{B}g-μ_{BS}M_{B}g=M_{B}a_B$$

Given the masses and the coefficients of friction, and the general constraints 1-3 above, solve for the acceleration.

Now consider your scenarios

1) If all the surfaces have friction in between them

Case 1) same frictional co-efficients

Case 2) different friction co-efficients

, then will we able to use constraint relations in them to find acceleration

There are only two surfaces involving friction. The surfaces of A and B in contact with one another and the surface of B in contact with the supporting surface. If there is friction between both surfaces, whether the coefficients are the same or different, Figure 1 applies and you simply use the applicable coefficients. Given these, knowledge of the masses, and the general constraints 1-3 above, you can solve for the acceleration.

2) If only some surfaces have friction between them and others are smooth.

Again, since there are only two surfaces involved, simply remove the friction force for the smooth surface from the free body diagram and keep the friction force for the other surface. There are two possible outcomes:

Figure 2 (no friction between A and B)-

If there is no friction between A and B but friction between B and the supporting surface, then we have the following:

$$T-μ_{BS}M_{B}g=M_{B}a_B$$

and

$$F-T=M_{A}a_A$$

Given the two masses, the coefficient of friction between B and the supporting surface, and the general constraints 1-3 given above, solve for the acceleration.

Figure 3 (friction between A and B)-

There is friction between A and B but no friction between B and its supporting surface.

$$T-μ_{AB}M_{B}g=M_{B}a_B$$

and

$$F-μ_{AB}M_{A}g-T=M_{A}a_A$$

Given the two masses, the coefficient of friction between A and B, and the general constraints 1-3, calculate the acceleration.

Hope this helps.

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  • $\begingroup$ Thanks a lot Bob but my question was that can constraint relations be used in these above cases?@Bob D $\endgroup$ – user235632 Jul 3 at 5:39
  • $\begingroup$ Can you clarify what you mean by “constraint relations” if they’re not the three i mentioned? $\endgroup$ – Bob D Jul 3 at 5:43
  • $\begingroup$ Oh ! I see , I read it once again .Got it $\endgroup$ – user235632 Jul 3 at 5:46
  • $\begingroup$ Thanks a lot @Bob D $\endgroup$ – user235632 Jul 3 at 5:46
  • $\begingroup$ Then the answer is acceptable? $\endgroup$ – Bob D Jul 3 at 5:50