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I'm a bit confused about how friction force is applied. The answer to this question is that the ground applies 75N (kinetic) friction on block B (in the negative direction) and block A applies 5N (static) friction in the negative direction. So the result is that both the block move with a common acceleration of 1.

What I don't get is why can't the (static) friction by ground on block B be 65N in the negative direction and by block A on B be 25N (kinetic) friction in the negative direction. In this case, block B will remain at rest and block A will move with acceleration of 5. I know this is not true by 'common sense' but what is the logic behind it? Will blocks tend to move with a common acceleration whenever possible?

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  • $\begingroup$ Solution method: 1) draw a free body diagram for each block; 2) determine whether or not the 10 kg block will accelerate (it will); 3) determine the acceleration of both blocks together; 4) the top block is being accelerated by friction force, so use $F=ma$ to determine that force. If that force is less than the maximum friction force allowed, the top block will not slip. $\endgroup$ Jun 27, 2023 at 15:01
  • $\begingroup$ What is u' ? Static or kinetic? $\endgroup$
    – Bob D
    Jun 27, 2023 at 15:16
  • $\begingroup$ If it is 9000N, the top block will stay basically stationary while the bottom block flies out. We can only know which thing happens when we plug in the numbers for each case, and determine if the required inequalities are satisfied or not. $\endgroup$ Jun 28, 2023 at 4:37

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Will blocks tend to move with a common acceleration whenever possible?

Exactly this. The way you have to think about the static friction is that the equation

$$F_s = \mu_s \cdot F_n$$

does not define the actual static friction, but its maximum magnitude. So the actual static friction will be anywhere between zero and $F_s$ as defined by the above equation.

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Lemme first tell you the definition of friction: Friction is a force which opposes relative motion The formula of limiting friction is UsN (Us=coefficient of static friction and N=Normal force (perpendicular to the system) In this case N=(M+m)g the static friction is adjustable hence it can change but kinetic friction is the friction which is constant while the motion of the object.

In this question the lower block is moving hence the upper block will experience friction force in the direction of the motion of the lower block. This is because the upper block moves backwards with respect to the lower block and hence friction force will act opposite to that. so hence that's why the two blocks will move together whenever the limiting friction between the block can hold the force.

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What I don't get is why can't the (static) friction by ground on block B be 65N in the negative direction

As my college mechanics professor would say, "First draw a free body diagram (FBD)". In this case it makes sense to draw an FBD of the combination of the two blocks, as shown in FIG 1 below. The reason is there can be no friction, static or kinetic, on the upper block unless the lower block moves. And the lower block cannot move unless the 90N applied force exceeds the maximum possible static friction force between the lower block and the ground, which is (Assuming $\mu'$ is the coefficient of static friction):

$$f_{BG}=\mu' (M_{A}+M_{B})g$$

Taking g = 10

$$f_{BG}=75N$$

Since the applied force of 90N exceeds the maximum possible static friction force, the friction force must be kinetic and not static friction. The problem states that the kinetic friction force is 75N. Thus in this example the kinetic friction force equals the maximum possible static friction force (in most cases it is less than the maximum static friction force).

Next we need to determine the friction force between the blocks and determine whether it it is static or kinetic. To do this we will assume it is static friction and the blocks move together and then determine if that is correct. From Newton's 2nd law and FIG 1 the acceleration of the two blocks moving together is

$$a=\frac{F_{net}}{(M_{A}+ M_{B})}=\frac{(90-75)}{(5+10)}= 1m/s^2$$

Now we draw a free body diagram of the upper block in FIG 2. The only horizontal force acting on the upper block is the friction force $f_{AB}$ applied by the lower block. If the upper block has an acceleration of $1m/s^2$ from Newtons 2nd law we have

$$f_{AB}=M_{A}a=5N$$

The maximum possible static friction force between the blocks is

$$f_{AB-max}=\mu M_{A}g=25 N$$

From this we see that the maximum possible static friction force between the blocks would not be exceeded confirming our assumption that the blocks move together (the top block does not slide on the bottom block).

Hope this helps.

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