A 4.0 kg block is stacked on top of a 12.0 kg block, whcih is accelerating along a horizontal table at a = 5.2 s^-2 m. Let Mu k = Mu s = Mu.

(a) What mininum coefficent of friction Mu between the two blocks will prevent the 4.0 kg block from sliding off?

(b) If Mu is only half this mininium vale, what is the acceleration of the 4.0 kg block with respect to the table, and

(c) with respect to the 12.0 kg block?

(d) What is the force that msut be applied to the 12.0 kg block in (a) and in (b), assuming that the table is frictionless?

ok...

what does it mean by "and in" in the last paragraph...

for now I just need help in part (a) ok on the top free body diagram...

force to the left of static friction

force to the right is the force... the force that the 12.0 kg block is experiancing to move 5.2 s^2 m ????

I would find that force by applying newtons second law which I know how to do but do I include friction on the bottom block???? Couldn't really tell from the way the problem was worded

ok once I know the force that the bottom block has applied to it to find the force of the block resting on top... the force that is opposing the force of firction... how do I find that is it the same as the one the bottom one experiances... Isn't the top block experiancing no opposing force to the force of friction because it's not moving? So how do I find the mininum...

I need help on part (a)

Friction is moving the top block. IT has to equal ma.

friction is 4g*mu
so 4*9.8*mu=4*5m/s^2 solve for mu. That is the min mu to keep the block from sliding off.

i don't see how to apply Newtons second law to the top block

there's a force pointing left the force of static friction

is there a force opposing this force pointin in the right direction?

How do I find the mininum coefficent of friction

Isn't

net force x direction = -Ffr sense it's in the left direction then I would get a negative Mu???

Were did 4.5 s^-2 m come from???

wouldn't it have a zero acceleration sense it's suppose to not be moving

ma=4kg*5.2m/s^2

I don't know your left/right sign conventions. However I do know if the top block is accelerating, the accelerating force must be in the direction of the motion. The bottom block is accelerating. Friction is the force that is moving the top block.

hmmmmmm interesting

THANKS YOU

also is the talbe frictionless or not I couldn't understand by the way it was worded

In part (a), you are asked to find the minimum coefficient of friction, μ, between the two blocks that will prevent the 4.0 kg block from sliding off. To solve this problem, let's break it down step by step:

Step 1: Draw a free body diagram for each block:
- For the 12.0 kg block: There are two forces acting on it - the force of gravity pulling it downwards (mg), and the force of friction (f) opposing its motion to the right.
- For the 4.0 kg block: There are two forces acting on it - the force of gravity pulling it downwards (mg), and the force of friction (F) between the two blocks.

Step 2: Apply Newton's second law to each block:
- For the 12.0 kg block: The net force acting on it is the force of friction (f) to the right, so we can write:
f = m₁ * a
where m₁ is the mass of the 12.0 kg block and a is its acceleration (given as 5.2 m/s^2).
- For the 4.0 kg block: The net force acting on it is the difference between the force of friction (F) to the left (opposing motion) and the force of gravity (mg) to the right, so we can write:
F - mg = m₂ * a
where m₂ is the mass of the 4.0 kg block and a is its acceleration (which is unknown).

Step 3: Express the force of friction in terms of the coefficient of friction:
- The force of friction between the two blocks can be expressed as:
f = μ * normal force
where μ is the coefficient of friction and the normal force is the force exerted on the 12.0 kg block due to the weight of the 4.0 kg block stacked on top of it.

Step 4: Substitute the force of friction into the equations:
- For the 12.0 kg block:
μ * normal force = m₁ * a
- For the 4.0 kg block:
F - mg = m₂ * a

Step 5: Solve for the minimum coefficient of friction:
- In order to find the minimum coefficient of friction, we need to consider the condition where the 4.0 kg block is on the verge of sliding off the 12.0 kg block. This happens when the force of friction is at its maximum value, which is given by:
f_max = μ * normal force_max
where normal force_max is equal to the weight of the 4.0 kg block.

- Setting f = f_max, we have:
μ * normal force_max = m₁ * a

- Rearranging the equation and solving for μ, we get:
μ = m₁ * a / normal force_max
normal force_max = m₂ * g, where g is the acceleration due to gravity.

By substituting the given values for m₁, a, and m₂ into the equation, you can calculate the minimum coefficient of friction, μ, that will prevent the 4.0 kg block from sliding off.

I hope this explanation helps you understand how to approach part (a) of the problem. Let me know if you need assistance with other parts or have any further questions.