Minimum force before a system of two blocks will move

[marksyncm]

Homework Statement

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$m=3kg$
$M=2kg$
Coefficient of static friction between m and M $\mu_1 = 0.6$
Coefficient of static friction between M and the ground $\mu_2 = 0.8$.

The rope is weightless, does not stretch, and the wheel is frictionless.

What is the maximum force $F$ that can be exerted before the system moves?

Homework Equations

$Friction = \mu F_N$

The Attempt at a Solution

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My reasoning: because the two blocks are attached together, if one of them moves, then so does the other. This means that the only way the blocks will move is if the larger of the two friction forces is overcome. In this case, the larger friction force is the one between block M and the ground, which equals $F_N * \mu_2 = 9.8 * (3+2) * 0.8 = 39.2 N$. Meaning we must apply $39.2 N$ of force before the blocks will move.

Is this correct?

 

[kuruman]

You seem to have ignored that the top block exerts a force on the bottom block.

 

[marksyncm]

As we apply a force to object $M$, the rope pulls on object $m$ towards the left, causing a force of friction from block $M$ onto block $m$ pointing towards the right, and an equal and opposite friction force from block $m$ onto block $M$. As a result, to move the system, we need to overcome the two combined friction forces:

$$3g\mu_1 + 5g\mu_2 = 17.64 + 39.2 = 56.84 N$$

Is this correct?

 

[kuruman]

It looks like you ignored the tension. Draw a free body diagram of the combined two-mass system and you will see what I mean.

 

[marksyncm]

Draw a free body diagram of the combined two-mass system and you will see what I mean.

Thank you. I should always start with this step; for some reason, I assumed the problem was trivial and decided to avoid the diagram. I doubt I'll do this again.

Having drawn the diagram, I can see the following forces acting on the block of mass $M$ as we apply force $F$ to it in the direction shown on the image:

• The force of friction exerted by the ground - $39.2 N$
• The force of friction exerted by block $m$ - $17.64 N$
• The force of tension in the rope

The force of tension acting on block $M$ is equal to the force of tension in the rope acting on block $m$, which in turn is equal to the force of friction exerted by block $M$ onto block $m$. So in total, the three forces counteracting the pulling force $F$ are $17.64 + 17.64 + 39.2 = 74.48 N$. Applying a force greater than $74.48 N$ will move the blocks.

Is that correct?

 

[haruspex]

Thank you. I should always start with this step; for some reason, I assumed the problem was trivial and decided to avoid the diagram. I doubt I'll do this again.

Having drawn the diagram, I can see the following forces acting on the block of mass $M$ as we apply force $F$ to it in the direction shown on the image:

• The force of friction exerted by the ground - $39.2 N$
• The force of friction exerted by block $m$ - $17.64 N$
• The force of tension in the rope

The force of tension acting on block $M$ is equal to the force of tension in the rope acting on block $m$, which in turn is equal to the force of friction exerted by block $M$ onto block $m$. So in total, the three forces counteracting the pulling force $F$ are $17.64 + 17.64 + 39.2 = 74.48 N$. Applying a force greater than $74.48 N$ will move the blocks.

Is that correct?

Right.
You can make the arithmetic a bit easier by keeping g out as a factor until the end.

 

[marksyncm]

Thank you.