How Does Doubling the Force Affect Kinetic Friction Between Two Blocks?

[toesockshoe]

Homework Statement

Block M1 is on top of block M2 which is on top of a horizontal frictionless surface. A horizontal force, F, is applied to M1. The coefficient of static friction is us and the coefficient of kinetic friction isuk. Find, a) the maximum force that can be applied to M1, such that the position of M1 relative to M2 will not change. (M1 will not slip on top of M2). Then, if the force acting on M1 is doubled, or F' = 2F, find the acceleration of each mass.

Homework Equations

F=ma

The Attempt at a Solution

I think i got part a correct, double check that for me, but I am stumped on part b. if you are increasing the force, then now there is kinetic friction. would M2 (the mass on the bottom ) be moving at all? wouldn't the acceleration be 0 because there is so much applied force that it would overcome the static force. also do we assume that the second applied force (twice the size of the first one) is applied after the first one is applied or do we disregard the first force completely? part a is done in the first image, and I started part b by drawing the free body diagram in the second picture.[/B]

 

[x86]

You seem to be confused about some concepts. (The exact same stuff I was confused about less than 4 months ago!). Hopefully my explanation fixes things for you.

Lets walkthrough part a,

What do we know? There is no friction between the ground and m2, but friction between m1 and m2. Think about this as if you were to glue two boxes together and then push them on ice.

There is a friction force on m1 opposite in direction to Fa. And by Newtons third law, on m2, there is the same friction force but opposite direction.

The question tells us to find the force applied to m1 given m1 and m2 do not slip (a fancy way of saying m1 and m2 move with the same acceleration, a1=a2).

Draw a FBD (always do this) for m1 using the standard x/y coordinate system

In the y-axis we have: gravity, normal force
In the x-axis we have: applied force (say +x), friction (-x)

From F=ma in the y direction, we know N = mg

The mass doesn't slip ==> static friction
Ff = u_s N = u_s mg

F=ma in the x direction, Fa = m1gu_s + m1a1 = m1gu_s + m1a (1)

Now for the second mass, draw the FBD. From Newtons third law, we know Ff (between m1 and m2) acts in the +x direction, as well as gravity and normal force.

F=ma in x direction, Ff = m2a2 ==> a2 = a1 = a = m1gu_s/m2 (2)

Now sub (2) into (1) and get Fa = m1gu_s(1 + m1/m2) (the same thing you got).

I think i got part a correct, double check that for me,

Yes, you did.

but I am stumped on part b. if you are increasing the force, then now there is kinetic friction. would M2 (the mass on the bottom ) be moving at all? wouldn't the acceleration be 0 because there is so much applied force that it would overcome the static force.

As you can see from my solution of part a, the mass would be moving! Draw a separate FBD. The friction from m1 on m2 makes it move in the +x direction. The acceleration is NOT zero. (But the acceleration of m1 with respect to m2 is indeed zero, they move as if they were one object IN PART A ONLY)

also do we assume that the second applied force (twice the size of the first one) is applied after the first one is applied or do we disregard the first force completely? part a is done in the first image, and I started part b by drawing the free body diagram in the second picture.

You can pretty much ignore part a, the only information you need is that the two masses slip and move with different accelerations!

Read my solution of part a, fix your misconceptions, and you'll see part b is just as easy as part a.

Note that for part a we found Fa with the MAXIMUM static friction force. In part 2, since Fa is doubled, you know for sure it slips, and thus kinetic friction is used.

 

[haruspex]

do we assume that the second applied force (twice the size of the first one) is applied after the first one is applied

It's twice the force found in (a), yes.

You can pretty much ignore part a

Not so. The question says:

Find, a) the maximum force that can be applied...Then, if the force acting on M1 is doubled,

So the force applied in (b) is double the force found in (a).

 

[x86]

It's twice the force found in (a), yes.

Not so. The question says:

So the force applied in (b) is double the force found in (a).

Yes, of course. I apologize for the bad wording. As for ignoring part A, I mean't ignoring the process of part A and just knowing the force, the given info (slipping, etc). It is a two part question, but you have to split it up to solve it. Also, that question wasnt mine, I wasn't sure how to quote the OP's questions so I just used bold/underline/italic instead. (

 

[toesockshoe]

It's twice the force found in (a), yes.

Not so. The question says:

So the force applied in (b) is double the force found in (a).

ok so i have something. can you check my part b? the first pic is the acceleration for m1 (the mass on top) and the second pic has the acceleration for m2 (the mass on the bottom)

 

[x86]

It looks good to me

 

[haruspex]

Doesn't the a₁ expression simplify somewhat? (Cancel out some M₁s)