Maximum force on 11kg block before 5.13kg block slips

[Hypnos_16]

Homework Statement

A 5.13-kg block is placed on top of a 11.0-kg block that rests on a frictionless table. The coefficient of static friction between the two blocks is 0.675. What is the maximum horizontal force that can be applied before the 5.13-kg block begins to slip relative to the 11.0-kg block, if the force is applied to the more massive block?

m1 = 5.13kg
m2 = 11.0kg
µ = 0.675

Homework Equations

i tried getting the Frictional force of the block sitting on the other block and came up with 34 Newtons. i figured since the Frictional Force of one block on the other is 34 then then that would be the greatest force that could be applied, after attempting this i came up with the wrong answer... I'm stumped

The Attempt at a Solution

Ff = µFn
Ff = (0.675)(5.13 x 9.8)
Ff = 34.0N

 

[kjohnson]

What acceleration would what 34N frictional force create for the smaller block? How does that acceleration relate to that of the larger block?

 

[Hypnos_16]

the 34 N Frictional Force doesn't apply any acceleration since no forces are acting on the block, it's just from one block sitting on the other, so i assumed then that since that force is 34 N that the force required to move it would be 34 N. Or would the friction force between the two need to be 34 N in order to move it?

 

[kjohnson]

34N is the max static friction force. Static friction only uses what is necessary up to the max at which case it will start to slide and be kinetic friction. That being said the static friction acting on the small block will accelerate it, but the velocity of the small block relative to the large block will be zero (i.e. they move as one).

You are correct in thinking that 34N applied to the SMALL block would be the minimum force to move it. But it wants to know what the max force applied to the larger block would be so that the two blocks still move as one..

 

[Hypnos_16]

Okay, so i need to find the max amount of force that can be applied to the larger block before the smaller one starts to slide off, so then how would i go about doing that?

 

[kjohnson]

Well you know that 34N of friction is the max that can be applied by the surface. Since the friction is the only force in the horizontal direction for the small block it will cause it to accelerate... find that acceleration and you have the acceleration of the system.

Next analyze the larger block and find the unknown force 'F' required to accelerate it to that value.

p.s. don't forget about Newtons third law (forces equal and opposite)

 

[Hypnos_16]

So i think i found acceleration,
Fx = ma
34 = 5.13a
a = 6.63m/s

however I'm not sure if the mass has to be both blocks added together or just the top block.

 

[kjohnson]

For the first part your simply analyzing the top block. Friction is the only thing causing the top block to move and you know the magnitude of 34N so you can find the acceleration.

An easier way for the second part is to consider the two blocks as a single block of mass=m1+m2 moving with the calculated acceleration (since they are moving as one). What force would cause the total mass to move with given acceleration? That is your answer.

 

[Hypnos_16]

Okay so i took a stab at it, using the acceleration i just calculated and came up with this

F = ma
F - Ff = ma
F - µFn = ma
F - µ(m1 + m2)g = ma
F = µ(m1 + m2)g + ma
F = (0.675)(5.13 + 11.0)(9.81) + (11)(6.63)
F = 373.7 + 72.9
F = 446.6

 

[kjohnson]

Okay so i took a stab at it, using the acceleration i just calculated and came up with this

F = ma
F - Ff = ma
F - µFn = ma

Up to this step you are correct. If you go about it this way the normal force is the force applied by the small block on the large block (i.e. just m_small*9.8) and m on the right hand side is just the mass of the large block (since that's the block were applying Newtons laws to in the second part)

The other way I said you can go about part 2 is to consider both blocks as one with the calculated acceleration, or simply:

F=(m1+m2)a


Both of these two ways give you the same answer and it is not a bad idea to understand why they both apply.

 

[Hypnos_16]

Alright! so i tried it both ways and got the same answer, and it was the right answer, and i understand how they both work. Thanks for all your help man. I appreciate it.