# Two forces questions - Extremely difficult

Two forces questions - Extremely difficult:(

## Homework Statement

Problem A: A stack of dinner plates on a kitchen counter is accelerating horizontally at 2.7m/s^2. Determine the smallest coefficient of static friction between the dinner plates that will prevent slippage.
Problem B: A 24kg box is tied to a 14kg box with a horizontal rope. The coefficient of friction between the boxes and floor is 0.32. You pull the larger box forward with a force of 180N at an angle 25deg above the horizontal. Find the acceleration of the boxes and the tension of the rope.

## Homework Equations

Fnet=ma=sum of all forces
Friction=(u)(Fn)

## The Attempt at a Solution

For problem A, the wording makes no sense. If they are accelerating, are they not slipping? Static friction only applies when the object is at rest, and the plates are not. And I assumed this is how I would solve it:
ma=Ff+Fa
ma-Fa=uFn
(Ma-Fa)/mg
Masses cancel:
a-Fa divided by g = u. But I don't know the applied force. This question is just weird.

For problem B, I did this:
The following is for the vertical direction:
Fnet=0=ma=Fn+Fg+Fa
But I get confused as to which masses to use :/. I originally did this:
Fnet=0=mg=Fn-mg+180sin25 (up taken as positive direction)
Fn=mg-180sin25 WHERE M IS EQUAL TO THE SUM OF BOTH MASSES (they can be treated as a single object)

X direction:
Fnet=ma= some non-zero value = Fa-Ff
Ma=180cos25-u(mg-180sin25)
Divide both sides by mass and solve
Therefore a=1.8m/s^2
And this is awkward because I just got the correct answer... Idk perhaps I experienced a calculator error before. But calculating the tension in the rope, I don't even know where to begin. Isn't it just equal to the net force of the smaller mass?

ehild
Homework Helper

## Homework Statement

Problem A: A stack of dinner plates on a kitchen counter is accelerating horizontally at 2.7m/s^2. Determine the smallest coefficient of static friction between the dinner plates that will prevent slippage.

## Homework Equations

Fnet=ma=sum of all forces
Friction=(u)(Fn)

## The Attempt at a Solution

For problem A, the wording makes no sense. If they are accelerating, are they not slipping? Static friction only applies when the object is at rest, and the plates are not.

The problem means the plates do not slip on each other, that is the whole stack accelerates together. What force accelerates the topmost plate?

ehild

#### Attachments

• stackplates.JPG
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The force of static friction does, right? Because its friction on the top plate keeps it on top of the stack.

ehild
Homework Helper
Yes. And what should be the magnitude of the static friction so as the plate accelerate with 2.7 m/s2?

ehild

Fnet=Ff(static)? :)

Ff=ma
Then solve

ehild
Homework Helper
Yes, knowing that Ff(static)≤μgm...

ehild

haruspex