Force Problem with coefficient of static friction

In summary, the maximum horizontal force F that can be applied to the lower block is 20.25N. This is found by setting the two equations of motion for the blocks equal to each other and solving for F, taking into account the opposite directions of the frictional force for each block.
  • #1
Nanuven
11
0
[SOLVED] Force Problem

A 4.0kg block is put on top of a 5.0 kg block. The coefficient of static friction between the two blocks is u = 0.23 . The assembly of blocks is placed on a horizontal, frictionless table. What is the magnitude of the maximum horizontal force F that can be applied to the lower block so that the blocks will move together without any slippage?


The only relevant equations I can think of are
F = ma and
F = uN

I assumed that if the blocks are moving together than the fictional force between the blocks is greater than or equal to the forward force on the bottom block. I got that the maximum static frictional force is F = (.23)(4.0 * 9.8) = 9.016N but I don't know where to go from there, any ideas? Thanks
 
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  • #2
That's a good start. You also have to take acceleration into account. The acceleration is clearly nonzero, as the fricitonal force is the only force acting on the top block. So you need to use Newton's second law to write down 2 equations, one for the horizontal motion of each block. Use the fact that the two blocks move together to conclude that they have a common acceleration.
 
  • #3
(gone - my apologies)
 
Last edited:
  • #4
Ok so the acceleration of both blocks is equal. Also the only force acting on the top block is the static frictional force. Therefore:

The force of the bigger block is

F = umg = ma

F = 9.016 = ma

a = 2.254m/s2

Then since the acceleration is equal I plug the acceleration into the bottom equation of F = ma.

F = (4 + 5) * 2.254 = 20.286N.

The answer key says that it is 20.25N though...any ideas where I went wrong?
 
  • #5
Mr. Vodka,

1. Your solution is an absolute mess. You're using 2 different symbols to represent the same acceleration, and you're using the same symbol (F) to represent 2 different forces.

2. We forbid homework helpers from posting complete solutions around here anyway. It's up to the student to do that.

Please keep that in mind when posting in the future.

Nanuven,

The force of the bigger block is

F = umg = ma

That's wrong. There are two forces acting on the bigger block in the horizontal direction. One is friction, and the other is the applied force.
 
  • #6
Ok now I think I'm getting a little closer and yet still further away at the same time.

So the equation for the top block is:

Fs = Frictional Force
F = applied force
m1 = mass of top block
m2 = mass of bottom block

Fs - F = ma

while the equation for the bottom block is:

F = ma

then I substituted in the variables and solved for acceleration in each one.

a = (Fs - F) / m1

a = F / (m1 + m2)

then set them equal to each other and cross multiply

(m1 + m2)(Fs - F) = (m1)(F)

(4 + 5)(9.016 - F) = 4F

81.144 - 9F = 4F

13F = 81.144

F = 6.24

but I'm still wrong...
 
  • #7
Nanuven said:
Fs - F = ma

This is wrong. It says that the acceleration points in the opposite direction of the applied force F. Does that make sense to you?
 
  • #8
Shouldn't both blocks be accelerating in the positive x axis?

Or are you saying that the equation is:

F - Fs = ma ??

shouldn't the frictional force be positive since it is pointing towards the acceleration, similar to how a car works?
 
  • #9
I get 20.286N as well Nanuven.

if you reduce the precision on a = 2.254m/s^2 to a = 2.25m/s^2 see what happens.
 
  • #10
Nanuven said:
Shouldn't both blocks be accelerating in the positive x axis?

Yes, they should.

Or are you saying that the equation is:

F - Fs = ma ??

Exactly.

shouldn't the frictional force be positive since it is pointing towards the acceleration, similar to how a car works?

No, certainly not. The frictional force on on the top block points in the direction of acceleration. It points in the opposite direction for the bottom block.

Think about it for a minute. Imagine you've got the bottom block all by itself, and you pull on it. Now imagine that you put the smaller block on top, and pull it again. Does it make sense to you that the frictional force exerted on the bottom block by the top block is actually going to help you?
 
  • #11
Ok, I got it. Thanks everyone!
 

1. What is a force problem with coefficient of static friction?

A force problem with coefficient of static friction is a type of physics problem that involves calculating the minimum force required to move an object on a surface with a specific coefficient of static friction. The coefficient of static friction is a value that represents the amount of friction between two surfaces when they are not moving relative to each other.

2. How do you calculate the coefficient of static friction?

The coefficient of static friction can be calculated by dividing the maximum frictional force that can be applied to an object before it starts to move by the normal force acting on the object. This can be represented by the equation μs = Fmax / N, where μs is the coefficient of static friction, Fmax is the maximum frictional force, and N is the normal force.

3. What factors affect the coefficient of static friction?

The coefficient of static friction is affected by the nature of the two surfaces in contact, the roughness of the surfaces, and the force pressing the surfaces together. It also depends on external factors such as temperature and humidity. A higher coefficient of static friction indicates a greater resistance to motion between the two surfaces.

4. How does the coefficient of static friction affect an object's motion?

If the coefficient of static friction is greater than the applied force, the object will remain stationary. However, if the applied force is greater than the coefficient of static friction, the object will start to move. This is because the coefficient of static friction represents the maximum force that can be applied before the object begins to move.

5. How is a force problem with coefficient of static friction solved?

To solve a force problem with coefficient of static friction, you must first identify the forces acting on the object and their magnitudes. Then, use the equation μs = Fmax / N to calculate the coefficient of static friction. Finally, use the coefficient of static friction in equations such as Newton's second law (F = ma) to determine the minimum force required to move the object.

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