Determine the kinetic coefficient of friction

In summary: Remember that all the forces stay the same as if the pulley were weightless. Remember that all the forces stay the same as if the pulley were weightless.
  • #1
EP
76
0
Block A of mass 2m hangs from a cord that passes over a pulley and is connected to block B of mass 3m. Block B is free to move on a frictionless horizontal surface. The pulley is a disk with frictionless bearings, having radius R and moment of inertia 3MR^2. Block C of mass 4m is on top of block B. The surface between blocks B and C is not frictionless. shortly after the system is released from rest, block A moves with a downward acceleration a, and the two blocks on the table move relative to each other.

a.) In terms of the given quantities and g, find the tensions, T1 and T2 for each section of string.

For this I have the Sum of Fy= 2m(g-a)=T1 for Fx= T2= 7ma which the anwer given is different. The t1 is correct! But is this a typo or am I wrong?b.) If a=2m/s^2, determine the kinetic coefficient of friction between blocks B and C.

I was thinking the force of friction would be equal to the tension in the x direction but when I work that out I don't get the right answer. Is that the right concept?

c.) What is the acceleration of block C at this time?

I don't have any work for this part yet but I'll put it on here just in case.
 
Physics news on Phys.org
  • #2
a) I agree with you on T1. But for T2 block B is moving wrt block C, therefore B will experience a kinetic frictional force, [itex]f_K[/itex] coming from C. C will experiences a similar force on it (action-reaction pair), but in the opposite direction. You need to clarify the direction of [itex]f_K[/itex] for yourself as far as B and C is concerned.
 
  • #3
Well the Fk would be going in the opposite direction of the system. The equation for the Fx would then be Ft2-Fk=Fx. But then there's no coeffecient in the answer given which is 2mg-5ma. I'm lost on what that frictional force is going to be equal too. I just can't seem to get that answer.
 
  • #4
EP said:
The equation for the Fx would then be Ft2-Fk=Fx.
That's right. What you have now is the resultant force on block C (block B has only Fk in the x direction and it is not accelerating at a since it is slipping).
 
  • #5
When you work it out do you get the answer given? I have Coeffecient of friction in my answer. I ended up getting Ft=7ma+u4mg which is not the answer given.
 
  • #6
EP said:
The equation for the Fx would then be Ft2-Fk=Fx
You have the the resultant force, Fx, for block B (sorry,I changed B and C inadvertently in my previous posts!) here. Which according to the symbols in the drawing is
[tex]T_2 - f\prime = F_x[/tex]
now since this is the resultant force for B
[tex]F_x=m_B a[/tex]
which will therefore give
[tex]T_2=3ma + f\prime[/tex]
and since
[tex]T_1=2m(g-a)[/tex]
one can get [itex]f\prime[/itex] in terms of m,g and a by setting [itex]T_2=T_1[/itex]
 

Attachments

  • sliding blocks 2.jpg
    sliding blocks 2.jpg
    12.8 KB · Views: 486
Last edited:
  • #7
andrevdh said:
You have the the resultant force, Fx, for block B (sorry,I changed B and C inadvertently in my previous posts!) here. Which according to the symbols in the drawing is
[tex]T_2 - f\prime = F_x[/tex]
now since this is the resultant force for B
[tex]F_x=m_B a[/tex]
which will therefore give
[tex]T_2=3ma + f\prime[/tex]
and since
[tex]T_1=2m(g-a)[/tex]
one can get [itex]f\prime[/itex] in terms of m,g and a by setting [itex]T_2=T_1[/itex]

Sorry to bring up an old thread but I have this same exact problem and stuck on part b. I don't think you can set T1 = T2 because the pulley isn't massless, wouldn't you need to torque equations to solve this?
 
  • #8
Feldoh said:
Sorry to bring up an old thread but I have this same exact problem and stuck on part b. I don't think you can set T1 = T2 because the pulley isn't massless, wouldn't you need to torque equations to solve this?

As long as the string is assumed to be massless (or not accelerating), which is normally the case in textbook problems, the tension throughout the string is the same.
 
  • #9
So it doesn't matter that the pulley isn't massless?
 
  • #10
Feldoh said:
So it doesn't matter that the pulley isn't massless?

It does matter if a pulley has mass or not, but I didn't see where the problem states that it has a mass. So it is assumed to be massless unless otherwise stated.
 
  • #11
Well after re-reading the original post it appears that it does have a mass associated with it, so yes you have to consider the torque.
 
  • #12
mass of the pulley adds more inertia to the whole system.
Inertia always slows the effects caused by the applied forces.
But all the forces here are the same whether the pulley has mass or none at all.
(hope that makes sense)

Remember that all the forces stay the same as if the pulley were weightless.
It's just the acceleration that's affected by the mass of the pulley

They've given you the moment of inertia which is the circular equivalent of mass
It's a hint, And they obviously want you to factor that in.
 

1. What is the kinetic coefficient of friction?

The kinetic coefficient of friction is a measure of the amount of resistance between two surfaces in contact when one is moving relative to the other. It is represented by the symbol μk and is unitless.

2. How do you determine the kinetic coefficient of friction?

The kinetic coefficient of friction can be determined by conducting an experiment in which one surface is dragged or pushed against another surface at a constant speed. The force required to maintain this constant speed is then measured and divided by the weight of the object to obtain the kinetic coefficient of friction.

3. What factors affect the kinetic coefficient of friction?

The kinetic coefficient of friction can be affected by various factors such as the nature and roughness of the two surfaces in contact, the presence of any lubricants or contaminants, the temperature, and the applied force or weight.

4. How does the kinetic coefficient of friction differ from the static coefficient of friction?

The kinetic coefficient of friction is the measure of resistance between two surfaces in contact when one is already in motion, while the static coefficient of friction is the measure of resistance when the two surfaces are at rest. The kinetic coefficient of friction is often lower than the static coefficient of friction.

5. Why is the kinetic coefficient of friction important?

The kinetic coefficient of friction is important in various fields of science and engineering as it helps determine the amount of force required to keep objects in motion, the efficiency of machines, and the design of materials and surfaces for specific applications. It also plays a crucial role in understanding the physical properties of materials and their interactions with each other.

Similar threads

  • Introductory Physics Homework Help
Replies
13
Views
857
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
16
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
803
  • Introductory Physics Homework Help
Replies
16
Views
3K
  • Introductory Physics Homework Help
Replies
3
Views
703
  • Introductory Physics Homework Help
3
Replies
102
Views
4K
  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
2K
Back
Top