Equilibrium and Tension questions - PLEASE HELP ME

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  • #1
nperk7288
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I have two questions I'm stuck on:

Question 1:

1 block rests upon a horizontal surface with an identical second block resting on top of the first block. The coefficient of static friction between the blocks equals the coefficient of static friction between the lower block and the surface. A horizontal force is applied to the top block and when the force reaches 47.0 N the upper block begins to slide. The force is then removed and the blocks are put back in their original positions. What is the magnitude of the horizontal force that should be applied to the lower block so it just begins to slide out from under the upper block?


Question 2:

A damp washcloth is hung over the edge of a table and so part of it (mass = m on) is on the table and part (mass = m off) is hanging over the edge. The coefficient of static friction between table and wash cloth is 0.40. What is the maximum fraction ( m off / (m off + m on)) of the cloth that can hang over the edge without causing the cloth to fall?

Please, any insight would be appreciated...
 

Answers and Replies

  • #2
Pyrrhus
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What have you done? Have you draw your Free Body diagrams?
 
  • #3
nperk7288
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Yes, I did Free Body diagrams but I don't know what to do, I don't want a solution, just a start, like what formulas to use first etc.
 
  • #4
nperk7288
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I'm really struggling in Physics right now because I'm doing an 8-unit full year course in 8 weeks...It's hard to understand everything when we go as quick as we have been.
 
  • #5
Pyrrhus
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Well, show me some of the equations you came up with.
 
  • #6
nperk7288
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For #1 I had 4 forces acting on the top box, FN pointing in the positive Y direction, FW or the Weight pointing in the negative Y direction, fs (the static friction force) pointing in the negative X direction, and FP = 47.0 N pointing in the positive X direction. On the bottom box I have 6 forces acting. They are ones similar to those mentioned above (a normal force pointing up, a friction forces pointing left, a force from the person pointing right and a force from the box's weight pointing down) plus an additional downward force FB2 from the weight of the top box, and an additional upward force FN pushing against the upper box. What do I do now? and is that correct?
 
Last edited:
  • #7
nperk7288
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I really need help quick; I still have a big lab to do for tomorrow as well so I will have to leave to do that in a couple minutes.
 
  • #8
Pyrrhus
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Well you got 2 cases, the first case when the 47 N force is applied, you should be able to get the coeffiecient of friction, by using Newton's 1st Law. For the 2nd case, you should calculate the horizontal force on the lower block.

About your FBD
On your first case you don't got the applied force on the lower block, and the friction force on the top of the bottom block is pointing right.
 
  • #9
nperk7288
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How does Newton's frist law help to get the coefficient of Friction?
 
  • #10
Pyrrhus
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Are you familiar with Newton's Laws? Pick up the textbook and read the chapter,
 
  • #11
OlderDan
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The statement of the first problem does not make sense. Nothing is said about applying a force to the lower block in the first part. If the upper block slides relative to the lower block without applying any additional force (it will with equal coefficients), it can also be slid back without any additional force on the lower block.

For number 2, the normal force is constant, so the friction force is constant. The amount of cloth hanging over the edge determines the force opposing the frictional force acting on the part of the cloth on the table. You can treat the cloth like a string with tension in the cloth acting between the two sections of cloth.
 
  • #12
nperk7288
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I know the laws but that has to do with an object staying at rest or in motion unless acted on by a net force
 
  • #13
nperk7288
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OlderDan said:
The statement of the first problem does not make sense. Nothing is said about applying a force to the lower block in the first part. If the upper block slides relative to the lower block without applying any additional force (it will with equal coefficients), it can also be slid back without any additional force on the lower block.

For number 2, the normal force is constant, so the friction force is constant. The amount of cloth hanging over the edge determines the force opposing the frictional force acting on the part of the cloth on the table. You can treat the cloth like a string with tension in the cloth acting between the two sections of cloth.

There is no force applied to the lower block in the first part. Only to top block is moving. The problem doesn't want to slide the top block back, the person just resets it manually and then wants to slide the lower one...

---------
-----------
---------> | Box 1 |
---------
-----------
---------
| Box 2 |
---------------------------floor

part two ... boxes are returned to normal...
-----------
| Box 1 |
-----------
------> | Box 2 |
---------------------------floor
Now the person wants to ONLY move the bottom box
 
  • #14
Pyrrhus
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Yes, now analyze both cases, with Newton's 1st Law, the first case should give you the coefficient of static friction, and with it, you can find the force applied on the 2nd case.
 
  • #15
nperk7288
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I am still completely lost... I really don't understand this... I'm about to break down!
 
  • #16
OlderDan
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part two ... boxes are returned to normal...
-----------
| Box 1 |
-----------
------> | Box 2 |
---------------------------floor
Now the person wants to ONLY move the bottom box

OK Sorry. Now I get it.
 
  • #17
Pyrrhus
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Here, start off by the top block

[tex] \sum F_{x} = ... = 0 [/tex]
 
  • #18
nperk7288
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That equals fs + FP
 
  • #19
nperk7288
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How does that give me the coefficient though?
 
  • #20
Pyrrhus
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Remember the signs.

and also

[tex] F_{f} = \mu N [/tex]
 
  • #21
nperk7288
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So 47.0 N = fs right? but I don't know what N is...
 
  • #22
Pyrrhus
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The N in my equation is the normal force.
 
  • #23
nperk7288
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But we don't know the normal force
 
  • #24
Pyrrhus
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Continue with

[tex] \sum F_{y} = ... = 0 [/tex]
 
  • #25
nperk7288
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ok that would be FN + Fmg
how can I figure that out without knowing the mass of the box...?
 
  • #26
Pyrrhus
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Well, Boxes are identical so their [itex] \mu m [/itex] is the same.
 
  • #27
OlderDan
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OlderDan said:
For number 2, the normal force is constant, so the friction force is constant. The amount of cloth hanging over the edge determines the force opposing the frictional force acting on the part of the cloth on the table. You can treat the cloth like a string with tension in the cloth acting between the two sections of cloth.

I should allow for the possibility that another assumption is made by your text in approaching this problem. It is really a very unrealistic problem, in my opinion. All the force supporting the cloth is supplied by the table, so the normal force must be constant if it is not sliding, as I stated before. However, for a limp cloth hanging over the edge of the table the normal force is nowhere near uniformly applied to the cloth. It is conceivable that the text will approach the problem by analogy to a mass on the table connected by a string to a mass hanging over a pulley, in which case they may assume a normal force that depends only on the mass still on the table. That is not correct in, my opinion, but the real solution more likely falls somewhere between the two approaches I have mentioned. The most likely thing, it seems to me, is that the edge of the table where most of the normal force is applied will have a reduced coefficient of friction compared to the larger surace area of the cloth on the table. Treating the sitution with constant normal force, and with normal force proportional to the amount oif cloth still on the table will probably give values that surround the realistic solution to the problem. Your text probably does one of the two simple approaches: either constant normal force, or normal force proportional to the mass still on the table.
 
  • #28
nperk7288
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I am still lost...
 
  • #29
OlderDan
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nperk7288 said:
I am still lost...
In post #21 you had
nperk7288 said:
So 47.0 N = fs right? but I don't know what N is...
You don't know the mass of the block, but in terms of that unknown mass you can write

[tex] 47N = f_s = \mu N = \mu Mg [/tex]

where g is the gravitational acceleration constant. You don't need to, but you could solve for

[tex] \mu = \frac{47N}{Mg} [/tex]

In order to get the bottom block to slide under the top block you must apply enough force to accelerate the bottom block slightly more than the top block. When you achieve this condition, the top block will be accelerated by the frictional force of 47N (slightly less when slipping starts because of the difference between static and kinetic friction; you need not worry about this), and by Newton's third law the top block will exert a force of 47N opposing the motion of the bottom block. The bottom block will have a normal force at its bottom equal to the weight of two blocks. This is found by considering the vertical forces acting on the bottom block, including its weight and the normal force pressing down on its top from the weight of the top block. The applied force must exceed the frictional forces opposing it by enough to accelerate the bottom block by slightly more than the acceleration of the top block. A free body diagram of the two blocks will give you these equations.

Tob block
[tex] Ma_{top} = 47N = \mu Mg [/tex]

Bottom block vertical
[tex] N_{bottom} = Mg + N_{top} = 2Mg [/tex]

Bottom block horizontal
[tex] Ma_{bottom} = F_{applied} - \mu N_{top} - \mu N_{bottom} [/tex]

[tex] F_{applied} = Ma_{bottom} + \mu N_{top} + \mu N_{bottom} [/tex]

[tex] F_{applied} = Ma_{bottom} + \mu Mg + 2\mu Mg [/tex]

The condition that [tex] a_{bottom} [/tex] must be slightly greater than [tex] a_{top} [/tex] combined with the above is all you need to find the minimum applied force.
 

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