Force & Friction Between 2 Prisms: Homework Solution

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In summary: And i just realize that i should not use F in the horizontal components, it should be F.sinx. Right?Yes, that's right. :smile:And similarly for the vertical components, you should get …R + f.sinx + mg.sinx.sinx = 0… which gives you R in terms of everything else.ok, it is R = mg(sin^2x - cos^2x) + F.sinxR = mg.sin2x + F.sinxThen i don't know what to do next. Can you explain it to me?You need to use the fact that the two prisms have the same acceleration.What does that tell you about the forces on the
  • #1

Homework Statement


A prism has the mass m with the angle x = 45°(look at the picture) placed on the ground without friction. The other prism, which has the same mass m, is placed above the first prism(the first prism is bigger). A horizontal force F is given to the second prism so the displacement with respect to the first prism is zero( or not moving). What is the friction between the two prism ( in F, m, and g). g = Earth gravitational acceleration

The picture -> http://i28.tinypic.com/nycih4.jpg

Homework Equations


[tex]F = m.a[/tex]

The Attempt at a Solution


[tex]F.cosx - f - m.g.sinx = m.a[/tex]

Since the prism is not moving, so a = 0

[tex]F.cosx - \mu.m.g.cosx - m.g.sinx = 0[/tex]
[tex]F.cosx = \mu.m.g.cosx - m.g.sinx[/tex]
[tex]\mu = \fric{F}{m.g} + tanx[/tex]


Then i have no idea what to do next. I haven't even used the mass of the first prism. I also don't quite understand how the forces work so can somebody help me please?
 
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  • #2
"Parable of the Surveyors"

Hi Vermillion! :smile:

The big prism is "placed on the ground without friction".

So both prisms are accelerating, with the same acceleration.

Start again! :smile:
 
  • #3
Thanks Tim,

I corrected the plus-minus mistake. It should be
[tex]F = mg(\mu + tanx)[/tex]

Then the acceleration of the prisms is

[tex]F = (m + m)a[/tex]
[tex]F = 2ma[/tex]
[tex]2ma = mg(\mu + tanx)[/tex]
[tex]a = \frac{g} {2}(\mu + tanx)[/tex]

Finally, i don't know how to relate the acceleration to the friction. Can you explain it to me?
 
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  • #4
Vermillion said:
[tex]F = (m + m)a[/tex]
[tex]F = 2ma[/tex]
[tex]2ma = mg(\mu + tanx)[/tex]
[tex]a = \frac{g} {2}(\mu + tanx)[/tex]

Finally, i don't know how to relate the acceleration to the friction. Can you explain it to me?

Hi Vermillion! :smile:

You're making the very common mistake of assuming that the friction force is µ times the normal force.

For static friction, it usually isn't!

µ times the normal force is the maximum for static friction.

It only applies if the question specifically tells you that the object is "on the point of moving" (or "just about to move", or something like that).

Here, the question not only doesn't say that … but it doesn't even bother to tell you what µ is (which should have been a clue, shouldn't it? :rolleyes:).

You have to use Newton's first and third laws, to find the reaction force, R.

Forget everything you know about friction.

Draw a new diagram, put in R, and take horizontal and vertical components. :smile:
 
  • #5
Do you mean, i should throw in F.Sinx to my friction equation? Which i forgot to do.
 
  • #6
Vermillion said:
Do you mean, i should throw in F.Sinx to my friction equation? Which i forgot to do.

No :smile: … I meant exactly what I said, which was …
tiny-tim said:
Forget everything you know about friction.

Draw a new diagram, put in R, and take horizontal and vertical components. :smile:
 
  • #7
Just for quick clarification, you mean like this?

Horizontal Components
[tex]\SigmaFx = 0[/tex]
[tex]F - f.cosx - mg.sinx.cosx = 0[/tex]

Sorry, I'm not used to talk physics in english.
 
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1. What is the definition of force?

Force is a physical quantity that is used to measure the amount of interaction between two objects. It can be described as a push or pull on an object that causes it to change its motion or shape.

2. How is force related to the friction between two prisms?

Force plays a crucial role in determining the amount of friction between two prisms. The greater the force applied, the greater the friction between the surfaces of the prisms. This is because the force creates more contact between the surfaces, leading to more resistance and friction.

3. What factors affect the amount of friction between two prisms?

There are several factors that can affect the amount of friction between two prisms, including the force applied, the surface area and texture of the prisms, and the type of material they are made of. Other factors such as temperature and lubrication can also play a role.

4. How can the amount of friction between two prisms be reduced?

To reduce the amount of friction between two prisms, one can decrease the force applied, use smoother surfaces, or apply a lubricant between the surfaces. Reducing the surface area of contact between the prisms can also decrease friction.

5. How is the coefficient of friction calculated?

The coefficient of friction is calculated by dividing the force of friction by the normal force (the force exerted by the surface on the object). This value is dependent on the materials of the surfaces in contact and can be experimentally determined.

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