Calculating Contact, Weight & Frictional Forces on a Moving Platform

[PhysicsIdiot]

http://img213.exs.cx/img213/852/sita2ju.jpg

Look at the link above for reference (sorry, I am not very skilled at paint)

It's on a horizontal platform moving with a constant acceleration.

Question 1: Calculate the contact forces between objects 1 and 2.

My Answer: It's a simple F=mg, surface area doesn't matter with forces, only pressure does. But do i use both masses or just mass1? (the force of mass1 ON mass2)

Question 2: Calculate the weight that would show on the scale.

My Answer: W = (m1 + m2)g

Question 3: Calculate the frictional forces assuming that the platform, scale and objects 1 and 2 are made out of the same material of coefficient of static friction "mew sub s".

My Answer: I'm still working this one out so any help would be greatly appreciated. I'm just writing/drawing out the forces on each object; haven't come to anything definate yet.

 

[Andrew Mason]

Question 1: Calculate the contact forces between objects 1 and 2.

My Answer: It's a simple F=mg, surface area doesn't matter with forces, only pressure does. But do i use both masses or just mass1? (the force of mass1 ON mass2)

m1 exerts a force of m1g on m2. m2 exerts an upward force on m1 equal to m1g because m1 does not accelerate in the vertical direction.

Question 2: Calculate the weight that would show on the scale.

My Answer: W = (m1 + m2)g

I can't tell where the scale is. Is it under m2? If so, you are right.

Question 3: Calculate the frictional forces assuming that the platform, scale and objects 1 and 2 are made out of the same material of coefficient of static friction "mew sub s".

The frictional force is a function of acceleration and the masses m1 and m2. It does not depend on \mu_s. \mu_smg determines the maximum frictional force that the surfaces can provide but they only provide that when a force of that magnitude is applied to the masses.

AM

 

[PhysicsIdiot]

m1 exerts a force of m1g on m2. m2 exerts an upward force on m1 equal to m1g because m1 does not accelerate in the vertical direction.

Isn't that the other way around? m2 exerts a force (down) of (m2)g on m1 and m1 exerts the same (up) onto m2? And does m1 exert a force of (m1)g up on m2 or does it exert a force equal to m2's force (i.e. m2 * g)

I can't tell where the scale is. Is it under m2? If so, you are right.

yeah, it's right under m2.

The frictional force is a function of acceleration and the masses m1 and m2. It does not depend on \mu_s. \mu_smg determines the maximum frictional force that the surfaces can provide but they only provide that when a force of that magnitude is applied to the masses.

AM

Sooo... it's F= ma ; F = f(friction) ; ma = f

So frictional force of m2 on m1 is just (m1 * g)?

And frictional force of m1 on the platform is (m1 + m2)g?

-----

Also, the platform is accelerating at that angle? It doesn't matter because it's constant right?

 

[Andrew Mason]

Isn't that the other way around? m2 exerts a force (down) of (m2)g on m1 and m1 exerts the same (up) onto m2? And does m1 exert a force of (m1)g up on m2 or does it exert a force equal to m2's force (i.e. m2 * g)

Sorry for the confusion. I had m1 and m2 mixed up in my mind. I was thinking of m2 on the bottom.

yeah, it's right under m2.

I meant m1. If it is between the floor and m1, then the scale reads m1g + m2g.

Sooo... it's F= ma ; F = f(friction) ; ma = f

So frictional force of m2 on m1 is just (m1 * g)?

Right. That is the force that m2 applies to m1. This is because the only force that is acting on m1 is this friction force and m1's acceleration is a. Therefore, F_f = m1a

And frictional force of m1 on the platform is (m1 + m2)g?

Correct. The friction force of the platform on m1 (and of m1 on the platform) is (m1 + m2)g.

Also, the platform is accelerating at that angle? It doesn't matter because it's constant right?

I am not sure what angle you mean. It is accelerating along the direction of the x axis.

AM

 

[PhysicsIdiot]

I am not sure what angle you mean. It is accelerating along the direction of the x axis.

AM

If you look at the picture, the platform is accelerating at an angle (not straight, otherwise the arrow would be straight) (look at the bottom right)

Thanks for all the help

 

[PhysicsIdiot]

wouldnt it be: m1 + m2 (g + Asin"theta")? my friend said he got that.. i don't know

"A" being the magnitude of the acceleration

this is for question # 2

 

[Andrew Mason]

wouldnt it be: m1 + m2 (g + Asin"theta")? my friend said he got that.. i don't know

"A" being the magnitude of the acceleration

this is for question # 2

Acceleration is in the x direction only. The angle is 0.

AM

 

[PhysicsIdiot]

Acceleration is in the x direction only. The angle is 0.

AM

i guess my drawing was off so i made a nother picture; the acceleration is done at an angle:

http://img37.exs.cx/img37/3628/sitb0yi.jpg

However, it's constant so would it matter for question #2 (or any of the other questions for that matter)?

 

[PhysicsIdiot]

Anyone?

 

[PhysicsIdiot]

Bump, I am still stuck on this.

 

[whozum]

If the scale (and entire system) is accelerating at an angle then the scale will read (m1+m2)*g+(m1+m2)*g*sin(theta). Only the y-component of the acceleration is affecting the scale's reading, because with acceleration comes force. Since the system is being accelerated there is an extra force on the scale (equal to (m1+m2)*g*sin(theta). ).

Think about standing on a scale in an elevator, as the elevator accelerates upwards, your scale will read a higher weight than when it is not accelerating. Accordingly, a lower weight as it accelerates downwards.

 

[PhysicsIdiot]

does this acceleration affect questions 1 and 3?

 

[whozum]

I'm not sure but I believe so, each mass acts as a scale to all other objects it is contacting, so a larger force on the entire system equates to a larger force on each object, so the forces between the objects will increase.