Mass Sliding at An Angle; find mass of one

AI Thread Summary
The discussion centers on calculating the mass M1 in a system involving two connected masses, M1 and M2, with M2 moving downwards while accelerating. Key concepts include the need to resolve forces acting on M1, particularly the components of gravitational force along and perpendicular to the inclined plane, as well as the effects of tension and friction. The normal force for M1 is not simply its weight but must account for the incline's angle and friction. The calculations provided indicate that the friction force must be included, as it is derived from the normal reaction force multiplied by the coefficient of friction. Ultimately, the participants emphasize the importance of correctly applying Newton's laws and understanding force components in this physics problem.
cassienoelle
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Homework Statement


M1 and M2 are two masses connected as shown.
(M1 on hypotenuse of triangle, M2 on right side, (adjacent to right angle)) (Pulley at corner of them, Angle Theta is on the opposite corner)
The pulley is light and frictionless. Find the mass M1, given that M2 (4.50 kg) is moving downwards and accelerates downwards at 3.09 m/s2, that θ is 20.0°, and that μk is 0.450.



Homework Equations


The total force (which is the total mass times the acceleration) is equal to the sum of the forces. The normal force is NOT simply "m1g"--WHY?



The Attempt at a Solution

 
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It would be better if you could show us a diagram of what you are talking about. :smile:
 
cassienoelle said:

Homework Statement


M1 and M2 are two masses connected as shown.

I couldn't see the "as shown"?
 
cassienoelle said:

Homework Statement


M1 and M2 are two masses connected as shown.
(M1 on hypotenuse of triangle, M2 on right side, (adjacent to right angle)) (Pulley at corner of them, Angle Theta is on the opposite corner)
The pulley is light and frictionless. Find the mass M1, given that M2 (4.50 kg) is moving downwards and accelerates downwards at 3.09 m/s2, that θ is 20.0°, and that μk is 0.450.



Homework Equations


The total force (which is the total mass times the acceleration) is equal to the sum of the forces. The normal force is NOT simply "m1g"--WHY?



The Attempt at a Solution


OK, I was expecting "as shown" to be a diagram.

The Normal force is perpendicular to the slope [that is what the "Normal" means - as in Tangent and normal ... Tangent, parallel to surface; Normal, perpendicular to surface.]
mg is vertically down, so you need to use some components to calculate the Normal force.
 
cassienoelle said:

Homework Statement


M1 and M2 are two masses connected as shown.
(M1 on hypotenuse of triangle, M2 on right side, (adjacent to right angle)) (Pulley at corner of them, Angle Theta is on the opposite corner)
The pulley is light and frictionless. Find the mass M1, given that M2 (4.50 kg) is moving downwards and accelerates downwards at 3.09 m/s2, that θ is 20.0°, and that μk is 0.450.



Homework Equations


The total force (which is the total mass times the acceleration) is equal to the sum of the forces. The normal force is NOT simply "m1g"--WHY?



The Attempt at a Solution


Without the string, M2 would accelerate down at g - either 9.8, 9.81 or 10 depending which value you have been advised to use.
The fact it accelerates at only 3.09 means the string is providing a force trying to prevent the acceleration. This force is due to two contributions.
Part of it will be friction between M1 and the slope.
Part of it will be the component of the weight force of M1 acting down the slope.
 
yes, but what components?
 
IMAGE:
https://s3.lite.msu.edu/res/msu/physicslib/msuphysicslib/09_Force_and_Motion/graphics/prob75_fricpullplane.gif
 
Last edited by a moderator:
cassienoelle said:
IMAGE:
https://s3.lite.msu.edu/res/msu/physicslib/msuphysicslib/09_Force_and_Motion/graphics/prob75_fricpullplane.gif

No image present.
 
Last edited by a moderator:
cassienoelle said:
yes, but what components?

We always take components parallel to and perpendicular to something convenient,
Here that is parallel to the slope and perpendicular to the slope.

You want those components of M1g.
 
  • #10
The picture should be attached now.
 

Attachments

  • masssliding.gif
    masssliding.gif
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  • #11
cassienoelle said:
IMAGE:
https://s3.lite.msu.edu/res/msu/physicslib/msuphysicslib/09_Force_and_Motion/graphics/prob75_fricpullplane.gif

You have to work out to get the image. We can only see it if we log-in and we certainly don't want your username and password posted here.
 
Last edited by a moderator:
  • #12
Ok, first let's make the equation for the M2 block.
What are the forces acting on the M2 block? :smile:
 
  • #13
On the M2 block there is gravity and normal force ?
 
  • #14
cassienoelle said:
On the M2 block there is gravity and normal force ?

Read post 5 and think about it.
 
  • #15
cassienoelle said:
On the M2 block there is gravity and normal force ?

You are forgetting one force "Tension". :smile:
Now form the equation for M2 block.
 
  • #16
M2 : T - mg = ma
?
 
  • #17
cassienoelle said:
M2 : T - mg = ma
?

Yep, that's right.
Now plug in the values and find tension. :)
 
  • #18
Pranav-Arora said:
Yep, that's right.
Now plug in the values and find tension. :)

is the acceleration negative because the box is falling down?
 
  • #19
cassienoelle said:
is the acceleration negative because the box is falling down?

No. :)
 
  • #20
k, so T = 58.05N
 
  • #21
I haven't calculated it so i am assuming it as right. :)

So for the M1 block, what are the forces acting on it? Think about this one carefully and form the equation. :smile:
 
  • #22
Tension, weight, and normal force?
 
  • #23
cassienoelle said:
Tension, weight, and normal force?

You again missed one "Friction" :)
So now form the equation.
 
  • #24
Honsestly, I have zero clue how to form the equation.
 
  • #25
cassienoelle said:
Honsestly, I have zero clue how to form the equation.

Ok, i give you a clue.
We have weight, normal, friction and tension.

The tension acts upwards along the string. We have two components for weight. Can you find them?
 
  • #26
well i know that : weight = mass * gravity
 
  • #27
cassienoelle said:
well i know that : weight = mass * gravity

Weight is acting downwards but we need to find what effect does it have along the inclined plane. For that you need to take component of mg along the inclined plane. Can you do that? :smile:
 
  • #28
W = cos20 * T
?
 
  • #29
cassienoelle said:
W = cos20 * T
?

That's completely absurd.
I think you are not able to picture out what i am talking about. Ok here's a picture:-

2m7yals.png


Can you find the component (the yellow arrow)?
 
  • #30
No, I can't.
 
  • #31
cassienoelle said:
No, I can't.

Aren't you knowing about vectors? Are you trying to do these questions yourself without any guidance?
If not, then this question should not bother you much. :)
 
  • #32
I know about vectors. But yes, I am trying to do these with almost no guidance. It's an online course and I have no idea what's going on. I would appreciate it if you would stop making me feel like an idiot. I came onto this website for the guidance.
 
  • #33
cassienoelle said:
I know about vectors. But yes, I am trying to do these with almost no guidance. It's an online course and I have no idea what's going on. I would appreciate it if you would stop making me feel like an idiot. I came onto this website for the guidance.

Really? Did i make you feel like that. I am very sorry if i did so.:frown:

Ok i will tell you how to do so in a minute. I am making pictures to make you understand. :)
 
  • #34
I want to just tell you that these question are categorized under Laws of Motion. :)
If you are not familiar with these, then i suggest you to complete the Khan Academy videos on Laws of Motion. :)

Here's the explanation:-
Let us consider an object of mass m kept on an inclined plane as shown in the figure.
16iwirp.png


The blue arrows indicate forces. Now to find the component along the inclined plane, draw a straight line as shown in the figure. The straight line is along the force mg.
21d4u2q.png


Now in triangle ABC, angle BAC is 90-\theta. Now its easy to have the components. The components are shown in the figure below.

i3bdza.png


So we have found out the components.
So now can you tell me what’s the normal reaction force here?

(Ask if you have any doubts, because this is a very important concept, don't miss this one)
 
Last edited:
  • #35
cassienoelle said:
On the M2 block there is gravity and normal force ?

I am not happy with your terminology use here.
You use the term Normal force - you should be using the full name Normal Reaction Force.
this reminds you it is referring to a force, perpendicular to a surface, which comes about as the action-reaction pair of something applying a force to the surface.

For M2, there is no surface involved, so there will be no Normal Reaction Force.

M2 does have a string attached. That means M2 will pull on the string, and the string will pull on the mass [that is an action reaction pair]. One of that pair acts on M2 - the string pulling the mass. We can say that force is due to the Tension in the string if you like.

So the forces acting on M2 are Weight [due to Gravity] and Tension.

M1 can have a Normal Reaction Force, since it is touching a surface.
Because the surface is rough there may be a Friction Force.
There is also, of course, the weight of M1.

Weight acts vertically down.
Friction acts parallel to the surface.
The Normal Reaction force acts at right angles to the surface [of course it does! That is why it is called the Normal Reaction Force].
 
  • #36
Pranav-Arora said:
Really? Did i make you feel like that. I am very sorry if i did so.:frown:

Ok i will tell you how to do so in a minute. I am making pictures to make you understand. :)

You are being given some wrong information by Pranav-Arora.

In post 16 you said

M2 : T - mg = ma

The fact you used T positive, and g as negative [mg actually, but m is not a vector] means you have defined Down as negative, and Up as positive, which is fine - you can define positive in any direction you like, provided you remain consistent throughout the problem.

You then asked, a couple of posts later, whether the acceleration would be negative since it was directed down.

The answer to that question should have been YES.

back to Mass M1

The weight Force M1.g can be resolved into two components, one parallel to the slope and one perpendicular to the slope.

They are M1.g.sin20 and M1.g.cos20 respectively.

NOTE: if you can't remember which one is which, consider the following: if the slope was nearly vertical, the parallel component would be almost equal to the weight force, while the perpendicular component would be almost zero. COsine is the function that approached zero for angles close to 90o, so the component with the cos must be the perpendicular component.

The parallel component, M1.g.sin20, will tend to make the mass accelerate down the slope.
The perpendicular component M1.g.cos20 will be balanced by the Normal Reaction Force - so FN = M1.g.cos20
There will be a friction force trying to stop M1 from moving in either direction
There is also the Tension in the string trying to accelerate the box UP the slope.

[We know Tension "wins" because we were told M2 accelerates down, and M1 is tied to it.

So the net force on M1 is M1.g.sin20 + Friction - T

The size of friction is FN* coefficient of friction.

*** I have followed your sign convention. You said that for M2, down was negative. M1 is tied to M2 so M1 up the slope is negative.

This net force will accelerate the M1 up the slope with the same acceleration that M2 falls. {they are tied together!]

Try evaluating some of these values and see how you get on.
 
  • #37
Sorry for the mistake by me. I would take care of that from now onwards.
Thanks PeterO for pointing out. :smile:
 
  • #38
So if
M2: T-mg=ma
Then T-(4.5)(9.81) = (4.5)(3.09)
Then T=58.05 ?
And if
M1: mgsin20+u-T = ma
then
m(9.81)(sin20) + .450 - 58.05 = 3.09m
so m = 217.35kg

but the homework says that's not right.
 
  • #39
cassienoelle said:
So if
M2: T-mg=ma
Then T-(4.5)(9.81) = (4.5)(3.09)
Then T=58.05 ?
And if
M1: mgsin20+u-T = ma
then
m(9.81)(sin20) + .450 - 58.05 = 3.09m
so m = 217.35kg

but the homework says that's not right.

Because you missed friction here. :smile:

And did you visit Khan Academy as i said?
 
  • #40
cassienoelle said:
So if
M2: T-mg=ma
Then T-(4.5)(9.81) = (4.5)(3.09)
Then T=58.05 ?
And if
M1: mgsin20+u-T = ma
then
m(9.81)(sin20) + .450 - 58.05 = 3.09m
so m = 217.35kg

but the homework says that's not right.

You have the coefficient of friction only [highlite in red above]. You have to multiply by the Normal Reaction force to get the total friction force.
 

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