Mass Sliding at An Angle; find mass of one

In summary: Friction and normal?Friction and normal?Weight and normal force are perpendicular to the string. Friction and tension are parallel to the string.
  • #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.
 
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  • #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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