Mass Sliding at An Angle; find mass of one

[PeterO]

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

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 90^o, 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 F_N = 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 F_N* 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.

 

[Saitama]

Sorry for the mistake by me. I would take care of that from now onwards.
Thanks PeterO for pointing out.

 

[cassienoelle]

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.

 

[Saitama]

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.

And did you visit Khan Academy as i said?

 

[PeterO]

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.