Find Mass M1 Given Acceleration, Angle, and Friction

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Homework Help Overview

The problem involves two masses, M1 and M2, connected by a pulley system, where M2 is hanging and M1 is on an incline. Given the mass of M2, its acceleration, the angle of the incline, and the coefficient of kinetic friction, the goal is to find the mass M1.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the use of free body diagrams and the application of Newton's second law. There are attempts to calculate tension and consider the effects of gravity and friction. Questions arise about the correct interpretation of forces and the setup of equations.

Discussion Status

Participants are actively engaging with the problem, sharing their diagrams and equations. Some have provided feedback on the correctness of the setups, and there is an ongoing exploration of the relationships between the forces involved. A participant has reported a calculation leading to a potential solution, but the discussion remains open for further verification and exploration.

Contextual Notes

There is a mention of a diagram that is not provided, which may affect the clarity of the discussion. Additionally, participants are navigating through assumptions about the forces acting on the masses and the definitions of variables used in their equations.

thatguy101
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Homework Statement


M1 and M2 are two masses connected as shown (M2 hangs over the table). The pulley is light (massless) and frictionless. Find the mass M1, given that M2 (3.0 kg) accelerates downwards at 2.31 m/s^2, q is 20°, and μk is 0.41.


Homework Equations


F=ma
μn=friction


The Attempt at a Solution


Well I thought I could find tension using F=3*2.31 but that didn't get me anywhere. I drew a force diagram for m1, but it looks like there's too much missing to be able to continue. Is the tension part at least right?
 
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hi thatguy101! welcome to pf! :smile:

(we usually use "T" for tension :wink:)
thatguy101 said:
Well I thought I could find tension using F=3*2.31

is that Ftotal = ma?

what about gravity? :wink:
 
so then would I subtract gravity?
Ft=3*2.31-3*9.81?
 
Why not give us the diagram so that we know what q = 20^{o} refers to?
 
Sorry. It's suppose to say θ.
m1 is on an incline of 20°, and m2 is hanging off the incline.
 
Ok so my diagram looks like this.
For m1 I have forces in the y direction as y: Fg*cosθ-Fn=0
And the forces in the x direction as x:Ft-Fg*sinθ-Ff=a.
And then for m2 there are none in the x direction but for y I have y: Fg-Ft=ma.
Am I right so far?
 
thatguy101 said:
Ok so my diagram looks like this.
For m1 I have forces in the y direction as y: Fg*cosθ-Fn=0
And the forces in the x direction as x:Ft-Fg*sinθ-Ff=a.
And then for m2 there are none in the x direction but for y I have y: Fg-Ft=ma.
Am I right so far?

yes, except in your second equation, you need to check the signs (and you forgot to write the m before the a) :wink:
 
Thank you. I figured it out.
So I had Ft-mg*sinθ+μ*mg*cosθ=ma
so Ft=mg*sinθ+μ*mg*cosθ+ma
factor out the m, Ft=m(g*sinθ+μ*g*cosθ+a)
then solved for m. and since Ft= m2g-m2a, I just put in the numbers and came up with 22.47 N amd just plugged in everything else into the equation above and got 2.38 kg
 

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