Find Mass M1 Given Acceleration, Angle, and Friction

AI Thread Summary
To find the mass M1 given the acceleration of M2, the angle of incline, and the coefficient of friction, the tension in the system must be calculated. The equations of motion for both masses involve gravitational forces, tension, and friction, with M1 on a 20° incline and M2 hanging vertically. After establishing the correct force diagrams and equations, the relationship between tension and the forces acting on M1 and M2 is clarified. The final calculations yield a mass M1 of approximately 2.38 kg. This approach effectively combines Newton's laws with the specifics of the problem to arrive at the solution.
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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