# Forces help

## 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.

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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tiny-tim
Homework Helper
hi thatguy101! welcome to pf!

(we usually use "T" for tension )
Well I thought I could find tension using F=3*2.31
is that Ftotal = ma?

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?

tiny-tim
Homework Helper
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)

Thank you. I figured it out.
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