Forces help

  • Thread starter thatguy101
  • Start date
  • #1
14
0

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?
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
25,832
251
hi thatguy101! welcome to pf! :smile:

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

what about gravity? :wink:
 
  • #3
14
0
so then would I subtract gravity?
Ft=3*2.31-3*9.81?
 
  • #4
993
13
Why not give us the diagram so that we know what q = 20[itex]^{o}[/itex] refers to?
 
  • #5
14
0
Sorry. It's suppose to say θ.
m1 is on an incline of 20°, and m2 is hanging off the incline.
 
  • #6
14
0
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?
 
  • #7
tiny-tim
Science Advisor
Homework Helper
25,832
251
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:
 
  • #8
14
0
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
 

Related Threads on Forces help

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
857
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
17
Views
2K
  • Last Post
Replies
6
Views
4K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
992
Top