Three blocks, two pulleys, on a incline

  • Thread starter gap0063
  • Start date
  • #1
65
0

Homework Statement


The suspended 2.29 kg mass on the right is
moving up, the 1.3 kg mass slides down the
ramp, and the suspended 7.7 kg mass on the
left is moving down. There is friction between
the block and the ramp.
The acceleration of gravity is 9.8 m/s2 . The
pulleys are massless and frictionless.
[PLAIN]http://img816.imageshack.us/img816/6190/009q.png [Broken]

What is the acceleration of the three block
system?
Answer in units of m/s2


Homework Equations


The sum of Fx= N-mgcostheta=o => N=mgcostheta
The sum of Fy= mgsintheta-usN=0=> mgsintheta-us(mgcostheta)


The Attempt at a Solution



are these the correct eqations for this?
 
Last edited by a moderator:

Answers and Replies

  • #2
jhae2.718
Gold Member
1,161
20
You'll want to look at each block in isolation. The system will have the same acceleration.

What forces act on the two hanging blocks?
What forces act on the block sliding down the incline?

What direction is the net acceleration?

Hint: there are two tension forces.
 
  • #3
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,167
507
For the mass on the incline, choose the horizontal axis as the x axis, and the axis perpendicular to the incline as the y axis. In which case you have your Fx and Fy equations reversed. Then once corected , this will give you the normal force acting on that one block on the incline. That's step one......then its free body, free body, free body diagrams! Answer jhae2.718 questions. Look at each block separately, identify alll forces acting, and apply Newton's laws.
 
  • #4
65
0
You'll want to look at each block in isolation. The system will have the same acceleration.

What forces act on the two hanging blocks?
What forces act on the block sliding down the incline?

What direction is the net acceleration?

Hint: there are two tension forces.

For block on the left... the mass I'll call m1... the forces acting on it are T1 acting up and m1g acting down

For block on incline... the mass I'll call m2... the forces act on it in the y direction of N and m2g and in the x directuon fk and T2

For the block on the right... the mass I'll call m3... the forces acting on it are T1 acting up and m3g acting down

Since m1=7.7 and is bigger than 2.29 the acceleration is pulling m3 up and m1 down...

So if I use the equations I gave before for the block on the incline and just use
sum of Fy=T1-mg=0 => T1=mg

but where does the acceleration come in for m1 and m3?
 

Related Threads on Three blocks, two pulleys, on a incline

  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
12
Views
10K
Replies
4
Views
2K
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
2
Views
6K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
11
Views
6K
Top