Friction/spring/inclined plane problem

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
7 replies · 3K views
Alex.malh
Messages
17
Reaction score
0

Homework Statement


There is a block of 1kg (m1) resting frictionless on another block of 5kg (m2). m1 is connected by a horizontal spring to m2. m2 is resting on an inclined plane of 45°. Between m2 and the plane there is friction.
A force of 200N is applied on m2, pushing it upwards.

F=200N, m1=1kg, m2=5kg, f=0.2, k=100N/m, initial spring length= 0.20m, g=10m/s²

What is the spring length at equilibrium?

Homework Equations


F=m*a
F=k*spring length
Friction W= (reaction force R)*f

The Attempt at a Solution


Increase in spring length = l
I start with 4 equations:
F - w*cos45 - R*cos45 - k*l = (m1+m2)*a
m1*g + m2*g - w*sin45 - R*sin45 = 0
w=0.2*R
k*l=m1*a

Solving this i get a=20m/s² which can't be right.
 

Attachments

  • Problem.PNG
    Problem.PNG
    2.3 KB · Views: 635
Physics news on Phys.org
Why do you add k*l in 1st equation ??

You should be writing 1st equation considering both blocks together as one system . You will then not get k*l in the equation .

The rest seems fine .
 
Alex.malh said:
F - w*cos45 - R*cos45 - k*l = (m1+m2)*a
m1*g + m2*g - w*sin45 - R*sin45 = 0
As qwertywerty notes, that kl term should not be there. That is internal to the two-mass system.
But the second equation looks wrong too. Check the signs, and consider what the vertical acceleration is.
 
You're right - my bad . So you would have to consider some net acceleration and write horizontal and vertical accelarations in it's terms .
 
haruspex said:
As qwertywerty notes, that kl term should not be there. That is internal to the two-mass system.
But the second equation looks wrong too. Check the signs, and consider what the vertical acceleration is.
Thanks for the quick reply.
New equations:
F - w*cos45 - R*cos45 = (m1+m2)*ah
(m1+m2)*g + w*sin45 - R*sin45 = (m1+m2) *av
w=0.2*R

Now to solve this I'm still missing a relationship between ah and av.
I've taken av = ah ( angle is 45°, if the block has moved 1m to the right, it will have moved the same distance vertically, xv = xh -> av = ah)
Now i end up with R=495N and a=36.7m/s²

Not possible imo.
Do you guys see where i go wrong?

thanks!

regards,
 
Would you check your calculations ?
 
Chestermiller said:
In your second equation, you have the wrong sign on the ma term. The blocks are accelerating vertically upward. The way you have it, av = - ah

Chet

Ah yes of course. Well, I've got it now. 7cm spring elongation :)
Thx a lot!