Friction/spring/inclined plane problem

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Homework Help Overview

The problem involves a block system where a 1kg block rests on a 5kg block, with the latter on an inclined plane. A force is applied to the lower block, and the system includes a spring connecting the two blocks. The discussion centers around determining the spring length at equilibrium, considering forces, friction, and acceleration.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the equations of motion for the system, questioning the inclusion of certain terms and the overall setup. There are attempts to derive relationships between horizontal and vertical accelerations, and some participants express confusion over the signs in their equations.

Discussion Status

There is ongoing dialogue about the correctness of the equations being used, with some participants providing feedback on each other's reasoning. Multiple interpretations of the problem setup are being explored, particularly regarding the treatment of the spring force and the acceleration components.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or assumptions they can make. The problem involves friction and inclined planes, adding complexity to the analysis.

Alex.malh
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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.
 

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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 ?
 
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
 
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!
 

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