Two blocks on a plane with a pulley

In summary: Thanks for answering my question.In your equation for m1, you have effectively defined the acceleration a as positive up the slope. That means it must be positive downwards for m2, but you wrote your m2 equation as though it is still positive up.
  • #1
jeff12
40
2

Homework Statement


A block of mass m1 = 3.70 kg on a frictionless inclined plane of angle 30.0° is connected by a cord over a massless, frictionless pulley to a second block of mass m2= 2.30 kg hanging vertically. What are (a) the magnitude of the acceleration of each block and (b) the direction of the acceleration of m2? (c) What is the tension in the cord?

upload_2015-10-21_22-40-44.png

Homework Equations


F=ma

The Attempt at a Solution


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I set the coordinate system according to the diagram above.

System: m1
Ft+Fg=m1a
Ft-m1gsinθ=m1a

System: m2
Ft+Fg=m2a
Ft-m2g=m2a
Ft=m2g+m2a

Then I plug in both of them.
m2g+m2a+m1gsinθ=m1a
m2a-m1a=-m1gsinθ-m2g
Then I sub in the values
a=3.15?

I don't know what I did wrong. The correct answer is 0.735.
 

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  • #2
You've mad a mistake with the signs. Consider the cord as your coordinate and define its positive direction from m1 to m2. Now redo your force balances with all forces in each direction with the same sign.
 
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  • #3
stockzahn said:
You've mad a mistake with the signs. Consider the cord as your coordinate and define its positive direction from m1 to m2. Now redo your force balances with all forces in each direction with the same sign.

Why are all forces positive? That doesn't sound right.
 
  • #4
jeff12 said:
Why are all forces positive? That doesn't sound right.

Sorry, maybe I didn't express myself clearly. Of course you have to distinguish the direction of the forces by using the correct signs. After defining the positive direction, you know all forces pointing in this direction have positive signs and the rest has negative signs.
 
  • #5
stockzahn said:
Sorry, maybe I didn't express myself clearly. Of course you have to distinguish the direction of the forces by using the correct signs. After defining the positive direction, you know all forces pointing in this direction have positive signs and the rest has negative signs.

So how would you define it? Which part of my signs is wrong?
 
  • #6
It's up to you, but as already proposed in my previous post: m1 → m2: positive
 
  • #7
stockzahn said:
It's up to you, but as already proposed in my previous post: m1 → m2: positive

So if to the right of M1, which is on the x-axis to the right was positive then the M2 going up would be positive too?
 
  • #8
jeff12 said:
So if to the right of M1, which is on the x-axis to the right was positive then the M2 going up would be positive too?

If your coordinate s along the chord is positive from m1 → m2, then for

m1: sgn(Ft) = + 1 & sgn(Fg) = - 1 ... according to the x-axis you defined
m2: sgn(Ft) = - 1 & sgn(Fg) = + 1 ... as s is positive along the chord in direction m1 → m2
 
  • #9
stockzahn said:
If your coordinate s along the chord is positive from m1 → m2, then for

m1: sgn(Ft) = + 1 & sgn(Fg) = - 1 ... according to the x-axis you defined
m2: sgn(Ft) = - 1 & sgn(Fg) = + 1 ... as s is positive along the chord in direction m1 → m2

Why is Fg positive for m2 Fg? Can you draw it out on the whiteboard which sign should be where?
 
  • #10
Sorry, whiteboard doesn't work on my computer. To see it just draw an arrow parallel to the chord direction pointing from m1 to the pulley and a second one after the pulley pointing in direction of m2. These are the positive directions of your cordinate s. Now compare the direction of each force at m1 with the first arrow: If the force points in the same direction the sign is +. Same for m2 and the second arrow.
 
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  • #11
jeff12 said:
So how would you define it? Which part of my signs is wrong?
In your equation for m1, you have effectively defined the acceleration a as positive up the slope. That means it must be positive downwards for m2, but you wrote your m2 equation as though it is still positive up.
 
  • #12
Nice post
 

Related to Two blocks on a plane with a pulley

1. How does the pulley affect the movement of the blocks?

The pulley changes the direction of the force applied to the blocks. As one block moves down, the other block moves up, allowing the blocks to move in opposite directions.

2. What is the purpose of using a pulley in this setup?

The pulley reduces the amount of force needed to move the blocks. It also allows for the blocks to move in opposite directions, which is useful in certain scenarios.

3. How does the weight of the blocks affect the system?

The weight of the blocks determines the amount of force needed to move them and the speed at which they move. Heavier blocks will require more force to move and will move slower than lighter blocks.

4. Can the blocks move at different speeds?

Yes, the blocks can move at different speeds depending on their weight and the force applied to them. The lighter block will typically move faster than the heavier block.

5. What are the factors that affect the efficiency of the pulley system?

The efficiency of the pulley system depends on the weight of the blocks, the angle of the pulley, and the friction between the pulley and the rope. A larger angle and lower friction will result in a more efficient system.

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