Two objects connected by a cord, with a pulley.

  • Thread starter budd99
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  • #1
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Homework Statement



The system shown (attached image) is released from rest and moves 50 cm in 1.0 s. What is the value of M? All surfaces are frictionless.

a. 0.42 kg
b. 0.34 kg
c. 0.50 kg
d. 0.59 kg
e. 0.68 kg


Homework Equations



Since the incline is 90 degress, I am trying to model this as an Atwood machine.

I don't understand how there can be any acceleration if the mass that we need to find is smaller than 3kg.

The Attempt at a Solution



moves 50cm in 1s., which means a is 0.5m/s2

So, using the formula
a = (m2 -m1)/(m1 + m2)g

0.5 = (m2 - 3)(3 + m2)9.80

m2 = 3.32

Which is not in the solution

Homework Statement





Homework Equations





The Attempt at a Solution

 

Attachments

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Answers and Replies

  • #2
Doc Al
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Since the incline is 90 degress, I am trying to model this as an Atwood machine.
Not a good idea, since an Atwood machine has both ends of the string vertical.

I don't understand how there can be any acceleration if the mass that we need to find is smaller than 3kg.
Why do you think this?


moves 50cm in 1s., which means a is 0.5m/s2
No. Use kinematics to determine the acceleration.

So, using the formula
a = (m2 -m1)/(m1 + m2)g
Instead of using a canned formula (which happens not to apply in this case), derive your own from first principles. Apply Newton's 2nd law to each mass.
 
  • #3
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Ok, I used 0.5 = 0 + 0.5a12 to find a, which gives 1.

So, if F = ma, then the F on the 3kg block is 3(1) which is 3.

On the block suspended in air, the F = mg.

Since we already found F = 3, then 3 = m(9.8), and finally m = .31

This still doesn't make sense, maybe I'm forgetting a T somewhere?
 
  • #4
Doc Al
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This still doesn't make sense, maybe I'm forgetting a T somewhere?
Exactly. Two forces act on the hanging mass.

(Also, from the diagram I can't read the mass of the sliding block. Is it 3 kg or 30 kg?)
 

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