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Pulley problem

  • Thread starter Coco12
  • Start date
  • #1
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Homework Statement



The mass of the two objects on both sides if the pulley is given. So is the acceleration to start moving. It's asking to find the applied force. How do I do that?
Mass of 1- 4.5kg
Mass if 2- 3.0 kg
How much force must u exert to get M1 to accelerate at .25m/s2

Homework Equations


Fnet= ma
-F1g+ f2g= (m1+m2)a


The Attempt at a Solution


I tried putting Fa into the fnet equation and solving for Fa but. It's not working??
 
Last edited:

Answers and Replies

  • #2
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4.5(a) = 4.5(9.8) - 3(9.8) - F
if a = .25

F = 13.575 Newtons added to mass 2 in order to slow the decent of mass 1 to .25m/s^2. Assuming frictionless pulley.
 
  • #3
272
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The answer is supposed to be 17N
 
  • #4
272
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4.5(a) = 4.5(9.8) - 3(9.8) - F
if a = .25

F = 13.575 Newtons added to mass 2 in order to slow the decent of mass 1 to .25m/s^2. Assuming frictionless pulley.
It's just asking the applied force needed to open the window which is M1
 
  • #5
The other person who replied said that you need to add what they got (13.575) to M2, which is 3
Rounding that gives you 17
 
  • #6
14
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-F
-3(9.8) + T = 3a
4.5(9.8) - T = 4.5a
-F -3(9.8) +4.5(9.8) = 7.5a

solve for F the force.
12.825 Newtons
 
  • #7
PhanthomJay
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Solutions should be attempted by the requestor, not the responder.
Draw free body diagrams of each mass, and apply Newton's Laws to each mass.
The book answer looks correct.
 
  • #8
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Just wondering why is the force F- and why the 4.5(9.8) isn't negative since that's on the left side of the pulley?? The applied force is going up the pulley on the left so..
 
  • #9
PhanthomJay
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Just wondering why is the force F- and why the 4.5(9.8) isn't negative since that's on the left side of the pulley?? The applied force is going up the pulley on the left so..
Ok so m1 is the 4.5 Kg window on the left, its weight acts down on it, you apply a force F upward on it to open it, and together with the cord tension acting up on it, it accelerates up at 0.25 m/s^2. Apply Newton's 2nd law, and continue on , next identifying the net force acting on m2 which accelerates m2 downward at 0.25 m/s^2.....
 
  • #10
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Ok so m1 is the 4.5 Kg window on the left, its weight acts down on it, you apply a force F upward on it to open it, and together with the cord tension acting up on it, it accelerates up at 0.25 m/s^2. Apply Newton's 2nd law, and continue on , next identifying the net force acting on m2 which accelerates m2 downward at 0.25 m/s^2.....
Ok I understand it now however why do u have to add the mass to the 12.825N ? How can u add mass to a N force?
 
  • #11
PhanthomJay
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Ok I understand it now however why do u have to add the mass to the 12.825N ? How can u add mass to a N force?
You can't. You should attempt the problem yourself rather than follow someone elses work.

Let's look again at the window, which has a mass m1 = 4.5 kg. Draw a free body diagram. The windows weight, m1(g) acts down. You apply a Force F, up. The tension, T, in the cord acts up (tension forces always pull away from the objects on which they act).

Now apply Newton's 2nd law, F_net = ma, thus

F + T -(m1)g = m1(a)

That is your first equation. You know m1, g, and a, but you still have 2 unknowns, F and T, so you need a second equation. You can get that second equation by drawing a free body diagram of m2, the 3 kg mass, and identifying the forces acting on it, and applying Newton's 2nd law. What do you get?
 

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