Newtons laws of motion help!

  • Thread starter mezmorize
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  • #1
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Hello everyone, i need help with these 2 question from physics. Im a year 11 physics student.

Joey and Caitlin hang from two ends of a wire, looped through an ideal pulley. Joey has a mass of 75.0Kg, whilst Caitlin weighs 637N.

a) calculate the tension of the wire?

b) If Caitlin is not able to reach out and touch anything, is she able to change the motion of the system? Why?
 

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  • #2
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a) calculate the tension of the wire?

What forces act on which of them? In this case, it's easy to determine directions of their motion and calculate acceleration.

b) If Caitlin is not able to reach out and touch anything, is she able to change the motion of the system? Why?

Some kind of force is required to change the motion of the system. Now, is there any way to obtain that force if Caitlin can't touch anything (just remember the definition of force)?
 
  • #3
Delta2
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a) From newton 2nd law applied to catlin and joey we get the following two equations.
[tex]W_C-T=m_Ca_C[/tex]
[tex]W_J-T=m_Ja_J[/tex] where T is the tension of the wire. All you need to do is solve this system of equations for T, given that [tex]a_C=-a_J[/tex] where minus sign comes from fact that they ll move in opposite direction(if one is going up the other will go down).

b) If caitlin is not able to reach out and touch anything except the rope, then the only force that caitlin can exert is to Joey through the rope. But then due to newtons 3rd Law, Joey will exert an equal but opposite force to caitlin. Because caitlin and joey are connected with the rope, this internal force cannot change the motion of neither of them, neither the motion of the rope, it just increases the tension of the rope.
 
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  • #4
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What forces act on which of them? In this case, it's easy to determine directions of their motion and calculate acceleration.



Some kind of force is required to change the motion of the system. Now, is there any way to obtain that force if Caitlin can't touch anything (just remember the definition of force)?

thanks :), ive calculated the acceleration which is 0.7ms-2 but i can still seem to find out how to find the tension :S
 
  • #5
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ahhhh dont worry delta explained it xD
 

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