Finding the tension of a string

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The discussion revolves around calculating the tension in string #1, with the user arriving at 108N while the book states it should be 118N. The user’s equation is questioned, suggesting that the acceleration is already in the direction of motion, negating the need for trigonometric functions. Another participant clarifies that the radius of the pulley is irrelevant due to its negligible mass and that the moment of inertia can be ignored in this case. Ultimately, the correct tension can be derived directly from the forces acting on the incline's block. The conversation highlights the importance of understanding the principles of motion and force balance in solving such problems.
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picure is http://home.earthlink.net/~suburban-xrisis/clip2.jpg
I need to find the Tension on string #1
The radius of the pully is 0.25m of negligible mass

I tried and always got 108N however the book says its 118N

(sin37*15*9.8)+(sin37*15*2)=108N

I don't see how it is 118??
 
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It doesn't make any sense...Is the string ideal...?If so,then the tension in it is constant...Why do they give you the radius of the pulley...?Definitely weird...

Anyway,even so,assuming that the wire is not ideal (it would make sense),the II-nd principle for the body on the incline should lead you to the answer...

Daniel.
 
UrbanXrisis said:
picure is http://home.earthlink.net/~suburban-xrisis/clip2.jpg
I need to find the Tension on string #1
The radius of the pully is 0.25m of negligible mass

I tried and always got 108N however the book says its 118N

(sin37*15*9.8)+(sin37*15*2)=108N

I don't see how it is 118??

You equation is wrong. How did you get: (sin37*15*9.8)+(sin37*15*2)
 
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learningphysics said:
You equation is wrong. How did you get: (sin37*15*9.8)+(sin37*15*2)
Indeed. You have to realize that you're given an acceleration that is already in the direction of motion. There is no need to use sines or cosines on that magnitude, since it's already in component form.

You should wind up with an equation that describes the net forces on the incline's block, which (thanks to Newton) will equate to the mass and acceleration of the block (both of which you know).

Plug and Chug :smile:
 
dextercioby said:
It doesn't make any sense...Is the string ideal...?If so,then the tension in it is constant...Why do they give you the radius of the pulley...?Definitely weird...

Anyway,even so,assuming that the wire is not ideal (it would make sense),the II-nd principle for the body on the incline should lead you to the answer...

Daniel.

the r is to find the moment of inertia of the pully. I actually have some trouble with this. Is there an equation that I can use? I have tried many and don't know why it's not giving the answer of 1.2kgm^2

any ideas?
 
UrbanXrisis said:
the r is to find the moment of inertia of the pully. I actually have some trouble with this. Is there an equation that I can use? I have tried many and don't know why it's not giving the answer of 1.2kgm^2

any ideas?
If there's negligiable mass, then the moment of inertia of the pully isn't significant enough to worry about. I got 118 N using only the information provided in the diagram itself.

(PS - the moment of inertia of a solid cylinder is \frac{1}{2}MR^2 ...again, though, in this case you don't really need it.)

^^ That gap is the strangest thing... :confused:
 
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Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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