Solving for Acceleration on an Inclined Ramp

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The discussion revolves around determining the correct sign for acceleration (a) in the context of a 1kg weight on a 20-degree inclined ramp being pulled upwards and to the left. The key equations presented are -a = t - mgsin20 and a = t - mgsin20, with confusion about whether acceleration should be considered -ma or +ma. The resolution hinges on the definition of the coordinate system and the direction of forces acting in both the X and Y directions. Ultimately, the sign of acceleration is contingent on how the origin is defined and the orientation of the vectors involved. The conversation highlights the need for clarity in coordinate systems when analyzing motion under gravity.
SS2006
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I have a 1kg weight on a rmap inclined 20degrees being pulled upwards and to the left
*******-
***********-
************** -
******************-
in that motion above(look at dashes only), is it -ma or +ma?
so the equation is

-a = t - mgsin20
or
a = t - mgsin20
I am just wondering if a leeft movement has anythign to do with gravity (A) being + or-
thanks
 
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try looking at it in terms of the forces acting in the X direction and the forces acting in the Y direction. Consider where you place your origin.

Then think about which vectors will be positive and which will be negative.
 
no no i just want a simple anaswre
i just meant weather acceleration should be -ma or +ma
 
It has to do with how you define coordinate system.
 
i have an exam tomorrow and i just wanted to know , are you saying there's no way to tell from the info i sent?
*sigh* nvm..
 
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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