Calculating Acceleration in a Block and Tackle System

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In a block and tackle system with a 5.7 kg block on a 34° incline and a 2.4 kg hanging block, the acceleration of the hanging block can be calculated using free body diagrams. The forces acting on the blocks must be resolved into components along the inclined plane (x-axis) and perpendicular to it (y-axis). The equations ƩFx = max and ƩFy = may are essential for solving the problem. The weight of the hanging block and the x-component of the weight of the block on the incline are key forces to consider. Ultimately, the correct calculation yields an acceleration of +0.95 m/s² for the hanging block.
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Two blocks are connected by a string, as shown in the figure. The smooth inclined surface makes an angle of = 34° with the horizontal, and the block on the incline has a mass of 5.7·kg. The mass of the hanging block is 2.4·kg.

What is the magnitude of the hanging block's acceleration?I really don't know how to start with this problem. If you could give me some starting equations that would be really helpful. Thank you

Edit: Here's the link! Sorry about that http://imgur.com/PH12mmc
 
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missing picture...
 
nil1996 said:
missing picture...

fixed. sorry about that!
 
First try to draw the free body diagram.Find the components of forces along the inclined plane(x-axis) and perpendicular to the inclined plane(y-axis)
Use ƩFx=max and ƩFy=may
Solve these equations you will definitely get the answer
 
nil1996 said:
First try to draw the free body diagram.Find the components of forces along the inclined plane(x-axis) and perpendicular to the inclined plane(y-axis)
Use ƩFx=max and ƩFy=may
Solve these equations you will definitely get the answer

Alright, so I just went back to this problem. When doing the forces along the X axis, would I use the Force of Weight of the hanging block (2.4*9.8=23.52N) as the positive X force and the x component of the force of weight on the bigger block (55.86sin34=31.24N) as the negative x force, and that's it for the forces on the x axis? What would the Y forces be?
 
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Ok, just figured it out and I'm annoyed at it. It's +.95, and I had -.95 to begin with. Not sure why it's positive but it is. Thanks for the help.
 
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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