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A 2-kg body rests on frictionless wedge that has an inclination of 60º and an acceleration a to the right such that the mass remains stationary relative to the wedge. Find a.

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Other answers:

Still don't understand the force...
FBD: |dw:1349586646724:dw| Not sure if it's right, though.
acos60=gcos30 (Balancing forces along the wedge in the reference frame of the wedge)
a= gtan60
How do you set the equation for y? Not sure if I'm doing it right...\[\Sigma F_y = Nsin60º = ma\]
Ncos60 - mg =0
Hold on, give me time.
Why Ncos60, not sin?
look at the sketch
Nsin30 - mg =0 (if you like)
Yeah, silly me. Got it. Just give me a little more time.
Can you explain me why Nsin60º = ma?
because there are only two forces acting on the mass... N and mg. if you think about it, it's magic otherwise... because if the block is being accelerated it can only be because the ramp is 'pushing' it... ie the Normal force...
if the block were somehow accelerated otherwise, it would be some magical force:)
Now you're making it a little clear. If the wedge was given a greater cceleration, the block will slip upward, right?
it's like sitting in a car seat when someone stomps on the gas... you feel pushed back but the actuality is that the seat's Normal force has increased and is acting to accelerate you...
Why not?
N would get longer, but it would still be perp. to the ramp face...
you're thinking maybe of a more "real-life" situation wherein air resistance is a factor... then yes, the block would slide up the ramp at some point...
Now you're mention about air resistance, it does make block slide up. Thanks.
hmm actually, maybe you're right about that...
because mg is fixed so... if N stays perp. to the ramp face and ma gets longer...I guess N would have a y component larger than mg....
Hmm I see. It does make sense.
I answered too quickly... sorry :P
It's ok. You helped me a lot. Thanks. I wish I could give you more medals, lol.

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