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- Thread starter chronos98
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- #27

rl.bhat

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In the problem it is clearly stated that the wedge is accelerating to the right. So there must be a force acting on the wedge.

- #28

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The weight of the 2.0 kg block is PERPENDICULAR to the direction of motion, but points downwards, so 'mg' is in the -ve y-direction.

If you follow the co-ordinate system mentioned here, you will have to resolve just the normal force Fn into its components.

Fn (along +ve-x axis) = Fn (cos 60)

Fn (along +ve-y axis) = Fn (sin 60)

Now, trying to sum up the forces along the +ve x-co-ordinate system,

{EQUATION 1} Total F(x) = Fn (cos 60) = Force applied F (ma)

Also, summing up the forces along the +ve y-co-ordinate system,

{EQUATION 2} Total F(y) = Fn (sin 60) - mg = 0

This F(y) is EQUAL to ZERO, since there is no acceleration in the y-direction.

On dividing equation 1 by equation 2, you get:

1/ [tan (60)] = (a/g)

The final answer comes out to be a = 5.66 m/s^2

I think it makes more sense if you draw out the FBD ... I could be wrong, but that is the only way it made some sense to me. :)

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