Calculating Acceleration and Direction of Objects on a Ramp

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The discussion centers on calculating the acceleration and direction of two blocks on a ramp, with specific emphasis on the relationship between their movements. Block B is initially calculated to have an acceleration of 3.35 m/s^2 down the ramp, with components determined for both X and Y directions. However, the calculations for block A reveal an error, as the relationship between the two blocks must be considered due to the normal force acting between them. The need to find the acceleration of block B relative to block A is highlighted, emphasizing that both accelerations are interconnected and cannot be treated independently. The correct approach involves using the normal force as an unknown in the equations of motion for both blocks.
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Homework Statement


upload_2016-7-24_19-30-38.png


Homework Equations

The Attempt at a Solution



First, I find the accelerations of each block, separate the accelerations into Y and X components and then add them as vectors.

for B:
upload_2016-7-24_19-53-31.png


B goes down the ramp with an acceleration of 3.35 m/s^2
X component : (3.35)(cos 20) = 3.15 m/s^2 to the right
Y component : (3.35)(sin 20) = 1.145 m/s^2 downwards
upload_2016-7-24_20-17-53.png


for A :
I add the Y component of the Normal force of B on A, to the weight of wedge A.

Y component of the Normal force of B on A = 80 N
Weight of Wedge A = 215.6 N

adding both : 80 N + 215.6 N = 295 N
but the slope is 30 degrees, so the force along the slope is : (295 N) ( sin 30) = 148 N

to find the acceleration of A down the ramp : F = ma

148 N = ( 22 kg) (a)
a = 6.72 m/s^2
x component : 5.811 m/s^2 to the left
y component : 3.36 m/s^2 downwards

now adding components , total x component is 2.66 m/s^2 to the left
and 4.505 m/s^2 downwards

therefore acceleration of B is 5.23 m/s^2 with direction 59.5 degrees below the the horizontal, directed to the left.Now, all this is wrong because the answer aB =5.94 m/s^2, 75.7 degrees below the horizontal, directed to the right.

Where did I go wrong? Did I have the right idea but miscalculated?
 
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goonking said:
B goes down the ramp with an acceleration of 3.35 m/s^2
That would be true if the ramp were static. Are you sure it is true if the ramp is itself accelerating?
You need to be clear when discussing B's acceleration whether you mean absolute or relative to A.
 
haruspex said:
That would be true if the ramp were static. Are you sure it is true if the ramp is itself accelerating?
You need to be clear when discussing B's acceleration whether you mean absolute or relative to A.
yes, that was my error, I thought I could take the a point as the origin, and just divide the problem into two separate ones, find the acceleration of both relative to the origin and just add up the vectors but that clearly isn't working out.

I would need to find the acceleration of B relative to A, and in order to do that, I would need to find the acceleration of A (with respect to the origin) first, correct?
 
goonking said:
I would need to find the acceleration of B relative to A, and in order to do that, I would need to find the acceleration of A (with respect to the origin) first, correct?
You can't do either independently of the other. They are linked by the normal force between them. Introduce that as an unknown and write out the usual F=ma equation for each body.
 
goonking said:
Y component of the Normal force of B on A = 80 N
may i know how do you get this?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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