Cart, Incline, Pulley, and a messed up Acceleration.

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The discussion revolves around a physics problem involving a cart on an incline with a pulley, where the calculated acceleration is approximately 1, while the expected answer is 0.77. The correct approach involves using the angle of 26 degrees for calculations, specifically applying the formula Fgx = mass x gravity x sin(26). After recalculating, the force due to gravity on the incline is determined to be around 15.5 N. The final equation for acceleration is clarified as a = (m1gsin26 - m2g) / (m1 + m2), yielding an accurate acceleration of approximately 0.7728. Properly applying the trigonometric relationships and forces involved is crucial for obtaining the correct result.
Some_Thing
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The question and my work can be seen in the image below:

http://img145.imageshack.us/img145/550/scan0001fgu.jpg

The problem is that I am getting around 1 for acceleration, while the answer is 0.77.

What am I doing wrong?
 
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rock.freak667 said:
Upload the image here [URL]http://imageshack.us and paste the forum code so that we can see it and try to help you.

Oops, I was using a url from hotmail. It showed up fine for me, but I guess that was because I was logged in at the time.

Thanks.
 
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The angle remains 26 and not 64. So Fgx = mass x gravity x sin theta where theta is 26 degrees.
 
math_04 said:
The angle remains 26 and not 64. So Fgx = mass x gravity x sin theta where theta is 26 degrees.

Fgx = mass x gravity x sin 26

Fgx = 3.6 x 9.8 x sin 26

Fgx = 15.465

This rounds off to 15.5 N, which is what I got.


NOTE: cos 64 = sin 26
 
You have got another force in the rope m2a.
Your equation should look something like this:

m2g+m2a+m1a=m1gsin26

a=(m1gsin26-m2g)/(m1+m2)
a=0.7728
 
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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