Arrrgh sliding friction problem

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The discussion revolves around a sliding friction problem that a user finds confusing due to an incorrect answer. Key points include the importance of accurately drawing the force diagram, particularly the yellow triangle representing angles and forces. The normal force should be correctly identified, as it can exceed the weight of the block if miscalculated. Users are advised to establish a new coordinate system to simplify the analysis of forces, using trigonometric functions for calculations. Overall, careful attention to force identification, calculations, and units is crucial for solving the problem correctly.
alberto
Ok, I have been going over this problem and the answer I got just doesn't make sense...help!

It's a really easy problem (apparently) but the answer that I keep getting doesn't make sense to me...please let me know what I'm doing wrong or forgetting or whatever...

I thought it'd be easier to answer if you could see the problem I was given so I put it up with my interpretation as to how to answer it on a yahoo! website :

http://www.geocities.com/janguera/rampfrictionproblem.html

Thanks in advance!
 
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Welcome to the forums!

Your mistake lies with how you drew your yellow triangle. If you call the 40 degree angle A, the one on the 100N side B and the one on the R side C, then the 90 degree angle needs to be at point C.

Your tipoff that you made a mistake is this: The normal force is pushing the block with a force larger than the block weighs.

The easiest way I've found to do these is to draw a new coordinate system:

Make the line which the block is sliding on the X axis and the direction which N points (it should be pointing toward the block, being the normal force...) the +y axis. Now, when you break any forces which are at an angle into their component parts, you know that they will form perpendicular angles with that coordinate system.

So, you will have 100cos(40) for the Y and 100sin(40) for the X.

Make sense?
 


Hi there, I completely understand your frustration with this sliding friction problem. It can be really frustrating when the answer you're getting doesn't seem to make sense. I took a look at the problem on the website you provided and I think I can help you figure out where you may have gone wrong.

First, make sure you are correctly identifying the forces acting on the object. In this case, it seems like there is a force pushing the object up the ramp and a force of friction pushing the object down the ramp. Remember that the force of friction is always in the opposite direction of motion.

Next, make sure you are using the correct formula for calculating friction. The formula for friction is μN, where μ is the coefficient of friction and N is the normal force. The normal force is equal to the weight of the object in this case, so make sure you are using the correct weight for the object.

Lastly, double check your calculations and units. Sometimes a small mistake in calculation or using the wrong units can throw off your answer.

If you're still having trouble, don't hesitate to reach out for more help. Sometimes it just takes a fresh pair of eyes to spot the mistake. Good luck!
 
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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