Pulley and Mass on an Inclined Ramp

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A 20 kg mass is released while a 5 kg mass slides up an inclined ramp, and the string breaks when the 5 kg mass reaches the end. The weight of the masses is calculated as 49N for the mass on the ramp and 196N for the hanging mass. Friction is determined to be -11N, but there is uncertainty about the sign of the friction force. To find the tension in the string, it's suggested to write separate force equations for each mass and apply Newton's second law. The discussion emphasizes the need to consider all forces acting on the masses to accurately calculate tension.
theuniverse
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Homework Statement


A 20 kg mass is released, and the 5 kg mass on the ramp slides up the ramp from rest. As soon as the 5 kg mass gets to the end of the ramp, the string breaks down. What's the tension of the string?

http://img10.imageshack.us/img10/7603/figure1w.th.jpg

Homework Equations


I would use:
Fg=mg
Fnet = Fg1 + Fg2 + Ff
Fnet = ma
Fk = (uk)(Fn)

The question provies distances too, but I am not sure whether I actually need them...

The Attempt at a Solution


First I found the weight of both masses using Fg=mg
M1(on ramp) = 49N
M2 (hanging) = 196N

Then I found the components of Fg (and thus Fn), as shown in my diagram.
Fn = 42.4N
Therefore, using Fk = (uk)(Fn), I found the friction to be -11N.

I know that I have to calculate Fnet. So: Fnet = 196N -24.5N +11N = 182.5N
*Note: I am not very sure what sign does the friction have in this problem...

Not sure how to find tension from here... Fnet = Ft + Fgx + Ff?

Thank you for your time,
 
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I recommend that you write separate force equations for each mass, then combine them. For the mass on the incline, consider forces parallel to the incline.
 
Thanks for replying.
I am not quite sure what you mean by separate force equations for each mass.
And wouldn't Fgx = 24.5N be the force parallel to the incline?
 
theuniverse said:
I am not quite sure what you mean by separate force equations for each mass.
Consider the forces on mass 1. Apply Newton's 2nd law. Do the same for mass 2.
And wouldn't Fgx = 24.5N be the force parallel to the incline?
That's the component of gravity on mass 1 parallel to the incline, but other forces act on that mass.
 
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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