(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

[PLAIN]https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/phys2111/fall/homework/Ch-05-Newtons-Laws/blocks_on_pulleys/12.gif [Broken]

In the above picture, the masses are M1 = 10 kg, M2 = 15 kg, and M3 = 25 kg. The pulleys and strings are massless, there is no friction, and gravity points downward. The whole system is held fixed, then released at rest. What is the acceleration of the 25 kg block? Enter a positive answer if it goes up, a negative answer if it goes down, or 0 if the acceleration is zero.

Hint 1) Draw a free-body diagram for each of the three masses. Try to convince yourself that the tension in the strings supporting M1 and M2 are equal, and that tension is half of the tension in the string supporting M3.

Hint 2) Let's define the directions of all accelerations to be positive if they are up, negative if they are down. Consider the accelerations of the three blocks (a1, a2, and a3). The basic result is that a3 = -(a1 + a2)/2. How can we convince ourselves of that? This is not easy to derive, but we can make it plausible. Consider what happens if M1 = M2. Then we know that those two blocks will accelerate together just as though they were a single mass. They are essentially one half of a simple Atwood's machine, the other half of which is Block 3. Under those conditions, a1 = a2 and they are both equal to -a3. The negative sign arises because if Block 3 goes down, then the other two blocks must go up (and vice versa). This is exactly what our formula gives. In the general case, the formula is still valid. Using this information plus the hint about the tensions plus writing F = ma for each mass, we end up with 3 equations and 3 unknowns (1 tension and 2 accelerations). Solve these to find the answer. At this point the algebra gets a little messy, and it is better to plug the masses into the equations right away rather than keeping things algebraic.

2. Relevant equations

F = ma for each block

3. The attempt at a solution

I set up free body diagrams for each block and found:

(FT1 relates to the tension force for block 1 etc...)

FT1 = 10(g + a1)

FT2 = 15(g + a2)

FT3 = 25(g + a3)

I agree that FT1 = FT2 but I don't know why they are half the tension in FT3 or that their accelerations are a3 = -(a1 + a2)/2. Could someone please explain this a bit better than they did?

Also, just by using the hints/formulas they gave me I solved for a3 to be .2087 but this isn't the right answer. So I must be doing something wrong.

All help would be greatly appreciated

-Fox

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# Homework Help: Three Blocks On Two Pulleys

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