If you are having trouble with this problem but you managed to do the other force problems, you might want to take another look at your force diagrams (free body diagrams).
The smaller block hangs over the edge of a table (over a pulley), and is attached to a string. There is a tension T upward, a force due to gravity (Fg = m*g) downward, and the block's resulting acceleration is downward. You use Newton's 2nd law to get an expression for the tension: ∑F=Fg-T=ma so, rearranging and writing Fg=mg, T=mg-ma.
Now look at the larger block. It has a force of gravity downward, a normal force upward, and 2 ropes pulling to the right. That makes it T to the right and another T to the right, so there is 2T pulling to the right (look at the pulley: it makes the string pull twice). With "right" as the positive direction, Newton's 2nd law becomes ∑F=2T=ma. Plug in the expression you had above for T and solve for the acceleration. "The rest is just algebra."
Hope that helps. Good luck.
7 comments:
How do you figure out the acceleration of the smaller block without knowing the tension to begin with?
I tried to follow your directions, but I still can't seem to get it right and I know a couple other people can't either..any suggestions? Thanks!
When you are looking at the smaller block to find an expression for the tension in the string, you don't actually get a number. You just get an expression (equation) for the tension, and there is a letter "a" for the acceleration of the smaller block in the expression. The acceleration of the small block is twice the acceleration of the larger block (because of the pulley, the small block goes twice as far, has twice the acceleration, etc).
When you look at the forces on the larger block you get another expression (equation), and this expression also has a letter "a" in it. But the acceleration of the larger block is half the acceleration of the smaller block. They are related, but not exactly the same.
Maybe you should write the acceleration of the larger block as "a" and the acceleration of the smaller block as "2a" since its acceleration is twice as large. Or write the acceleration of the smaller block as "a" and the acceleration of the larger block as "(1/2)a".
I wonder if that helps. Good luck.
yes that helped so much, thank you!!
Hmmm, still not seeing it. Following what you said, you end up with a=2mg-2ma(of the smaller block which is still unknown) all over the mass of the larger block. So how do you get rid of the other a?
If you made the bigger acceleration =2a and the smaller acceleration =a then both a's mean the same thing. So you can solve for "a". Take all terms with "a" to one side of the equation and all other terms to the other side. Factor out the "a", divide by the (...) and solve for a. Should work. If not, there is probably a silly mistake somewhere so we would need more details to see where this got messed up.
Good luck.
Wow, can't believe I didn't notice that... It worked. Thank you!
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