Forces related question (it's a problem)

[Moontears]

Ok, I solved this forces problem and I keep getting a different answer than presented in the back of the book. Now I am almost certain that I am correct and I just want to see if anyone else may be able to answer this problem and don't even post your solution, only what you got for the acceleration of the coins. If in the mood, please attempt.

The problem is as follows:

You attach a loonie (mL = 6.99g) and a dime (mD = 2.09g) to the ends of a thread. You put the thread over a smooth horizontal bar and pull the thread taut. Finally, you release your hands, letting the loonie drop and the dime rise. Friction between the thread and the bar is negligible, and the magnitude of g = 9.80 m/s/s [down]. Determine the magnitude of the acceleration of the coins.


cheers

 

[FZ+]

Ok, let's deal with the loonie and the dime seperately as two systems sharing an acceleration and a tension.

For the loonie, we know that:

a = ((mL) * g - T)/mL

This rearranges to..

T = mL*g - mL * a

For the dime, we get similarly...

a = ((T - (mD) * g)/mD

T = mDa + mDg

Now, T = T (shock horror! so

mD*a + mDg = mL*g - mL * a
so a = (mL*g - mD*g)/(mD + mL)
= 5.29 ms^-2

(However, the value you have for g is wrong. g to 3 sf = 9.81, not 9.80.)

 

[HallsofIvy]

FZ+'s point about g is that, if you are using two significant figures, 9.8 is sufficient but that, if you are using three significant figures, the correct value is 9.81, not 9.80.

Now, FZ+, why would it be incorrect to ignore the thread connecting the two coins and just treat this as a single object with mass
6.99- 2.09= 4.90 grams? Of course, such an object would just accelerate downward with acceleration 9.81 cm/sec^2.

 

[Moontears]

Thank you FZ+, yeah you are getting the same answer as the back of the book. I just had another way of approaching the problem which made sense in my nogin but I guess was wrong. My way I kept getting 18.1 m/s/s, which was way off. Thank you for your time.