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View Full Version : Finding acceleration and tension


Karissa
Oct 20, 2009, 03:38 PM
So, there's a diagram, but I'm not sure if I can post one. So. It's basically
An object hanging from a string that's wrapped around a pulley with another object on the opposite end of the string.

M1 is 2kg
M2 is 4 kg

Find the acceleration and tension in the string.

PLEASE HELP ME!

ebaines
Oct 22, 2009, 06:08 AM
Helo Karissa. One way to do this problem is to start by considering what's happening at the pulley. It's holding up two weights, so it must be "feeling" the total weight of both masses. The way those forces are caried into the pully is through the string, which passes over the top of the pulley. So at the pulley the string is pulling down both on the left and the right side of the pulley, with equal force in both places. This force is the tension in the string. Hence the pulley must hold up against two times the tension in the string.

Can you take it from here?

Another way to approach this is to consider that the two masses are "out of balance" by 2 Kg. Hence the masses will accelerate - the heavy one falls toward the gtround and the light one is pulled upward. The two mases are tied together, so the rate of acceleration must be equal You have a total of 6 Kg being subected to a force of 2 Kg*g, and from this you can determine th acceleration (F=ma). What do you getr for a? Now consider just one of the masses, say the 2 Kg one. Now that you know its acceleration upward, you can determine the tension in the string that would cause it to overcome its weight and accelerate upward at that rate.

I suggest you try this problem both ways and see if you get the same result. Post back and let us know what you get.

JohnSenAs
Oct 19, 2010, 10:57 PM
f=ma
(4kg)(9.8m/s^2)-(2kg)(9.8m/s^2)=(4kg+2kg)a
39.2N-19.6N=6kga
a=3.27m/s^2

t-19.6=(2kg)(3.27m/s^2)
t-19.6=6.54N
t=26.14N