It would be pretty awesome if you'd help me out.?

Question

This mechanics problem is thoroughly kicking my ***. I would appreciate any help - preferably if you could take me step through step through how to solve it...

Two objects with masses of 2.00 kg and 5.00 kg are connected by a light string that passes over a frictionless pulley.
Determine the tension in the string in Newtons.
Determine the acceleration of each object.
(c) Determine the distance each object will move in the first second of motion if both objects start from rest.

Thanks much!

Answer 1

Let's say m1=m2
You have guest it right ... they would just hang there

Let m1= 2.00kg and m2=5.00 kg

And m2 is pulling down while m1 is desperately moving down on its own and opposing m2. In a tug of war there is tension force F

a )F= m2g - m1g= g(m2 - m1)
b )well F=ma is it not
So a= F/(m1+m2) makes sense since they move together.
a=g(m2 - m1)/(m1+m2)
c) S=0.5 a t^2 t=1 and a=g(m2 - m1)/(m1+m2)

Have fun

Answer 2

Atwood's machine.

Essentially, you have two unknowns: tension T and acceleration a. You can do this the conventional way, drawing a "free-body diagram" for each mass, writing a couple of equations, eliminate, by whatever method you choose, one of the unknowns, then solve for the other. There´s also a "fast-track" solution.

a) Let's try the conventional method first. There are two forces acting on the 5.00 kg mass: its weight, w, and string tension T. This object, being heavier than the other, moves downward. We can write

unbalanced force = ma = w − T = mg − T;

or, since m = 5.00 kg,

5 a = 5 g − T.

As this is written, it indicates that the prevailing force is weight, and the body accelerates downward, in the direction g points to. For the 2.00 kg body, likewise,

2 a = T − 2 g.

This is very much the same as the upper equation, but the order of terms in the right side is reversed to make clear that this body accelerates upward, in the same direction T has. Solve for a and substitute in the former equation:

a = T/2 − g
5 (T/2 − g) = 5 g − T
T (5/2 + 1) = 5 g(1 + 1)
T = (10/ 7/2) g = 20/7 g = 28.0 N.

This is not the only way these equations can be written. For instance, you could consider that weight should be negative, since it points downward, and tension positive, since in both cases it points upwards. This is perfectly acceptable, as long as the direction of resulting forces are tagged accordingly. It's a matter of convention, really; it would lead to the same result, as long as chosen convention is observed thoroughly.

b) Since a = T/2 − g, a = 14 − 9.8 = 4.2 m/s²

c) s = ½ at², where t = 1 s. Thus, s = ½ a = 2.1 m.


The fast way: regarding both bodies and the connecting string as a single body, or "system", it's clear that the unbalanced force acting on system is F = (5 g − 2g) N, while the joint mass of the system is 5 + 2 = 7 kg. Acceleration of system, a = F/m = 3g/7 = 4.2 m/s². Tension may be found from the FBD of any of the two bodies. For the 2.00 kg body, T = 2(a + g) = 2 ( 4.2 + 9.8) = 28.0 N.

If you repeat this same procedure with arbitrary masses m1 and m2, you should arrive at the following expressions:

a = [ (m1 – m2)/(m1 + m2) ] g
T = [ 2 m1 m2 / (m1 + m2) ] g.