m1 (2.35 kg) & m2 (5.55) kg are connected by a light string that passes over a light frictionless pulley?

Question

(a) Determine the tension in the string.
(b) Determine the acceleration of each object.
(c) Determine the distance each object will move in the first second of motion if they start from rest.


I am so confused as to where to begin. Any help would be greatly appreciated.

Answer 1

Your 1st and 3rd answers are wrong. Couldn't follow the 2nd. Might be right but needs explanation.

I picture m2 on the right. So obviously the pulley will rotate CW in my picture. But I won't make use of that knowledge. It'll come out that way at the end.

The forces on m2 are its weight (down) and the tension T (up) in the string. So the net force is:
Fnet2 = m2*g-T2
We can also say about Fnet2 that it will cause m2 to undergo acceleration a2 according to Newton's 2nd
Fnet2 = m2*g-T2 = m2*a2
So T2 = m2*(g-a2)

On the other side of the pulley
Fnet1 = m1*g-T1
and again expanding on that using Newton's 2nd
Fnet1 = m1*g-T1 = m1*a1
So T1 = m1*(g-a1)

Because the pulley is "light" and frictionless,
T1 = T2 So
m2*(g-a2) = m1*(g-a1)
And, because m1 and m2 are tied together with the string and if one goes down, the other must go up
a1 = -a2
Back to the previous equation - substitute -a2 for a1
m2*(g-a2) = m1*(g+a2)
m2*g-m2*a2 = m1*g+m1*a2
(m2-m1)*g = (m1+m2)*a2
a2 = (m2-m1)*g / (m1+m2)

OK, that's your question (b). Question (a) asks for the tension T (1 or 2, they're the same)
T2 = m2*(g-a2) (an equation from above) will give it to you.

(c) You know the acceleration now so the distance in the y direction is given by the motion equation (initial velocity is 0)
y = (1/2)*a*t^2
where a = a2 and t = 1 second. It'll be the same amount for both. Up for m1 and down for m2.

Answer 2

a. The tension on the string will be 2.35Kg on the side with that weight attached and 5.55 on the side with that weight attached (Only the support of the pulley itself would be the sum of both weights)
b. the acceleration will equal 1 G force
c. the distance of travel of each object will be 16 feet since the weight does not effect the rate of fall of any object (as long as air resistance is not a factor which it is normally considered not to be in the case of questions like this) all objects fall 16 feet in the first second and then accerate at the rate of 32 feet for each following second. in other words in 2 seconds it would fall 64 feet having fallen 16 feet in the first second, then accellerated an additional 32 feet making for 48 feet travelled during the 2nd second for a total of 64 feet of fall in 2 seconds

Answer 3

Well just imagine both masses on a pulley system. The tension on the string is m1*g. I can't explain it better without drawing a picture. The acceleration is g. And for the distance you know Vi =0. A=g.T=1s, Di=0 and you want Df. Plug it in. So Df=AT^2+ViT+Di. Df=g. (Check your book to verify that equation)

Answer 4

M1*a = T - M1*g, -M2*a = T - M2*g , solving these eqs. yields T = M1*g(((M1-M2)/(-M2-M1)) + 1). a) T = 32.39 N b) +- 3.97 m/s2 c) 1.99 m.

I started with a free-body diagram.